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:

$\begin{array}{cc}Numberofletters& Numberofsurnames\\ 1-4& 6\\ 4-7& 30\\ 7-10& 40\\ 10-13& 16\\ 13-16& 4\\ 16-19& 4\end{array}$

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

$\begin{array}{ccc}Numberof& Numberof& Cumulative\\ letters& surnames& frequency\\ 1-4& 6& 6\\ 4-7& 30& 36\\ 7-10& 40& 76\\ 10-13& 16& 92\\ 13-16& 4& 96\\ 16-19& 4& 100\end{array}$

Now, n = 100

So, $\frac{n}{2}=\frac{100}{2}=50$

This observation lies in the class 7 - 10. So, 7 - 10 is the median class.

Therefore, l = 7, h = 3, f = 40, cf = 36

$\therefore Median=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h=7+\left(\frac{50-36}{40}\right)\times 3$

$=7+\frac{21}{20}=7+1.05=8.05$

Hence, the median number of letters in the surnames is 8.05

Mean. Take a = 8.5, h = 3

$\begin{array}{cccccc}Numberof& Numberof& Class& d_i=x_i-8.5& u_i=\frac{x_i-8.5}{3}& f_i{u}_i\\ letters& surnames& mark& & & \\ & \left(f_i\right)& & & & \\ 1-4& 6& 2.5& -6& -2& -12\\ 4-7& 30& 5.5& -3& -1& -30\\ 7-10& 40& 8.5& 0& 0& 0\\ 10-13& 16& 11.5& 3& 1& 16\\ 13-16& 4& 14.5& 6& 2& 8\\ 16-19& 4& 17.5& 9& 3& 12\\ Total& \sum f_i=100& & & & \sum f_i{u}_i=-6\end{array}$

Using the step-deviation method.

$\overline{x}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h=8.5+\left(\frac{-6}{100}\right)\times 3$

$=8.5-0.18=8.32$

Hence, the mean number of letters in the surnames is 8.32

Mode: Since the maximum number of surnames have number of letters in the interval 7 - 10, the modal class is 7 - 10

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

$\therefore Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h=7+\left(\frac{40-30}{2\times 40-30-16}\right)\times 3$

$=7+\frac{30}{34}=7+0.88=7.88$

Hence, the modal size of the surnames is 7.88