Main derivation
S01 · definition
Before
D=(d_i)_{i=1}^{n},\ n>0After
\bar d=\tfrac{1}{n}\sum_{i=1}^{n}d_i,\ \tilde d,\ R=\max D-\min D,\ \operatorname{Mo}(D)Conditions: n>0
Rule: mko_descstat_def · Uses: G1, G2
Leaf basis: mko_descstat_def + def_unfold
State the MKO's four definitions: the mean is the sum divided by n; the median is taken after sorting (the middle value for odd n, the mean of the two middle values for even n); the range is the maximum minus the minimum; the mode set follows the convention of G2. Every computation is done on the list with repetitions kept.
S02 · implies
Before
D=(4,10,1,3,8,1,6,3)
After
n=8>0
Conditions: none
Rule: finite_counting_basics · Uses: G1
Leaf basis: finite_counting_basics
Counting positions, D has 8 entries; the repeated values 1 and 3 each occupy two positions, so n=8, not the number 6 of distinct values. Since n>0, the mean and the median are defined.
S03 · equivalent
Before
\sum_{i=1}^{8}d_iAfter
4+10+1+3+8+1+6+3=36
Conditions: none
Rule: finite_sum_rules · Uses: G1, finite_sum_rules
Leaf basis: finite_sum_rules + numeric_evaluation
Add the terms one at a time by the recursive rule for finite sums; the partial sums are 4, 14, 15, 18, 26, 27, 33, 36. The value 1 is added twice, and so is 3; each observation is added once.
S04 · equivalent
Before
\bar d=\tfrac{36}{8}After
\bar d=\tfrac{9}{2}Conditions: 8\neq 0
Rule: frac_rules · Uses: S01, S02, S03
Leaf basis: frac_rules + numeric_evaluation
The mean is 36/8. Dividing numerator and denominator by gcd(36,8)=4 (frac_rules, a/b=(ak)/(bk) with k≠0) gives 9/2; gcd(9,2)=1, so it is reduced (MKO assumption a2). Check: 8·(9/2)=36, so the deviations from the mean sum to 36−8·(9/2)=0.
S05 · implies
Before
(4,10,1,3,8,1,6,3)
After
(s_{(k)})_{k=1}^{8}=(1,1,3,3,4,6,8,10)Conditions: none
Rule: perm_invariance · Uses: G1, perm_invariance
Leaf basis: perm_invariance + finite_counting_basics
Arrange D in increasing order. Check that only the order changed: in both lists 1 and 3 occur twice each and 4, 6, 8, 10 once each, 8 entries in all, and the sum is still 36. By MKO assumption a1, sorting changes only the order and does not remove repeated values.
S06 · equivalent
Before
\tilde d,\ n=8\ \text{even}After
\tilde d=\tfrac{s_{(4)}+s_{(5)}}{2}=\tfrac{3+4}{2}=\tfrac{7}{2}Conditions: n\ \text{even}
Rule: mko_descstat_def · Uses: S01, S02, S05
Leaf basis: mko_descstat_def + frac_rules + numeric_evaluation
n=8 is even, so the two middle positions are n/2=4 and n/2+1=5. After sorting, s_(4)=3 and s_(5)=4, and the median is their mean (3+4)/2=7/2. It lies between 3 and 4 and is not one of the observations.
S07 · equivalent
Before
R=\max D-\min D
After
R=10-1=9
Conditions: none
Rule: mko_descstat_def · Uses: S01, S05
Leaf basis: mko_descstat_def + order_basics + numeric_evaluation
After sorting, the minimum is the first entry 1 and the maximum is the last entry 10; sorting does not change which observations are present, so these are also the minimum and maximum of D. The range is 10−1=9.
S08 · implies
Before
f(v)\ \text{for } v\in\{1,3,4,6,8,10\}After
f(1)=2,\ f(3)=2,\ f(4)=f(6)=f(8)=f(10)=1
Conditions: none
Rule: finite_counting_basics · Uses: G1, S05, finite_counting_basics
Leaf basis: finite_counting_basics + numeric_evaluation
Count how often each value occurs. Check: 2+2+1+1+1+1=8=n, so every observation is counted exactly once. Frequencies must be counted with repetitions kept (MKO second paragraph).
