Question

The mean of the following frequency distribution is $ 16 $ , find the missing frequency:

Class	0-4	4-8	8-12	12-16	16-20	20-24	24-28	28-32	32-36
Frequency	6	8	17	23	16	15	-	4	3

Answer 1

Hint: Here, we will take mid values of each class as $ \left( {{x_i}} \right) $ and frequency $ \left( {{f_i}} \right) $ . The mean value is equivalent to the fraction between the addition of a product of mid value with frequency and the total frequency.

Complete step-by-step answer:
Given:
The mean for the given frequency distribution is $ 16 $ .
The frequency distribution table for the given data is as follows:

Class	Frequency $ \left( {{f_i}} \right) $	Mid-value $ \left( {{x_i}} \right) $	$ {f_i}{x_i} $
0-4	6	2	12
4-8	8	6	48
8-12	17	10	170
12-16	23	14	322
16-20	16	18	288
20-24	15	22	330
	$ x $	26	$ 26x $
28-32	4	30	120
32-36	3	34	102
Total	$ \sum {{f_i}} = 92 + x $	-	$ \sum {{f_i}{x_i}} = 1392 + 26x $

Let us assume that x is the missing frequency between interval 24 to 28.
We know that the general formula to find the mean value is,
 $ {\rm{mean = }}\dfrac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }} $
Now, we will substitute the value for the sum of product of frequency and midpoint and the value for the sum of total frequency.
Therefore, the mean for the frequency distribution is calculated as:
 $ \begin{array}{c}
16 = \dfrac{{1392 + 26x}}{{92 + x}}\\
1472 + 16x = 1392 + 26x\\
10x = 80\\
x = 8
\end{array} $

The missing frequency in the range from \[{\rm{24}}\] to $ 28 $ is equal to $ 8 $ .

Note: In such types of problems, the class will not be taken only mid-point should be taken because the interval cannot be multiplied to the frequency. If we don’t remember the formula, we can multiply each midpoint with frequency and add all of them then divide it with the sum of frequency.