# Finite Probability Variance Foundations

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batch-20261004-advance-20261005-r07-variance-foundations · 2026-10-04 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation0/1

Compute the mean, then sum weighted squared deviations exactly to obtain 0/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0001/)

Variance in finite probability models · qf-variance-weighted · candidate





## 2. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation9/4

Compute the mean, then sum weighted squared deviations exactly to obtain 9/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0002/)

Variance in finite probability models · qf-variance-weighted · candidate





## 3. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation4/1

Compute the mean, then sum weighted squared deviations exactly to obtain 4/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0003/)

Variance in finite probability models · qf-variance-weighted · candidate





## 4. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation25/4

Compute the mean, then sum weighted squared deviations exactly to obtain 25/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0004/)

Variance in finite probability models · qf-variance-weighted · candidate





## 5. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation189/25

Compute the mean, then sum weighted squared deviations exactly to obtain 189/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0005/)

Variance in finite probability models · qf-variance-weighted · candidate





## 6. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation98/9

Compute the mean, then sum weighted squared deviations exactly to obtain 98/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0006/)

Variance in finite probability models · qf-variance-weighted · candidate





## 7. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation720/49

Compute the mean, then sum weighted squared deviations exactly to obtain 720/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0007/)

Variance in finite probability models · qf-variance-weighted · candidate





## 8. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1215/64

Compute the mean, then sum weighted squared deviations exactly to obtain 1215/64. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0008/)

Variance in finite probability models · qf-variance-weighted · candidate





## 9. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation825/49

Compute the mean, then sum weighted squared deviations exactly to obtain 825/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0009/)

Variance in finite probability models · qf-variance-weighted · candidate





## 10. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation363/16

Compute the mean, then sum weighted squared deviations exactly to obtain 363/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0010/)

Variance in finite probability models · qf-variance-weighted · candidate





## 11. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation260/9

Compute the mean, then sum weighted squared deviations exactly to obtain 260/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0011/)

Variance in finite probability models · qf-variance-weighted · candidate





## 12. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation3549/100

Compute the mean, then sum weighted squared deviations exactly to obtain 3549/100. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0012/)

Variance in finite probability models · qf-variance-weighted · candidate





## 13. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation245/9

Compute the mean, then sum weighted squared deviations exactly to obtain 245/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0013/)

Variance in finite probability models · qf-variance-weighted · candidate





## 14. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation36/1

Compute the mean, then sum weighted squared deviations exactly to obtain 36/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0014/)

Variance in finite probability models · qf-variance-weighted · candidate





## 15. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation5440/121

Compute the mean, then sum weighted squared deviations exactly to obtain 5440/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0015/)

Variance in finite probability models · qf-variance-weighted · candidate





## 16. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation867/16

Compute the mean, then sum weighted squared deviations exactly to obtain 867/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0016/)

Variance in finite probability models · qf-variance-weighted · candidate





## 17. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation4617/121

Compute the mean, then sum weighted squared deviations exactly to obtain 4617/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0017/)

Variance in finite probability models · qf-variance-weighted · candidate





## 18. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1805/36

Compute the mean, then sum weighted squared deviations exactly to obtain 1805/36. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0018/)

Variance in finite probability models · qf-variance-weighted · candidate





## 19. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation10500/169

Compute the mean, then sum weighted squared deviations exactly to obtain 10500/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0019/)

Variance in finite probability models · qf-variance-weighted · candidate





## 20. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation297/4

Compute the mean, then sum weighted squared deviations exactly to obtain 297/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0020/)

Variance in finite probability models · qf-variance-weighted · candidate





## 21. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation8349/169

Compute the mean, then sum weighted squared deviations exactly to obtain 8349/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0021/)

Variance in finite probability models · qf-variance-weighted · candidate





## 22. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation3174/49

Compute the mean, then sum weighted squared deviations exactly to obtain 3174/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0022/)

