Question

The frequency distribution table for the following will be

$\begin{array}{cccccc}\text{Class Mark}& 15& 25& 35& 45& 55\\ \text{Frequency}& 2& 16& 5& 14& 9\end{array}$

• A. Class mark
• B. $\begin{array}{cc}\text{CI}& \text{Frequency}\\ 15-25& 2\\ 25-35& 16\\ 35-45& 5\\ 45-55& 14\\ 55-65& 9\end{array}$
• C. $\begin{array}{cc}\text{CI}& \text{Frequency}\\ 10-20& 2\\ 20-30& 16\\ 30-40& 5\\ 40-50& 14\\ 50-60& 9\end{array}$
• D. $\begin{array}{cc}\text{CI}& \text{Frequency}\\ 0-10& 2\\ 10-20& 16\\ 20-30& 5\\ 30-40& 14\\ 40-50& 9\end{array}$

Answer 1

The correct option is C

$\begin{array}{cc}\text{CI}& \text{Frequency}\\ 10-20& 2\\ 20-30& 16\\ 30-40& 5\\ 40-50& 14\\ 50-60& 9\end{array}$

As the difference between the values of the consecutive class marks is 10,

therefore subtract $\frac{10}{2}=5$ from each class mark to get the lower limit

and add 5 to each class mark to get the upper limit.

Thus, the given frequency distribution will be of the form

$\begin{array}{cc}\text{CI}& \text{Frequency}\\ 10-20& 2\\ 20-30& 16\\ 30-40& 5\\ 40-50& 14\\ 50-60& 9\end{array}$