Question

Find the mean of the following distribution: 
\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Class-interval} & 11-13 & 13-15 & 15-17 & 17-19 & 19-21 & 21-23 & 23-25 \\ \hline \text{Frequency} & \text{7} & \text{6} & \text{9} & \text{13} & \text{20} & \text{5} & \text{4} \\ \hline \end{array}\] 
 

Hint:

For distributions where the class marks and frequencies are large, the 'Assumed Mean Method' or 'Step-Deviation Method' can simplify calculations. However, for manageable numbers like in this problem, the direct method is efficient and straightforward.

Answer 1

 

Step 1: Understanding the Concept: 
To find the mean of a grouped frequency distribution, we use the direct method. This involves finding the midpoint (or class mark) of each class interval, multiplying it by the corresponding frequency, summing up these products, and finally dividing by the total frequency. 
 

Step 2: Key Formula or Approach: 
The formula for the mean (\(\bar{x}\)) is: \[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} \] where \(f_i\) is the frequency of the i-th class and \(x_i\) is the class mark of the i-th class. The class mark \(x_i\) is calculated as \(\frac{\text{Upper limit} + \text{Lower limit}}{2}\). 
 

Step 3: Calculation Table: 
We can organize the calculations in a table:
\[\begin{array}{|c|c|c|c|} \hline \text{Class-interval} & \text{Frequency }(f_i) & \text{Class Mark }(x_i) & f_i x_i \\ \hline 11\!-\!13 & 7 & 12 & 84 \\ \hline 13\!-\!15 & 6 & 14 & 84 \\ \hline 15\!-\!17 & 9 & 16 & 144 \\ \hline 17\!-\!19 & 13 & 18 & 234 \\ \hline 19\!-\!21 & 20 & 20 & 400 \\ \hline 21\!-\!23 & 5 & 22 & 110 \\ \hline 23\!-\!25 & 4 & 24 & 96 \\ \hline \textbf{Total} & \sum f_i = 64 & & \sum f_i x_i = 1152 \\ \hline \end{array}\]
 

 

Step 4: Calculating the Mean: 
Sum of all frequencies, \(\sum f_i = 64\). 
Sum of the products, \(\sum f_i x_i = 1152\). 
Now, apply the formula for the mean: \[ \bar{x} = \frac{1152}{64} \] \[ \bar{x} = 18 \]

Step 5: Final Answer: 
The mean of the given distribution is 18.