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:

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

• A. 8.32
• B. 8.92
• C. 6.79
• D. 9.79

Answer 1

The correct option is A 8.32

	Number of letters	Number of surnames $f_1$	Cumulative frequency
	1-4	6	$6=6$
	4-7	30	$6+30=36$
Median class	7-10	40	$36+40=76$	$50=\frac{n}{2}$
	10-13	16	$76+16=92$
	13-16	4	$92+4=96$
	16-19	4	$96+4=100

Total $n=100$
(i) Here,
$l=7,n=100,f=40,cf=36,h=3$
$Median=l+\left\{\frac{\frac{n}{2}-cf}{f}\right\}\times h$
$=7+\left\{\frac{50-36}{50}\right\}\times 3=7+\frac{21}{20}=8.05$
(ii) Modal class is (7-10)
$l=7,f_m=40,f_1=30,f_2=16,h=3$
$Mode=l+\left\{\frac{f_m-f_1}{2f_m-f_1-f_2}\right\}\times h$
$=7+\left\{\frac{40-30}{80-30-16}\right\}\times 3=7+\frac{30}{34}=7.88$
(iii) Here, $a=8.5,h=3,n=100$ and $\sum f_i{u}_i=-6$

Number of letters	$f_i$	Class mark $x_i$	$u_i=\frac{x_i=8.5}{3}$	$f_i\times u_i$
1-4	6	2.5	-2	-12
4-7	30	5.5	-1	-30
7-10	40	8.5=a	0	0
10-13	16	11.5	1	16
13-16	4	14.5	2	8
16-19	4	17.5	3	12
Total	$n=100			-6

$Mean=a+h\times \frac{1}{n}\times \sum f_i{u}_i=8.5+3\times \frac{1}{100}\times (-6)$
$8.5-\frac{18}{100}=8.5-0.18=8.32$