## Elementary Technical Mathematics

The frequency distribution is as, $\begin{matrix} \text{interval} & \text{midpoint} & \text{frequency} \\ 5.5-10.5 & 8 & 7 \\ 10.5-15.5 & 13 & 11 \\ 15.5-20.5 & 18 & 10 \\ 20.5-25.5 & 23 & 6 \\ 25.5-30.5 & 28 & 1 \\ 30.5-35.5 & 33 & 1 \\ \end{matrix}$
\begin{align} & 15,12,10,9,15,22,7,23,12,8,18,22,11,30,14,18, \\ & 12,20,22,35,10,8,11,19,7,23,17,15,20,16,17,18, \\ & 22,15,20,13 \\ \end{align} 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{R}=\text{28} \\ \end{align} Since $\text{28}$ is close to $\text{30}$, let us choose 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 occurrence of defective pairs in each interval. Then the frequency distribution will be as, $\begin{matrix} \text{interval} & \text{midpoint} & \text{frequency} \\ 5.5-10.5 & 8 & 7 \\ 10.5-15.5 & 13 & 11 \\ 15.5-20.5 & 18 & 10 \\ 20.5-25.5 & 23 & 6 \\ 25.5-30.5 & 28 & 1 \\ 30.5-35.5 & 33 & 1 \\ \end{matrix}$