Variance in finite probability models · qf-variance-weighted · candidate





## 23. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation80/1

Compute the mean, then sum weighted squared deviations exactly to obtain 80/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0023/)

Variance in finite probability models · qf-variance-weighted · candidate





## 24. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation24375/256

Compute the mean, then sum weighted squared deviations exactly to obtain 24375/256. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0024/)

Variance in finite probability models · qf-variance-weighted · candidate





## 25. Exact Finite Probability Variance

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1521/25

Compute the mean, then sum weighted squared deviations exactly to obtain 1521/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0025/)

Variance in finite probability models · qf-variance-weighted · candidate





## 26. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Put Z=X+(-12). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation0/1

Compute the mean, then sum weighted squared deviations exactly to obtain 0/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0026/)

Variance in finite probability models · qf-variance-shift · candidate





## 27. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Put Z=X+(-11). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation9/4

Compute the mean, then sum weighted squared deviations exactly to obtain 9/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0027/)

Variance in finite probability models · qf-variance-shift · candidate





## 28. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Put Z=X+(-10). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation4/1

Compute the mean, then sum weighted squared deviations exactly to obtain 4/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0028/)

Variance in finite probability models · qf-variance-shift · candidate





## 29. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Put Z=X+(-9). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation25/4

Compute the mean, then sum weighted squared deviations exactly to obtain 25/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0029/)

Variance in finite probability models · qf-variance-shift · candidate





## 30. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Put Z=X+(-8). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation189/25

Compute the mean, then sum weighted squared deviations exactly to obtain 189/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0030/)

Variance in finite probability models · qf-variance-shift · candidate





## 31. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Put Z=X+(-7). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation98/9

Compute the mean, then sum weighted squared deviations exactly to obtain 98/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0031/)

Variance in finite probability models · qf-variance-shift · candidate





## 32. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Put Z=X+(-6). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation720/49

Compute the mean, then sum weighted squared deviations exactly to obtain 720/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0032/)

Variance in finite probability models · qf-variance-shift · candidate





## 33. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Put Z=X+(-5). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1215/64

Compute the mean, then sum weighted squared deviations exactly to obtain 1215/64. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0033/)

Variance in finite probability models · qf-variance-shift · candidate





## 34. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Put Z=X+(-4). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation825/49

Compute the mean, then sum weighted squared deviations exactly to obtain 825/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0034/)

Variance in finite probability models · qf-variance-shift · candidate





## 35. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Put Z=X+(-3). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation363/16

Compute the mean, then sum weighted squared deviations exactly to obtain 363/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0035/)

Variance in finite probability models · qf-variance-shift · candidate





## 36. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Put Z=X+(-2). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation260/9

Compute the mean, then sum weighted squared deviations exactly to obtain 260/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0036/)

Variance in finite probability models · qf-variance-shift · candidate





## 37. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Put Z=X+(-1). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation3549/100

Compute the mean, then sum weighted squared deviations exactly to obtain 3549/100. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0037/)

Variance in finite probability models · qf-variance-shift · candidate





## 38. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Put Z=X+(0). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation245/9

Compute the mean, then sum weighted squared deviations exactly to obtain 245/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0038/)

Variance in finite probability models · qf-variance-shift · candidate





## 39. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Put Z=X+(1). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation36/1

Compute the mean, then sum weighted squared deviations exactly to obtain 36/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0039/)

Variance in finite probability models · qf-variance-shift · candidate





## 40. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Put Z=X+(2). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation5440/121

Compute the mean, then sum weighted squared deviations exactly to obtain 5440/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0040/)

Variance in finite probability models · qf-variance-shift · candidate





## 41. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Put Z=X+(3). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation867/16

Compute the mean, then sum weighted squared deviations exactly to obtain 867/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0041/)

Variance in finite probability models · qf-variance-shift · candidate





## 42. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Put Z=X+(4). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation4617/121

Compute the mean, then sum weighted squared deviations exactly to obtain 4617/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0042/)

