Question

$100$ surnames were randomly picked up from a local telephone directry and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows

No. of letters	$1-4$	$4-7$	$7-10$	$10-13$	$13-16$	$16-19$
No. of surnames	$6$	$30$	$40$	$16$	$4$	$4$

Find the median and mean number of letters in the surnames.

• A. $Median=12.5,Mean=9.1$
• B. $Median=9.09,Mean=5.12$
• C. $Median=8.05,Mean=8.32$
• D. $Median=7.31,Mean=7.16$

Answer 1

The correct option is C $Median=8.05,Mean=8.32$

Answer:-

No. of letters	No. of surnames $\left(f_i\right)$	C.F.	$X_i$	$x_i{f}_i$
$1-4$	$6$	$6$	$2.5$	$15$
$4-7$	$30$	$36$	$5.5$	$165$
$7-10$	$40$	$76$	$8.5$	$340$
$10-13$	$16$	$92$	$11.5$	$184$
$133-16$	$4$	$96$	$14.5$	$58$
$16-19$	$4$4	$100$	$17.5$	$70$
	$\Sigma f_i=100$			$\Sigma x_i{f}_i=832$

Mean = $\frac{832}{100}=8.32$

$\frac{N}{2}=\frac{100}{2}=50$

76 is greater than 50. $\therefore$ Median class is $7-10.$

$\therefore Median=L+\frac{\frac{N}{2}-C}{f}\times w$

L = Lower limit of median class $=7$

C = C.F. of previous class $=36$

f = frequency of median class $=40$

w = width of median class = upper limit - lower limit $=10-7=3$

Median $=7+\frac{50-36}{40}\times 3$

$=7+\frac{14}{40}\times 3$

$=7+\frac{42}{40}=7+1.05=8.05$

C) Mean $=8.32$ $\&$ Median $=8.05$