The table below shows the distribution of marks obtained by 40 candidates in a computer-based recruitment test: | Mark Interval | Frequency (f) | |---|---| | 1 - 10 | 4 | | 11 - 20 | 8 | | 21 - 30 | 12 | | 31 - 40 | 6 | | 41 - 50 | 10 | What is the estimated mean mark of the candidates?

The table below shows the distribution of marks obtained by 40 candidates in a computer-ba… — JAMB UTME

The table below shows the distribution of marks obtained by $40$ candidates in a computer-based recruitment test:

What is the estimated mean mark of the candidates?

The estimated mean mark of the candidates is

28.0

To find the mean of grouped frequency data, each class interval must be represented by its midpoint (

x

). Finding the midpoints (

5.5, 15.5, 25.5, 35.5, 45.5

) and multiplying by their frequencies gives total products summing to

1120

. Dividing by the total number of candidates (

40

) yields

28.0

Calculate the midpoint (

x

) for each class interval

Class midpoints are:

5.5

15.5

25.5

35.5

, and

45.5

The class mark or midpoint is the representative value for data within a grouped interval.

Multiply each class midpoint (

x

) by its corresponding frequency (

f

) to find

f \cdot x

4 \times 5.5 = 22

8 \times 15.5 = 124

12 \times 25.5 = 306

6 \times 35.5 = 213

10 \times 45.5 = 455

Determines the total weighted value contribution of each class.

Sum all

f \cdot x

values and calculate the total frequency

\sum f

\sum f \cdot x = 22 + 124 + 306 + 213 + 455 = 1120

, and

\sum f = 4 + 8 + 12 + 6 + 10 = 40

Provides the overall sum of data estimates and total sample size needed for the grouped mean formula.

Apply the grouped mean formula

\bar{x} = \frac{\sum f \cdot x}{\sum f}

\bar{x} = \frac{1120}{40} = 28.0.

Computes the estimated mean for grouped frequency data.

Grouped Mean Calculation using Class Midpoints

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