Answer
The mean of the grouped data is $16$.
Work Step by Step
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 $16$.