Variance in finite probability models · qf-variance-shift · candidate





## 43. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Put Z=X+(5). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1805/36

Compute the mean, then sum weighted squared deviations exactly to obtain 1805/36. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0043/)

Variance in finite probability models · qf-variance-shift · candidate





## 44. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Put Z=X+(6). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation10500/169

Compute the mean, then sum weighted squared deviations exactly to obtain 10500/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0044/)

Variance in finite probability models · qf-variance-shift · candidate





## 45. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Put Z=X+(7). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation297/4

Compute the mean, then sum weighted squared deviations exactly to obtain 297/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0045/)

Variance in finite probability models · qf-variance-shift · candidate





## 46. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Put Z=X+(8). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation8349/169

Compute the mean, then sum weighted squared deviations exactly to obtain 8349/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0046/)

Variance in finite probability models · qf-variance-shift · candidate





## 47. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Put Z=X+(9). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation3174/49

Compute the mean, then sum weighted squared deviations exactly to obtain 3174/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0047/)

Variance in finite probability models · qf-variance-shift · candidate





## 48. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Put Z=X+(10). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation80/1

Compute the mean, then sum weighted squared deviations exactly to obtain 80/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0048/)

Variance in finite probability models · qf-variance-shift · candidate





## 49. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Put Z=X+(11). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation24375/256

Compute the mean, then sum weighted squared deviations exactly to obtain 24375/256. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0049/)

Variance in finite probability models · qf-variance-shift · candidate





## 50. Probability Variance after Translation

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Put Z=X+(12). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1521/25

Compute the mean, then sum weighted squared deviations exactly to obtain 1521/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0050/)

Variance in finite probability models · qf-variance-shift · candidate





## 51. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Put Z=(0)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation0/1

Compute the mean, then sum weighted squared deviations exactly to obtain 0/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0051/)

Variance in finite probability models · qf-variance-scale · candidate





## 52. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Put Z=(-2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation9/1

Compute the mean, then sum weighted squared deviations exactly to obtain 9/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0052/)

Variance in finite probability models · qf-variance-scale · candidate





## 53. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Put Z=(3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation36/1

Compute the mean, then sum weighted squared deviations exactly to obtain 36/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0053/)

Variance in finite probability models · qf-variance-scale · candidate





## 54. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Put Z=(-4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation100/1

Compute the mean, then sum weighted squared deviations exactly to obtain 100/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0054/)

Variance in finite probability models · qf-variance-scale · candidate





## 55. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Put Z=(5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation189/1

Compute the mean, then sum weighted squared deviations exactly to obtain 189/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0055/)

Variance in finite probability models · qf-variance-scale · candidate





## 56. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Put Z=(-1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation98/9

Compute the mean, then sum weighted squared deviations exactly to obtain 98/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0056/)

Variance in finite probability models · qf-variance-scale · candidate





## 57. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Put Z=(2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation2880/49

Compute the mean, then sum weighted squared deviations exactly to obtain 2880/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0057/)

Variance in finite probability models · qf-variance-scale · candidate





## 58. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Put Z=(-3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation10935/64

Compute the mean, then sum weighted squared deviations exactly to obtain 10935/64. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0058/)

Variance in finite probability models · qf-variance-scale · candidate





## 59. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Put Z=(4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation13200/49

Compute the mean, then sum weighted squared deviations exactly to obtain 13200/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0059/)

Variance in finite probability models · qf-variance-scale · candidate





## 60. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Put Z=(-5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation9075/16

Compute the mean, then sum weighted squared deviations exactly to obtain 9075/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0060/)

Variance in finite probability models · qf-variance-scale · candidate





## 61. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Put Z=(1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation260/9

Compute the mean, then sum weighted squared deviations exactly to obtain 260/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0061/)

Variance in finite probability models · qf-variance-scale · candidate





## 62. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Put Z=(-2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation3549/25

