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.