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

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

Cumulative frequency just greater $\frac{n}2 ( i.e., \frac{100}2 = 50)$ than is 76, belonging to class interval 7−10.
Median class = 7−10
Lower limit ($l$) of median class = 7
Frequency ($f$) of median class = 36
Cumulative frequency ($cf$) of median class = 40
Class size ($h$) = 3

Median = $l + (\frac{\frac{n}2 - cf}f \times h)$

Median = $7 + (\frac{50 - 36}{40} \times 3)$

Median = 7 + $\frac{40 \times 3}{40}$

Median = 8.05

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To find the class mark (xi) for each interval, the following relation is used.

Class mark $(x_i)$ = $\frac {\text{Upper \,limit + Lower \,limit}}{2}$

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	Frequency (f$_i$)	** $\bf{x_i}$ **	$\bf{d_i = x_i -11.5}$	$\bf{u_i = \frac{d_i}{3}}$	$\bf{f_iu_i}$
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, it can be observed that

$\sum f_i = 100$
$\sum f_iu_i = -106$

Mean, $\overset{-}{x} = a + (\frac{\sum f_iu_i}{\sum f_i})\times h$

$\overset{-}{x}$ = $11.5 + (\frac{-106 }{100})\times 3$

$\overset{-}{x}$ = 11.5 - 3.18
Mean, $\overset{-}{x}$ = 8.32

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The data in the given table can be written as

Number of letters	Frequency (f$_i$)
1 - 4	6
4 - 7	30
7 - 10	40
10 - 13	16
13 - 16	4
16 - 19	4
Total	100

From the data given above, it can be observed that the maximum class frequency is 40, belonging to class interval 7 - 10.

Therefore, modal class = 7 - 10
Lower limit ($l$) of modal class = 7
Frequency ($f_1$) of modal class = 40
Frequency ($f_0$) of class preceding the modal class = 30
Frequency ($f_2$) of class succeeding the modal class = 16
Class size ($h$) = 3

Mode = $l$ + $(\frac{f_1 - f_0 }{2f_1 - f_0 - f_2)} \times h$

Mode = $7 + (\frac{40 - 30 }{ 2(40) - 30 - 16}) \times3$

Mode =$7+ [\frac{10}{34}] \times 3$

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

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