If the mean of the following distribution $54$ , find the value of $p$ :
Class: $0 - 20$ $20 - 40$ $40 - 60$ $60 - 80$ $80 - 100$
Frequency: $7$ $p$ $10$ $9$ $13$

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Solution

Computation Arithematic Mean

Class Mid-values
( $x_{i}$ ) requency
( $f_{i}$ ) $d_{i} = x_{i} - 50$ $u_{i} = \frac{x_{i} - 50}{20}$ $f_{i} u_{i}$
0-20 10 7 -40 -2 -14
20-40 30 pF -20 -1 -p
40-60 50 10 0 0 0
60-80 70 9 20 1 9
80-100 90 13 40 2 26

$\sum f_{i} = 39 + p$ $\sum f_{i} u_{i} = 21 - p$
We have,
A= 50, N= 39 +p, h= 20, $\sum f_{i} u_{i} = 21 - p$ and $¯ ¯¯¯ ¯ X = 54$

$∴ M e a n = A + h (\frac{1}{N} \sum f_{i} u_{i})$

$\Rightarrow 54 = 50 + 20 \times (\frac{21 - p}{39 + p})$

$\Rightarrow 4 = 20 \times (\frac{21 - p}{39 + p})$
$\Rightarrow 1 = 5 (\frac{21 - p}{39 + p}) \Rightarrow 39 + p = 105 - 5 p \Rightarrow 6 p = 66 \Rightarrow p = 11$

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$C l a s s$	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
$F r e q u e n c y$	$15$	$18$	$21$	$p$	$17$

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C.I.	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
Frequency	$8$	$15$	$20$	$p$	$3$

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f_{1}

f_{2}

Class	0-20	20-40	40-60	60-80	80-100	Total
Frequency	17	28	$f_{1}$	24	$f_{2}$	120

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Class:	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
Frequency:	$7$	$p$	$10$	$9$	$13$