Question

If the mean of the given frequency distribution is 160, then the value of f is equal to

$\begin{array}{cc}\text{Classes}& \text{Frequency}\\ 100-120& 10\\ 120-140& 15\\ 140-160& 20\\ 160-180& f\\ 180-200& 15\\ 200-220& 12\end{array}$

• A. 25
• B. 15
• C. 10
• D. 5

Answer 1

The correct option is C 10

$\begin{array}{cccc}\text{Classes}& \text{Frequency (}{f}_i\text{)}& \text{Class Marks}{x}_i& f_i{x}_i\\ 100-120& 10& 110& 1100\\ 120-140& 15& 130& 1950\\ 140-160& 20& 150& 3000\\ 160-180& f& 170& 170f\\ 180-200& 15& 190& 2850\\ 200-220& 12& 210& 2520\\ \text{Total}& {\sum }_{i=1}^6{f}_i=72+f& & {\sum }_{i=1}^6{f}_i{x}_i=11420+170f\end{array}$
Thus $\overline{x}=\frac{{\sum }_{i=1}^6{f}_1{X}_1}{{\sum }_{i=1}^6{f}_1}=\frac{11420+270f}{72+f}=160$

$\Rightarrow 11420+170f=160(72+f)$
$\Rightarrow 11420+170f=11520+160f$
$\Rightarrow 10f=100$
$\Rightarrow f=10$
Hence the correct answer is option c.