The table below shows the frequency distribution of marks obtained by a group of students in a short test: | Mark (x) | 1 | 2 | 3 | 4 | | :--- | :--- | :--- | :--- | :--- | | Frequency (f) | 2 | 3 | 3 | 2 | What is the mean deviation of the distribution?

The table below shows the frequency distribution of marks obtained by a group of students … — JAMB UTME

The table below shows the frequency distribution of marks obtained by a group of students in a short test:

Mark ( $x$ )	1	2	3	4
Frequency ( $f$ )	2	3	3	2

What is the mean deviation of the distribution?

The mean deviation of the distribution is

0.9

The mean of the data is

\bar{x} = \frac{25}{10} = 2.5

. Calculating the sum of weighted absolute deviations yields

2(1.5) + 3(0.5) + 3(0.5) + 2(1.5) = 9.0

. Dividing by total frequency

10

gives the mean deviation as

0.9

Calculate the mean (

\bar{x}

) of the frequency distribution.

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

\sum fx = (2 \times 1) + (3 \times 2) + (3 \times 3) + (2 \times 4) = 2 + 6 + 9 + 8 = 25

. Therefore,

\bar{x} = \frac{25}{10} = 2.5

The mean is required to determine the deviations of each score value.

Find the absolute deviation

|x - \bar{x}|

for each score value.

For

x = 1

|1 - 2.5| = 1.5

; for

x = 2

|2 - 2.5| = 0.5

; for

x = 3

|3 - 2.5| = 0.5

; for

x = 4

|4 - 2.5| = 1.5

Mean deviation measures average distance from the mean, ignoring signs.

Multiply each absolute deviation by its corresponding frequency and sum them.

\sum f|x - \bar{x}| = (2 \times 1.5) + (3 \times 0.5) + (3 \times 0.5) + (2 \times 1.5) = 3 + 1.5 + 1.5 + 3 = 9.0

Frequencies reflect how many times each deviation occurs in the dataset.

Compute the mean deviation by dividing the weighted sum by the total frequency

\sum f

\text{Mean Deviation} = \frac{\sum f|x - \bar{x}|}{\sum f} = \frac{9.0}{10} = 0.9.

The mean deviation is the average of these absolute deviations across all observations.

Mean Deviation for Discrete Frequency Distribution