Find the value of $p$ , if the mean of the following distribution is $18$ .
Variate $(x)$ $13$ $15$ $17$ $19$ $20 + p$ $23$
Frequency $(f)$ $8$ $2$ $3$ $4$ $5 p$ $6$

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Solution

we know thta
Mean $= \frac{\sum f i x i}{\sum f i}$
$\Rightarrow \frac{104 + 30 + 51 + 76 + 5 p^{2} + 100 p + 138}{8 + 2 + 3 + 4 + 5 p + 6} = 18$
$\Rightarrow \frac{399 + 5 p^{2} + 100 p}{23 + 5 p} = 18$
$\Rightarrow 399 + 5 p^{2} + 100 p = 414 + 90 p$
$\Rightarrow 5 p^{2} + 10 p = 15$
$\Rightarrow p^{2} + 2 p = 3$
So, $p = 1$ or $p = 3$
Since, $p$ is frequency $t$ cannot be negative
So, $p = 1$

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p

20

$x$ :	$15$	$17$	$19$	$20 + p$	$23$
$f$ :	$2$	$3$	$4$	$5 p$	$6$

18

19 - 21

Class	$11 - 13$	$13 - 15$	$15 - 17$	$17 - 19$	$19 - 21$	$21 - 23$	$23 - 25$
Frequency	$3$	$6$	$9$	$13$	$f$	$5$	$4$

x:	15	17	19	20 + p	23
f:	2	3	4	5 p	6

x	15	17	19	20 + p	23
f	2	3	4	5p	6

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Variate $(x)$	$13$	$15$	$17$	$19$	$20 + p$	$23$
Frequency $(f)$	$8$	$2$	$3$	$4$	$5 p$	$6$