Question

The mean of the following frequency distribution is 41. The value of a is equal to

$\begin{array}{cc}\text{Class interval}& \text{Frequency}\\ 0-20& 12\\ 20-40& 8\\ 40-60& a\\ 60-80& 6\\ 80-100& 4\end{array}$

• A. 18
• B. 10
• C. 15
• D. 9

Answer 1

The correct option is B 10

Given that the mean of ($\overline{x}$) of the given distribution is 41.
$\begin{array}{cccc}\text{Class interval}& f_i& x_i& f_i{x}_i\\ 0-20& 12& 10& 120\\ 20-40& 8& 30& 240\\ 40-60& a& 50& 50a\\ 60-80& 6& 70& 420\\ 80-100& 4& 90& 360\\ \text{Total}& {\sum }_{i=1}^5=30+a& & {\sum }_{i=1}^5{f}_i{x}_i=50a+1140\end{array}$

Mean $\overline{x}=\frac{{\sum }_{i=1}^5{f}_i{X}_1}{{\sum }_{i=1}^5{f}_i}$

$\Rightarrow 41=\frac{50a+1140}{a+30}$

$\Rightarrow 41a+1230=50a+1140$
$\Rightarrow 9a=90$
$\Rightarrow a=10$
Hence the correct answer is option b.