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 explanation
0/1
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
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 explanation
9/4
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
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 explanation
4/1
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
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 explanation
25/4
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
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 explanation
189/25
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
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 explanation
98/9
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
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 explanation
720/49
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
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 explanation
1215/64
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
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 explanation
825/49
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
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 explanation
363/16
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
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 explanation
260/9
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
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 explanation
3549/100
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
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 explanation
245/9
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
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 explanation
36/1
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
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 explanation
5440/121
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
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 explanation
867/16
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
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 explanation
4617/121
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
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 explanation
1805/36
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
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 explanation
10500/169
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
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 explanation
297/4
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
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 explanation
8349/169
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
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 explanation
3174/49
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
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 explanation
80/1
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
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 explanation
24375/256
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
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 explanation
1521/25
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
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 explanation
0/1
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
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 explanation
9/4
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
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 explanation
4/1
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
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 explanation
25/4
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
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 explanation
189/25
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
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 explanation
98/9
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
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 explanation
720/49
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
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 explanation
1215/64
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
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 explanation
825/49
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
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 explanation
363/16
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
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 explanation
260/9
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
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 explanation
3549/100
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
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 explanation
245/9
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
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 explanation
36/1
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
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 explanation
5440/121
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
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 explanation
867/16
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
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 explanation
4617/121
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
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 explanation
1805/36
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
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 explanation
10500/169
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
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 explanation
297/4
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
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 explanation
8349/169
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
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 explanation
3174/49
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
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 explanation
80/1
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
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 explanation
24375/256
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
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 explanation
1521/25
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
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 explanation
0/1
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
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 explanation
9/1
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
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 explanation
36/1
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
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 explanation
100/1
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
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 explanation
189/1
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
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 explanation
98/9
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
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 explanation
2880/49
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
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 explanation
10935/64
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
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 explanation
13200/49
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
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 explanation
9075/16
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
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 explanation
260/9
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
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 explanation
3549/25
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
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 explanation
245/1
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
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 explanation
576/1
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
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 explanation
136000/121
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
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 explanation
867/16
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
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 explanation
18468/121
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
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 explanation
1805/4
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
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 explanation
168000/169
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
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 explanation
7425/4
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
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 explanation
8349/169
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
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 explanation
12696/49
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
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 explanation
720/1
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
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 explanation
24375/16
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
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 explanation
1521/1
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
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 explanation
No
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
Variance in finite probability models · qf-variance-zero-support · candidate
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