Question

Find out the mean profits from the following frequency distribution.

Profit per shop(in Rs)	Number of shops $ \left( {{f}_{i}} \right) $
100 – 200	10
200 – 300	18
300 – 400	20
400 – 500	26
500 – 600	30
600 – 700	28
700 - 800	18

Answer 1

Hint: Here, first make the table with the values given and create columns of midpoint $ \left( {{x}_{i}} \right) $ , deviation $ \left( {{d}_{i}}={{x}_{i}}-a \right) $ , and $ {{f}_{i}}{{d}_{i}} $ adjacent to their values. Then find the summation of all the values in the midpoint column and find the summation of all the values in the column of $ {{f}_{i}}{{d}_{i}} $ . Use the formula of assumed mean method, $ \overline{x}=a+\dfrac{\sum\limits_{i=1}^{k}{{{f}_{i}}{{d}_{i}}}}{\sum\limits_{i=1}^{k}{{{f}_{i}}}} $ and substitute the values and find the mean profit.

Complete step-by-step answer:
We will be using the assumed mean method, here we are going to first draw the given table and adjacent to it make columns for the midpoint $ \left( {{x}_{i}} \right) $ , deviation $ \left( {{d}_{i}} \right) $ , and $ {{f}_{i}}{{d}_{i}} $ . Here, we have columns for the profit per shop, so let us find the midpoint using the first column. The midpoint is the mean of the lower and upper limit.

For the limit, 100 – 200,
 $ \begin{align}
  & {{x}_{i}}=\dfrac{100+200}{2} \\
 & =\dfrac{300}{2} \\
 & =150
\end{align} $

Similarly, make a column of midpoint for all the values.

$ {{x}_{i}} $
150
250
350
450
550
650
750

Now, let us make a column for deviation $ \left( {{d}_{i}} \right) $ , where $ {{d}_{i}}={{x}_{i}}-450 $ . Here, we took 450, because it was the median number that is the 4th number from top to down and down to top and it is also known as assumed mean (a = 450). Also, for every adjacent row, the $ {{x}_{i}} $ changes accordingly.
For the limit, 100 – 200,
 $ \begin{align}
  & {{d}_{i}}={{x}_{i}}-450 \\
 & =150-450 \\
 & =-300
\end{align} $
Similarly, make a column for the deviation for all the remaining limits.

$ {{d}_{i}}={{x}_{i}}-450 $
- 300
- 200
- 100
0
100
200
300

Now, to find the next set of values, for the column $ {{f}_{i}}{{d}_{i}} $ , multiply the values in $ {{f}_{i}} $ adjacent to $ {{d}_{i}} $ .

$ {{f}_{i}}{{d}_{i}} $
- 3000
- 3600
- 2000
0
3000
5600
5400

Now, find the sum of all the number of shops, that is $ \sum\limits_{i=1}^{k}{{{f}_{i}}=}\,150 $ and find the sum of all the values in $ {{f}_{i}}{{d}_{i}} $ , $ \sum\limits_{i=1}^{k}{{{f}_{i}}{{d}_{i}}=5400} $ .
Therefore, the final table looks like this,

Profit per shop(in Rs)	Number of shops $ \left( {{f}_{i}} \right) $	$ {{x}_{i}} $	$ {{d}_{i}}={{x}_{i}}-450 $	$ {{f}_{i}}{{d}_{i}} $
100 – 200	10	150	- 300	- 3000
200 – 300	18	250	- 200	- 3600
300 – 400	20	350	- 100	- 2000
400 – 500	26	450	0	0
500 – 600	30	550	100	3000
600 – 700	28	650	200	5600
700 - 800	18	750	300	5400
	$ \sum\limits_{i=1}^{k}{{{f}_{i}}=}\,150 $			$ \sum\limits_{i=1}^{k}{{{f}_{i}}{{d}_{i}}=5400} $

Now, we can use the values we got from the table to find the mean profits.
We know,
By assumed mean method,
 $ \overline{x}=a+\dfrac{\sum\limits_{i=1}^{k}{{{f}_{i}}{{d}_{i}}}}{\sum\limits_{i=1}^{k}{{{f}_{i}}}} $
where, a = 450, $ \sum\limits_{i=1}^{k}{{{f}_{i}}=}\,150 $ and $ \sum\limits_{i=1}^{k}{{{f}_{i}}{{d}_{i}}=5400} $ . Substitute these values in the formula and find the mean profits.
 $ \begin{align}
  & \overline{x}=450+\dfrac{5400}{150} \\
 & =450+\dfrac{540}{15} \\
 & =450+36 \\
 & =486
\end{align} $
Hence, the mean profit is Rs. 486.

Note: Here, we can either solve by making the table and solve it right there or you could use the method we used. There can be many mistakes while solving such questions, hence check your steps thoroughly. When you select the assumed mean (a), for an odd number of cases you can easily use the median of all the cases but for an even number of cases, choose either one of the middle ones which means we have two assumed means in even number of cases, you can choose either one of them.