Compute the mean, then sum weighted squared deviations exactly to obtain 3549/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0062/)

Variance in finite probability models · qf-variance-scale · candidate





## 63. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Put Z=(3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation245/1

Compute the mean, then sum weighted squared deviations exactly to obtain 245/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0063/)

Variance in finite probability models · qf-variance-scale · candidate





## 64. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Put Z=(-4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation576/1

Compute the mean, then sum weighted squared deviations exactly to obtain 576/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0064/)

Variance in finite probability models · qf-variance-scale · candidate





## 65. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Put Z=(5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation136000/121

Compute the mean, then sum weighted squared deviations exactly to obtain 136000/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0065/)

Variance in finite probability models · qf-variance-scale · candidate





## 66. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Put Z=(-1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation867/16

Compute the mean, then sum weighted squared deviations exactly to obtain 867/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0066/)

Variance in finite probability models · qf-variance-scale · candidate





## 67. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Put Z=(2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation18468/121

Compute the mean, then sum weighted squared deviations exactly to obtain 18468/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0067/)

Variance in finite probability models · qf-variance-scale · candidate





## 68. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Put Z=(-3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1805/4

Compute the mean, then sum weighted squared deviations exactly to obtain 1805/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0068/)

Variance in finite probability models · qf-variance-scale · candidate





## 69. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Put Z=(4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation168000/169

Compute the mean, then sum weighted squared deviations exactly to obtain 168000/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0069/)

Variance in finite probability models · qf-variance-scale · candidate





## 70. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Put Z=(-5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation7425/4

Compute the mean, then sum weighted squared deviations exactly to obtain 7425/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0070/)

Variance in finite probability models · qf-variance-scale · candidate





## 71. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Put Z=(1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation8349/169

Compute the mean, then sum weighted squared deviations exactly to obtain 8349/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0071/)

Variance in finite probability models · qf-variance-scale · candidate





## 72. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Put Z=(-2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation12696/49

Compute the mean, then sum weighted squared deviations exactly to obtain 12696/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0072/)

Variance in finite probability models · qf-variance-scale · candidate





## 73. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Put Z=(3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation720/1

Compute the mean, then sum weighted squared deviations exactly to obtain 720/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0073/)

Variance in finite probability models · qf-variance-scale · candidate





## 74. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Put Z=(-4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation24375/16

Compute the mean, then sum weighted squared deviations exactly to obtain 24375/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0074/)

Variance in finite probability models · qf-variance-scale · candidate





## 75. Probability Variance after Scaling

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Put Z=(5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.

View answer and explanation1521/1

Compute the mean, then sum weighted squared deviations exactly to obtain 1521/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0075/)

Variance in finite probability models · qf-variance-scale · candidate





## 76. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationYes

The attained values on positive-weight support are {5}. Constancy there gives the answer Yes. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0076/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 77. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-7, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0077/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 78. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-8, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0078/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 79. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-9, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0079/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 80. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-6, 0}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0080/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 81. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-11, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0081/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 82. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-4, 4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0082/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 83. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-13, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0083/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 84. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-2, 8}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0084/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 85. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-15, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0085/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 86. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {0, 12}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0086/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 87. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-17, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0087/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 88. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {2, 16}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0088/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 89. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-19, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0089/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 90. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {4, 20}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0090/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 91. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-21, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0091/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 92. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {6, 24}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0092/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 93. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-23, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0093/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 94. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {8, 28}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0094/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 95. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-25, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0095/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 96. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {10, 32}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0096/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 97. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-27, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0097/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 98. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {12, 36}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0098/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 99. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {-29, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0099/)

Variance in finite probability models · qf-variance-zero-support · candidate





## 100. Zero Variance on Positive-Probability Support

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.

View answer and explanationNo

The attained values on positive-weight support are {14, 40}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261004-advance-20261005-r07-variance-0100/)

Variance in finite probability models · qf-variance-zero-support · candidate



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