The frequency distribution of scores x obtained by a group of candidates in an aptitude test is given in the table below: | Score (x) | 2 | 4 | 6 | 8 | 10 | |---|---|---|---|---|---| | Frequency (f) | 3 | k | 7 | 4 | 2 | If the mean score of the distribution is 5.8, find the value of k.

The frequency distribution of scores x obtained by a group of candidates in an aptitude te… — JAMB UTME

The frequency distribution of scores $x$ obtained by a group of candidates in an aptitude test is given in the table below:

Score ( $x$ )	2	4	6	8	10
Frequency ( $f$ )	3	$k$	7	4	2

If the mean score of the distribution is $5.8$ , find the value of $k$ .

Answer: 4

The value of

k

4

The mean score of an ungrouped frequency distribution is given by

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

. Evaluating the sum of frequencies gives

\sum f = 16 + k

and the weighted sum of scores gives

\sum fx = 100 + 4k

. Substituting the given mean of

5.8

, we get

\frac{100 + 4k}{16 + k} = 5.8

. Cross-multiplying yields

100 + 4k = 92.8 + 5.8k

, leading to

1.8k = 7.2

, so

k = 4

Express total frequency

\sum f

in terms of

k

\sum f = 16 + k

The total number of observations is the sum of all frequencies in the distribution.

Calculate the sum of weighted scores

\sum fx

in terms of

k

\sum fx = 100 + 4k

Each score value must be multiplied by its corresponding frequency and summed together.

Set up the mean equation using the formula

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

5.8 = \frac{100 + 4k}{16 + k}

The mean of ungrouped data presented in a frequency table is total sum divided by total frequency.

Solve the linear equation for

k

k = 4

Cross-multiplying gives

92.8 + 5.8k = 100 + 4k

, which simplifies to

1.8k = 7.2

, yielding

k = 4

Mean of Ungrouped Data from a Frequency Table