Question

100 surnames were randomly picked up from a telephone directory and the distribution of the number of letters of the English alphabet in the surnames are obtained as follows:

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

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

Answer 1

Number of letters	Number of surnames (fi)	commulative frequency
1-4	6	6
4-7	30	6+30=36
7-10	40	36+40=76
10-13	16	76+16=92
13-16	4	92+4 =96
16-9	4	96+4=100
	\(\sum\)fi = 100

We know that median = l + \(\frac{\frac{n}{2}-cf}{f} \times h.\)

Where, n = \(\sum\)f_{i = 100, therefore, \(\frac{n}{2}\) = \(\frac{100}{2}\) = 50.}

In the above distribution table, the commulative frequency 76 is nearest greater value of \(\frac{n}{2}\) = 50. 

Hence, the median class is 7 – 10. 

Now, l = Lower limit of median class = 7. 

Cf = Commulative frequency of the class before medium class = 36. 

f = frequency of median class = 40. 

h = Class interval = 10 – 7 = 3.

\(\therefore\) Median = \(7 + \frac{50-36}{40}\times3\) = \(7+ \frac{14}{40} \times 3\) = \(7+ \frac{21}{20}\) = 7+ 1.05 = 8.05.

Hence, the median number of letters is 8.05.

Mean :-

We Know that, Mean (\(\bar x\)) = a + h \(\frac{∑ f_iu_i}{∑ f_i}\) . 

Where a = assumed mean = 11.5 

And h = class interval = 4 – 1 = 3. 

And = \(\frac{x_i -a}{h}\) . 

By above distribution table, we get ∑f_i  = 100 & ∑f_iu_i  = -106. 

∴ Mean (\(\bar x\)) = 11.5 + 3 × \(\frac{-106}{100}\)= 11.5 – \(\frac{318}{100}\) = 11.5 − 3.18 = 8.32. 

Hence, the mean number of letters is 8.32.

Mode :-

We Know that mode = l + \(\frac{f_1 - f_0} {2f_1 - f_0 - f_2} \times h.\) 

In the above distribution table, the highest frequency is 40 and interval of highest frequency is 7 – 10. 

∴ The modal class = interval of highest frequency = 7 – 10. 

Now, = lower limit of modal class = 7, 

h = class interval = 10 – 7 = 3, 

f₁ = frequency of modal class = 40. 

f₀ = frequency of the class before modal class = 30, 

f₂ = frequency of the class after modal class = 16

Therefore, the mode = 7 + \(\frac{40-30}{2 \times 40 - 30-16} \times 3\) 

= \({7} + \frac{10}{80-46}\times3 = 7+\frac{10}{34} \times 3\) 

= \(7+\frac{30}{34} = 7 + \frac{15}{17} = 7+0.88 = 7.88\) 

Hence, the modal size of surnames is 7.88.