## Elementary Technical Mathematics

The mean of the grouped data is $16$.
Now the range of the numbers is $\text{Highest number}-\text{lowest number}$ $\text{Highest number }=\text{35}$ and $\text{lowest number}=\text{7}$ Therefore the range is, \begin{align} & \text{R}=\text{35}-\text{7} \\ & =\text{28} \end{align} Since $\text{28}$ is close to $\text{30}$, let us chose the odd number $\text{5}$ as the interval length. This means we will need $\frac{\text{30}}{5}=\text{6}$ group intervals. The first interval is $\text{5}\text{.5 }-\text{ 10}\text{.5}$ with midpoint as $\text{8}$. Here $\text{5}\text{.5}$ is the lower limit and $\text{10}\text{.5}$ is the upper limit of the interval, where the frequency is the number of occurrences of defective pairs in each interval. Then the frequency distribution will be as, $\begin{matrix} Interval & Midpoint(x) & Frequency(f) & xf & \\ 5.5-10.5 & 8& 7& 56& \\ 10.5-15.5 & 13& 11& 143& \\ 15.5-20.5 & 18& 10& 180& \\ 20.5-25.5 & 23& 6& 138& \\ 25.5-30.5 & 28& 1& 28& \\ 30.5-35.5 & 33& 1& 33& \\ \end{matrix}$ \begin{align} & \text{Sum of } xf=56+143+180+138+28+33 \\ & =578 \end{align} And, \begin{align} & \text{Sum of }f=7+11+10+6+1+1 \\ & =36 \end{align} Mean of the grouped data is$\frac{\text{Sum of }xf\text{ }}{\text{Sum of }f}$. Therefore, \begin{align} & \frac{\text{Sum of }xf\text{ }}{\text{Sum of }f}=\frac{578}{36} \\ & =16.05 \end{align} Mean is $16.05$. Hence the mean of the grouped data is $16$.