Question

The following table shows the frequency distribution of the diameters of 40 bottles. Find the mean of the data.
$\begin{array}{cccccc}\text{Diameter (mm)}& 35-39& 40-44& 45-49& 50-54& 55-60\\ \text{Frequency}& 6& 12& 15& 10& 7\end{array}$

• A. 45
• B. 37
• C. 47
• D. 55

Answer 1

The correct option is C 47

$\begin{array}{cccccc}\text{Diameter}\left(mm\right)& 35-39& 40-44& 45-49& 50-54& 55-60\\ \text{Frequency(f)}& 6& 12& 15& 10& 7\\ \text{Mid-Point(x)}& 37& 42& 47& 52& 57\\ fx& 222& 504& 705& 520& 399\end{array}$
Here, the mid-point of each interval =$\frac{\text{Lower class limit + Upper class limit}}{2}$
$=\frac{35+39}{2}=37$
Similarly, we can find mid-points of all the class interval.
$\text{Now Mean}=\frac{\sum f_i{x}_i}{\sum f_i}$
$\text{Mean}=\frac{222+504+705+520+399}{50}=\frac{2350}{50}$
$\therefore \text{Mean}=47$