Question

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

Number of letters	$1-4$	$4-7$	$7-10$	$10-13$	$13-16$	$16-19$
Number of surnames	$6$	$30$	$40$	$16$	$4$	$4$

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Answer 1

Let us prepare the following table to compute the median :

Number of letters	Number of surnames (Frequency)	Cumulative frequency
$1-4$	$6$	$6$
$4-7$	$30$	$36$
$7-10$	$40$	$76$
$10-13$	$16$	$92$
$13-16$	$4$	$96$
$16-19$	$4$	$100=n$

We have, $n=100$

$\Rightarrow \frac{n}{2}=50$

The cumulative frequency just greater than $\frac{n}{2}$ is $76$ and the corresponding class is $7–10$.

Thus, $7–10$ is the median class such that

$\frac{n}{2}=50,l=7,f=40,cf=36$ and $h=3$

Substitute these values in the formula

Median, $M=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$

$M=7+\left(\frac{50-36}{40}\right)\times 3$

$M=7+\frac{14}{40}\times 3=7+1.05=8.05$

Now, calculation of mean:

Number of letters	Mid-Point $\left(x_i\right)$	Frequency $\left(f_i\right)$	$f_i{x}_i$
$1-4$	$2.5$	$6$	$15$
$4-7$	$5.5$	$30$	$165$
$7-10$	$8.5$	$40$	$340$
$10-13$	$11.5$	$16$	$184$
$13-16$	$14.5$	$4$	$58$
$16-19$	$17.5$	$4$	$70$
Total		$100$	$832$

Therefore, Mean, $\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{832}{100}=8.32$

Calculation ofMode:

The class $7–10$ has the maximum frequency therefore, this is the modal class.

Here,

$l=7,h=3,f_1=40,f_0=30$ and $f_2=16$

Now, let us substitute these values in the formula

Mode $=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$

$=7+\frac{40-30}{80-30-16}\times 3$

$=7+\frac{10}{34}\times 3=7+0.88=7.88$

Hence, median $=8.05$, mean $=8.32$ and mode $=7.88$