The arithmetic mean of five observations 1, 3, x, 9, and 12 is 6. What is the variance of … — JAMB UTME

The arithmetic mean of five observations $1$ , $3$ , $x$ , $9$ , and $12$ is $6$ . What is the variance of this set of observations?

The variance of the set of observations is

16

First, the missing value is found by equating the sum of the five numbers to

5 \times 6 = 30

, which gives

x = 5

. The data set is

1, 3, 5, 9, 12

. The deviations from the mean

6

are

-5, -3, -1, 3, 6

, and their squares are

25, 9, 1, 9, 36

. Summing these squared deviations gives

80

. Dividing by the number of observations (

5

) yields the variance of

16

Find the missing value

x

using the arithmetic mean formula.

x = 5

The mean of

N = 5

numbers is

\bar{x} = 6

, so the total sum is

5 \times 6 = 30

. Summing the known numbers gives

1 + 3 + x + 9 + 12 = 30 \implies 25 + x = 30 \implies x = 5

Calculate the deviations

(x_i - \bar{x})

for each data value from the mean

\bar{x} = 6

Deviations are

-5, -3, -1, 3, 6

Subtracting

6

from each value

(1, 3, 5, 9, 12)

gives

1-6=-5

3-6=-3

5-6=-1

9-6=3

, and

12-6=6

Square each deviation.

Squared deviations are

25, 9, 1, 9, 36

Squaring each deviation yields

(-5)^2 = 25

(-3)^2 = 9

(-1)^2 = 1

3^2 = 9

, and

6^2 = 36

Compute the population variance

\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N}

\sigma^2 = 16

The sum of squared deviations is

25 + 9 + 1 + 9 + 36 = 80

. Dividing by

N = 5

gives

\frac{80}{5} = 16

Variance of Ungrouped Data with a Missing Observation

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