The table below shows the distribution of weekly overtime hours worked by 10 technicians in a manufacturing plant: | Overtime Hours (x) | 2 | 4 | 6 | 8 | |---|---|---|---|---| | Number of Technicians (f) | 4 | 3 | 2 | 1 | What is the variance of the overtime hours?

The table below shows the distribution of weekly overtime hours worked by 10 technicians i… — JAMB UTME

The table below shows the distribution of weekly overtime hours worked by $10$ technicians in a manufacturing plant:

Overtime Hours ( $x$ )	2	4	6	8
Number of Technicians ( $f$ )	4	3	2	1

What is the variance of the overtime hours?

The variance of the overtime hours is

4

The correct answer is

4

. First, compute the mean

\bar{x} = \frac{40}{10} = 4

. Then compute the sum of weighted squared deviations

\sum f(x - \bar{x})^2 = 4(2-4)^2 + 3(4-4)^2 + 2(6-4)^2 + 1(8-4)^2 = 16 + 0 + 8 + 16 = 40

. Dividing this by total frequency

\sum f = 10

yields

\sigma^2 = \frac{40}{10} = 4

Calculate the total frequency (

\sum f

) and the sum of the products of values and frequencies (

\sum fx

\sum f = 4 + 3 + 2 + 1 = 10

and

\sum fx = (2 \times 4) + (4 \times 3) + (6 \times 2) + (8 \times 1) = 8 + 12 + 12 + 8 = 40

These sums are required to compute the mean of the distribution.

Determine the mean (

\bar{x}

) of the distribution.

\bar{x} = \frac{\sum fx}{\sum f} = \frac{40}{10} = 4

The mean is used as the central reference point to find deviations.

Find the squared deviation

(x - \bar{x})^2

for each score and multiply by its respective frequency

f

For

x = 2

4(2 - 4)^2 = 16

.
For

x = 4

3(4 - 4)^2 = 0

.
For

x = 6

2(6 - 4)^2 = 8

.
For

x = 8

1(8 - 4)^2 = 16

.
Total sum

\sum f(x - \bar{x})^2 = 16 + 0 + 8 + 16 = 40

Variance evaluates the average of these weighted squared deviations.

Compute the variance (

\sigma^2

) by dividing the sum of weighted squared deviations by the total frequency.

\sigma^2 = \frac{\sum f(x - \bar{x})^2}{\sum f} = \frac{40}{10} = 4

This gives the population variance for the given frequency distribution.

Variance for Frequency Distributions

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