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	Number of surnames
1 - 4	6
4 − 7	30
7 - 10	40
10 - 13	6
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.

Answer 1

The cumulative frequencies with their respective class intervals are as follows:

It can be observed that the difference between two consecutive upper class limits is 2. The class marks with their respective frequencies are obtained as below:

Number of letters	Frequency (f)	Cumulative frequency
1 − 4	0	6
4 − 7	30	30 + 6 = 36
7 − 10	40	36 + 40 = 76
10 − 13	16	76 + 16 = 92
13 − 16	4	92 + 4 = 96
16 − 19	4	96 + 4 = 100
Total (n)	100

It can be observed that the cumulative frequency just greater than `n/2 (i.e 100/2 = 50)` is 76, belonging to class interval 7 − 10.

Median class = 7 − 10

Lower limit (l) of median class = 7

Cumulative frequency (cf) of class preceding median class = 36

Frequency (f) of median class = 40

Class size (h) = 3

Median = `l +((n/2-cf)/f) xxh`

= `7+((50-36)/40)xx3`

= `7+(14xx3)/40`

= 8.05

To find the class marks of the given class intervals, the following relation is used.

`"class Mark" = ("Upper class limit + Lower class limit")/2`

Taking 11.5 as assumed mean (a), d_i, u_i, and f_iu_i are calculated according to step deviation method as follows:

Number of letters	Number of surnames fi	xi	di = xi− 11.5	ui =di/3	fiui
1 − 4	6	2.5	− 9	− 3	−18
4 − 7	30	5.5	− 6	− 2	−60
7 − 10	40	8.5	− 3	− 1	−40
10 − 13	16	11.5	0	0	0
13 − 16	4	14.5	3	1	4
16 − 19	4	17.5	6	2	8
Total	100				−106

From the table, we obtain

`sumf_i = -106`

`sumf_iu_i = 100`

Mean `barx = a+ ((sumf_iu_i)/(sumf_i))xxh`

=  `11.5 + ((-106)/100)xx3`

= 11.5 − 3.18

= 8.32

The data in the given table can be written as:

Number of letters	Frequency (fi)
1 − 4	6
4 − 7	30
7 − 10	40
10 − 13	16
13 − 16	4
16 − 19	4
Total (n)	100

From the table, it can be observed that the maximum class frequency is 40, belonging to class interval 7 − 10.

Modal class = 7 − 10

Lower limit (l) of modal class = 7

Class size (h) = 3

Frequency (f₁) of modal class = 40

Frequency (f₀) of class preceding the modal class = 30

Frequency (f₂) of class succeeding the modal class = 16

Mode = `l+((f_1-f_0)/(2f_1-f_0-f_2))xxh`

= `7+[(40-30)/(2(40)-30-16)]xx3`

= `7 +  10/34 xx 3`

= `7+30/34 = 7.88`

Therefore, median number and mean number of letters in surnames are 8.05 and 8.32, respectively, while modal size of surnames is 7.88.