Question

Calculate the mean of weekly pocket money for students from the following data,

Weekly pocket money in rupees	Number of Students
25-30	2
30-35	3
35-40	5
40-45	7
45-50	21
50-55	10
55-60	2

If the weekly pocket money is increased by 8 for all students what will be the new mean of the weekly pocket money.\[\]

Answer 1

Hint: First calculate the midpoints of each interval and then draw the midpoint-frequency table to find the product of midpoints and frequencies. Use the mean formula of class interval- frequency data to find the mean. Change the limits and midpoints of the intervals to find the new mean.

Complete step-by-step solution:
If there are $n$ number of data points say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ are given then the mean of the data points is denoted as $m$ and is given by
\[m=\dfrac{{{x}_{1}}+{{x}_{2}}+...={{x}_{n}}}{n}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\]
We can see in the given data in the question that there are total 50 number of students who are distributed over 7 intervals of pocket money .
We find the mean of such class intervals using midpoint and frequency table. If there are $n$ intervals where say ${{I}_{1}},{{I}_{2}},...,{{I}_{n}}$ with midpoints ${{p}_{1}},{{p}_{2}},...,{{p}_{n}}$ is frequented by frequencies ${{f}_{1}},{{f}_{2}},...,{{f}_{n}}$ then the mean of such a distribution is given by
\[m=\dfrac{{{p}_{1}}{{f}_{1}}+{{p}_{2}}{{f}_{2}}+...+{{p}_{n}}{{f}_{n}}}{{{f}_{1}}+{{f}_{2}}+...+{{f}_{n}}}=\dfrac{\sum\limits_{i=1}^{n}{{{p}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}\]
So, first we find out the value of midpoint of each interval. Here the number of interval is 7. The midpoints of the interval 25-30 , 30-35, 35-40, 40-45, 45-50, 50-55 , 55-60 are 27.5, 32.5, 37.5, 42.5, 47.5, 52.5.57.5 which are the values of ${{p}_{1}},{{p}_{2}},{{p}_{3}},{{p}_{4}},{{p}_{6}},{{p}_{7}}$. Here the frequencies are number of students in each interval which are ${{f}_{1}}=2,{{f}_{2}}=3,{{f}_{3}}=5,{{f}_{4}}=7,{{f}_{5}}=21,{{f}_{6}}=10,{{f}_{7}}=2$. So sum of the frequencies is $\sum\limits_{i=1}^{7}{{{f}_{i}}=}{{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}+{{f}_{6}}+{{f}_{7}}=50$. We find sum of product of midpoints and frequency that is $\sum\limits_{i=1}^{n}{{{p}_{i}}{{f}_{i}}}$ in the table below.

Intervals	$f_i$	midpoints$(P_i)$	$f_i$$P_i$
25-30	2	27.5	55
30-35	3	32.5	97.5
35-40	5	37.5	187.5
40-45	7	42.5	297.5
45-50	21	47.5	997.5
50-55	10	52.5	525
55-60	2	57.5	115

We get the value of $\sum\limits_{i=1}^{n}{{{p}_{i}}{{f}_{i}}}=2275.$ . So the mean of the given data is
\[m=\dfrac{\sum\limits_{i=1}^{n}{{{p}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}=\dfrac{2275}{50}=45.5\]
The second part of the question increases the mean by 8 rupees. So the lower limit and upper limit of the intervals will increase by 8. The midpoints ${{p}_{i}}$ also will increase by 8. So the new mean will be
\[m=\dfrac{\sum\limits_{i=1}^{n}{\left( {{p}_{i}}+8 \right){{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}=\dfrac{\sum\limits_{i=1}^{n}{{{p}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}+\dfrac{\sum\limits_{i=1}^{n}{8{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}=\dfrac{2275}{50}+8\dfrac{50}{50}=45.5+8=53.5\]So the mean of the weekly pocket money is 45.5 and the new mean is 53.5 .

Note: The key concept here is to put the data and calculate it in the table properly. We should be careful of confusion between mean, median, and mode. The mode is the interval with the largest frequency. The median is the class that contains above half of the total frequency.