S09 · implies
Before
\max_v f(v)=2
After
\operatorname{Mo}(D)=\{1,3\}Conditions: none
Rule: mko_descstat_def · Uses: S08, G2
Leaf basis: mko_descstat_def + order_basics
The maximum frequency is 2, attained by 1 and 3. By the convention of G2 both belong to the mode set, listed in increasing order, each once: {1,3}. Omitting a tied value is a misconception the MKO lists.
S10 · theorem_application
Before
1\le d_i\le 10\ \text{for every } iAfter
1\le \tfrac{9}{2}\le 10Conditions: n=8>0
Rule: L_mean_bounds · Uses: S04, S07, L_mean_bounds
Leaf basis: L_mean_bounds + numeric_evaluation
Every observation lies between the minimum 1 and the maximum 10, so by Lemma L_mean_bounds (m=1, M=10) the mean 9/2 must lie between 1 and 10; indeed 1≤9/2≤10. This is a plausibility check of S04 and cannot replace the computation in S03 and S04. The lemma is proved in S10.1 to S10.3.
S11 · implies
Before
\bar d=\tfrac{9}{2},\ \tilde d=\tfrac{7}{2}After
\bar d\notin D,\ \tilde d\notin D
Conditions: none
Rule: numeric_evaluation · Uses: S04, S06, G1
Leaf basis: numeric_evaluation + divides_def
9/2 is in lowest terms with denominator 2 (9 leaves remainder 1 on division by 2), so it is not an integer; every observation in D is an integer, so the mean is none of the observations. Likewise 7/2 is not an integer, so the median is none of the observations either. MKO first paragraph: the mean need not be an observation, and for even n the median may lie between the two middle values.
S12 · implies
Before
\{1,3,4,6,8,10\}\ \text{(deduplicated)}After
\tfrac{32}{6}=\tfrac{16}{3}\neq\tfrac{9}{2},\quad \tfrac{4+6}{2}=5\neq\tfrac{7}{2}Conditions: none
Rule: mko_descstat_def · Uses: S04, S06, G1
Leaf basis: mko_descstat_def + numeric_evaluation
If D were first deduplicated to {1,3,4,6,8,10} (6 values), the mean would be (1+3+4+6+8+10)/6=32/6=16/3 and the median (4+6)/2=5, both different from S04 and S06. So deduplication really changes the results and violates MKO assumption a1; the correct answers are those for D with repetitions kept. Deduplication preserves the minimum and maximum of any nonempty finite observation list, so it always preserves the range (10−1=9 here); this does not make it harmless for summaries such as the mean, median, or frequencies.
S13 · implies
Before
\text{tasks of the goal}After
n=8,\ \bar d=\tfrac{9}{2},\ \tilde d=\tfrac{7}{2},\ R=9,\ \operatorname{Mo}(D)=\{1,3\}Conditions: none
Rule: and_intro_elim · Uses: S02, S04, S06, S07, S09, S10, G3
Leaf basis: and_intro_elim
Combine: n=8, mean 9/2, median 7/2, range 9, mode set {1,3}; the mean passes the check 1≤9/2≤10. By G3 these numbers describe D itself only and imply no population mean, probability model or causal conclusion.
Error controls and negative controls
Do not deduplicate first
One misconception the MKO names is deduplicating before computing. After deduplication the mean becomes 16/3 and the median 5 (S12), neither of which summarizes D. Repeated values are part of the data: two 1s mean two observations equal to 1.
Every tied value of maximal frequency belongs to the mode set
1 and 3 both occur twice, so the mode set is {1,3}, not {1}. If every value occurred once, under this entry's convention all distinct values would be in the mode set; other textbooks may call that 'no mode', which is a different convention, not a contradiction (MKO assumption a3).
Take the median after sorting
In the unsorted D the 4th and 5th entries are 3 and 8, whose mean 11/2 is not the median. One must first sort to (1,1,3,3,4,6,8,10) and then take the 4th and 5th entries 3 and 4, giving 7/2 (S06).
- The median is the mean of the 4th and 5th entries of D, (3+8)/2=11/2 →
reject — The median is taken after sorting; the sorted 4th and 5th entries are 3 and 4, so the median is 7/2. - The mean is (1+3+4+6+8+10)/6=16/3 →
reject — That is the mean after deduplication; with repetitions kept it is 36/8=9/2. - The mode set is {1} →
reject — 3 also occurs twice, the maximum frequency; the mode set is {1,3}. - The mean of D is 9/2, so the population that produced D has mean 9/2 →
reject — A descriptive summary describes D only; there is no population or sampling model here (MKO assumption a3).