Question

The mean of the following data is $18.75$ then the value of $p$ :

Class mark $\left(x_i\right)$	$10$	$15$	$p$	$25$	$30$
Frequency $\left(f_i\right)$	$5$	$10$	$7$	$8$	$2$

• A. $21$
• B. $20.6$
• C. $20$
• D. $22$

Answer 1

Answer: (C)

$20$

Frequency distribution table:

$x_i$	$f_i$	$f_i{x}_i$
$10$	$5$	$50$
$15$	$10$	$150$
$p$	$7$	$7p$
$25$	$8$	$200$
$30$	$2$	$60$

$N=\sum f_i=32$

$\sum f_i{x}_i=460+7p$

Mean $\overline{x}=\frac{\sum f_i{x}_i}{N}$

$\Rightarrow 18.75=\frac{460+7p}{32}$

$\Rightarrow 600=460+7p$

$\Rightarrow p=\frac{600-460}{7}$

$\Rightarrow p=20$

Hence, option $C$ is correct.