Variance

Find the (sample) variance of the list:
(67, 70, 74, 75, 77, 68, 72, 76, 69, 72)

The (sample) variance of a list of numbers X = {X_1, X_2, ..., X_n} with mean μ = (X_1 + X_2 + ... + X_n)/n is given by:
(abs(X_1 - μ)^2 + abs(X_2 - μ)^2 + ... + abs(X_n - μ)^2)/(n - 1)

There are n = 10 elements in the list X = {67, 70, 74, 75, 77, 68, 72, 76, 69, 72}:
(abs(X_1 - μ)^2 + abs(X_2 - μ)^2 + abs(X_3 - μ)^2 + abs(X_4 - μ)^2 + abs(X_5 - μ)^2 + abs(X_6 - μ)^2 + abs(X_7 - μ)^2 + abs(X_8 - μ)^2 + abs(X_9 - μ)^2 + abs(X_10 - μ)^2)/(10 - 1)

The elements X_i of the list X = {67, 70, 74, 75, 77, 68, 72, 76, 69, 72} are:
X_1 = 67
X_2 = 70
X_3 = 74
X_4 = 75
X_5 = 77
X_6 = 68
X_7 = 72
X_8 = 76
X_9 = 69
X_10 = 72
(abs(67 - μ)^2 + abs(70 - μ)^2 + abs(74 - μ)^2 + abs(75 - μ)^2 + abs(77 - μ)^2 + abs(68 - μ)^2 + abs(72 - μ)^2 + abs(76 - μ)^2 + abs(69 - μ)^2 + abs(72 - μ)^2)/(10 - 1)

The mean (μ) is given by

μ = (X_1 + X_2 + X_3 + X_4 + X_5 + X_6 + X_7 + X_8 + X_9 + X_10)/10 = (67 + 70 + 74 + 75 + 77 + 68 + 72 + 76 + 69 + 72)/10 = 72:
(abs(67 - 72)^2 + abs(70 - 72)^2 + abs(74 - 72)^2 + abs(75 - 72)^2 + abs(77 - 72)^2 + abs(68 - 72)^2 + abs(72 - 72)^2 + abs(76 - 72)^2 + abs(69 - 72)^2 + abs(72 - 72)^2)/(10 - 1)

The values of the differences are:
67 - 72 = -5
70 - 72 = -2
74 - 72 = 2
75 - 72 = 3
77 - 72 = 5
68 - 72 = -4
72 - 72 = 0
76 - 72 = 4
69 - 72 = -3
72 - 72 = 0
10 - 1 = 9
(abs(-5)^2 + abs(-2)^2 + abs(2)^2 + abs(3)^2 + abs(5)^2 + abs(-4)^2 + abs(0)^2 + abs(4)^2 + abs(-3)^2 + abs(0)^2)/9

The values of the terms in the numerator are:
abs(-5)^2 = 25
abs(-2)^2 = 4
abs(2)^2 = 4
abs(3)^2 = 9
abs(5)^2 = 25
abs(-4)^2 = 16
abs(0)^2 = 0
abs(4)^2 = 16
abs(-3)^2 = 9
abs(0)^2 = 0
(25 + 4 + 4 + 9 + 25 + 16 + 0 + 16 + 9 + 0)/9

25 + 4 + 4 + 9 + 25 + 16 + 0 + 16 + 9 + 0 = 108:
Answer: |12

What is the variance of 67, 70, 74, 75, 77, 68, 72, 76, 69, 72?

\(\large\sigma^2=\frac{\sum\limits_{i=1}^{n} (x_i-\overline x\ )^2 }{n-1} \)  

\(n=10\)

\({\color{blue}\overline x}=\frac{\sum(67, 70, 74, 75, 77, 68, 72, 76, 69, 72)}{10}\color{blue}=72\)

\(\sum\limits_{i=1}^{10} (x_i-\overline x\ )^2 \) = (67-72)²+(70-72)²+(74-72)²+(75-72)²+(77-72)²+(68-72)²+(72-72)²+(76-72)²+(69-72)²+(72-72)²

=25+4+4+9+25+16+0+16+9+0

=108

\(\sigma^2=\frac{108}{10-1}=12\)

\(\sigma=\sqrt{12}=3.4641\)

\(The \ variance \ is \ \sigma =3.4641.\)

   !

The variance of the given set of numbers is 108/10 = 10.8.

Division by 9 to get a value of 12 provides the best estimate for the variance of the parent population from which this set of numbers has been taken as a sample. We are not being told that this is what is required, we are simply being asked for the variance of this set.

You got it backwards:

Variance =12

Population variance =10.8