Question

100 surnames were randomly picked up from a local telephone directly 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	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.

Answer 1

Hint: First we will calculate mean using the formula $\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$, where $\sum {{f_i}{x_i}} $ is the total sum of the product of mid-value of class interval and frequency and $\sum {{f_i}} $ is the sum of frequency. After that calculate the median by the formula $l + \dfrac{{\dfrac{N}{2} - cf}}{f} \times h$. Then use the formula of mode $l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}$ to calculate the mode.

Complete step-by-step solution:
We are given that the monthly consumption is the confidence interval C.I.
Let us assume that ${f_i}$ represents the number of consumers and ${x_i}$ is the mid-value of the interval.
We will now form a table to find the value of the product ${f_i}{x_i}$ for the mean.

C.I.	${x_i}$	${f_i}$	${f_i}{x_i}$
1 – 4	2.5	6	15
4 – 7	5.5	30	165
7 – 10	8.5	40	340
10 – 13	11.5	16	184
13 – 16	14.5	4	58
16 – 19	17.5	4	70
Total		$\sum {{f_i}} = 100$	$\sum {{f_i}{x_i}} = 832$

We know that the formula to calculate mean using the formula $\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}$, where $\sum {{f_i}{x_i}} $ is the total sum of the product of mid-value of class interval and frequency and $\sum {{f_i}} $ is the sum of frequency.
Substitute the values to find the value of mean in the above formula, we get,
$ \Rightarrow $ Mean $ = \dfrac{{832}}{{100}}$
Divide the numerator by the denominator,
$ \Rightarrow $ Mean $ = 8.32$
Hence, the mean is 8.32.
We will now find the value of $cf$ from the above table for median and mode.
We know from the above table that the value of $n$ is 100.

C.I.	${x_i}$	${f_i}$	$cf$
1 – 4	2.5	6	6
4 – 7	5.5	30	36
7 – 10	8.5	40	76
10 – 13	11.5	16	92
13 – 16	14.5	4	96
16 – 19	17.5	4	100

The value of $\dfrac{n}{2}$ is,
$ \Rightarrow \dfrac{{100}}{2} = 50$
So, the median class is 7 – 10.
The lowest value of the median class is,
$ \Rightarrow l = 7$
The frequency of the median class is,
$ \Rightarrow f = 40$
The difference of interval is,
$ \Rightarrow h = 3$
The cumulative frequency above the median class is,
$ \Rightarrow cf = 36$
Substitute these values in the median formula,
$ \Rightarrow $ Median $ = 7 + \dfrac{{50 - 36}}{{40}} \times 3$
Simplify the terms,
$ \Rightarrow $ Median $ = 7 + 1.05$
Add the terms,
$ \Rightarrow $ Median $ = 8.05$
Hence, the median is 8.05.
Now the modal class is the class where ${f_i}$ is the highest, thus the modal class from the above table is,
$ \Rightarrow 7 - 10$
Then in the mode class, we have
$ \Rightarrow l = 7$
$ \Rightarrow {f_0} = 30$
$ \Rightarrow {f_1} = 40$
$ \Rightarrow {f_2} = 16$
$h = 3$
Substitute the values in mode formula,
$ \Rightarrow $ Mode $ = 7 + \dfrac{{40 - 30}}{{2\left( {40} \right) - 30 - 16}} \times 3$
Simplify the terms,
$ \Rightarrow $ Mode $ = 7 + \dfrac{{10}}{{34}} \times 3$
Multiply the numerator and then divide by denominator,
$ \Rightarrow $ Mode $ = 7 + 0.88$
Add the terms,
$ \Rightarrow $ Mode $ = 7.88$
Hence, the mode is 7.88.

Note: In solving these types of questions, students should know the formulae of mean, median, and mode. The question is really simple, students should note down the values from the problem carefully, else the answer can be wrong. The other possibility of a mistake in this problem is while calculating as it has a lot of mathematical calculations.