The arithmetic mean of the following frequency distribution is $53$ . Determine the value of $p$ .
Class-interval $0 - 20$ $20 - 40$ $40 - 60$ $60 - 80$ $80 - 100$
Frequency $12$ $15$ $32$ $p$ $13$

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Solution

Step 1: Finding mean:
Completing the table,
Class-interval Mid-value $x_{i}$ Frequency $f_{i}$ $f_{i} x_{i}$
$0 - 20$ $10$ $12$ $120$
$20 - 40$ $30$ $15$ $450$
$40 - 60$ $50$ $32$ $1600$
$60 - 80$ $70$ $p$ $70 p$
$80 - 100$ $90$ $13$ $1170$
$Σ f_{i} = 72 + p$ $Σ f_{i} x_{i} = 3340 + 70 p$
The formula for mean is $\bar{x} = \frac{Σ f_{i} x_{i}}{Σ f_{i}}$ .
where $Σ f_{i}$ is the total frequency and $x_{i}$ is the mid-value of class intervals.
From the question, we have $\bar{x} = 53$ .
Step 2: Finding the value of $p$ :
Substituting the values in the formula for mean,
$\Rightarrow 53 = \frac{3340 + 70 p}{72 + p} \Rightarrow 3816 + 53 p = 3340 + 70 p \Rightarrow 17 p = 476 ∴ p = 28$
Hence, the value of $p$ is obtained as $28$ .

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53

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