# 100 surnames were randomly picked up from a local telephone directory

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:

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.

Solution:

(i) Here,

$\ell=7, \mathrm{n}=100, \mathrm{f}=40, \mathrm{cf}=36, \mathrm{~h}=3$

Median $=\ell+\left\{\frac{\frac{\mathbf{n}}{\mathbf{2}}-\mathbf{c f}}{\mathbf{f}}\right\} \times \mathrm{h}$

$=7+\left\{\frac{\mathbf{5 0}-\mathbf{3 6}}{\mathbf{4 0}}\right\} \times 3=7+\frac{\mathbf{2 1}}{\mathbf{2 0}}=8.05$

(ii) Modal class is (7 – 10).

$\ell=7, \mathrm{f}_{\mathrm{m}}=40, \mathrm{f}_{1}=30, \mathrm{f}_{2}=16, \mathrm{~h}=3$

Mode $=\ell+\left\{\frac{\mathbf{f}_{\mathbf{m}}-\mathbf{f}}{\mathbf{2 f}_{\mathbf{m}}-\mathbf{f}-\mathbf{f}_{\mathbf{2}}}\right\} \times \mathbf{h}$

$=7+\left\{\frac{\mathbf{4 0}-\mathbf{3 0}}{\mathbf{8 0}-\mathbf{3 0}-\mathbf{1 6}}\right\} \times 3=7+\frac{\mathbf{3 0}}{\mathbf{3 4}}=7.88$

(iii) Here, $a=8.5, h=3, n=100$ and $\Sigma f_{i} u_{i}=-6$.

Mean $=a+h \times \frac{\mathbf{1}}{\mathbf{n}} \times \Sigma f_{1} u_{1}=8.5+3 \times \frac{\mathbf{1}}{\mathbf{1 0 0}} \times(-6)=8.5-\frac{\mathbf{1 8}}{\mathbf{1 0 0}}=8.5-0.18=8.32$