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 corresponding to their class intervals are presented below.

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

The cumulative frequency immediately exceeding \( \frac{n}{2} \) (i.e., \( \frac{100}{2} = 50 \)) is 76, which falls within the 7-10 class interval.
The median class is identified as 7-10.
The lower limit (\(l\)) of the median class is 7.
The frequency (\(f\)) of the median class is 30.
The cumulative frequency (\(cf\)) of the class preceding the median class is 36.
The class size (\(h\)) is 3.

The median is calculated using the formula: Median = \( l + (\frac{\frac{n}2 - cf}f \times h)\).

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

Median = 7 + \( \frac{14 \times 3}{40}\)

Median = 7 + \( \frac{42}{40}\)

Median = 7 + 1.05

Median = 8.05

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The class mark (\({x}_{i}\)) for each interval is determined by the following formula: Class mark (\({x}_{i}\)) = \( \frac{\text{Upper limit + Lower limit}}{2} \).

Assuming 11.5 as the assumed mean (a), the values for \({d}_{i}\), \({u}_{i}\), and \({f}_{i}{u}_{i}\) are calculated using the step deviation method as shown in the table below.

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 is observed that:

\(\sum f_i = 100\)
\(\sum f_iu_i = -106\)

The mean is calculated using the formula: 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 provided data can be represented 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

Observing the data, the highest class frequency is 40, which corresponds to the 7-10 class interval.

Thus, the modal class is 7-10.
The lower limit (\(l\)) of the modal class is 7.
The frequency (\(f_1\)) of the modal class is 40.
The frequency (\(f_0\)) of the class preceding the modal class is 30.
The frequency (\(f_2\)) of the class succeeding the modal class is 16.
The class size (\(h\)) is 3.

The mode is calculated using the formula: 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 = \( 7 + [\frac{10}{80 - 30 - 16}] \times 3 \)

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

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

Mode = 7 + 0.88

Mode = 7.88

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