91Ó°ÊÓ

Data are given on the number of defective frames used for child booster seats in cars for tensamples.

The size of each sample is 120.

The data is tabulated below:

The total number of samples is 10.

The total number of defects in all the samples is calculated below:

\[3 + 2 + 4 + 6 + 5 + 9 + 7 + 10 + 12 + 15 = 73\]

The mean number of defects is computed below:

\(\begin{aligned}{c}Mean = \frac{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{defects}}}}{{{\rm{Total}}\;{\rm{number}}\;{\rm{of}}\;{\rm{observations}}}}\\ = \frac{{73}}{{10}}\\ = 7.3\end{aligned}\)

Thus, the mean value isequal to7.3defective frames.

The data arranged in ascending order is tabulated below:

The number of samples is even.

Thus, the following formula is used to compute the median:

\(\begin{aligned}{c}Median = \frac{{{{\left( {\frac{n}{2}} \right)}^{th}}obs + {{\left( {\frac{n}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{{\left( {\frac{{10}}{2}} \right)}^{th}}obs + {{\left( {\frac{{10}}{2} + 1} \right)}^{th}}obs}}{2}\\ = \frac{{{5^{th}}obs + {6^{th}}obs}}{2}\\ = \frac{{6 + 7}}{2}\\ = 6.5\end{aligned}\)

Thus, the median value is equal to 6.5 defective frames.

The standard deviation (S.D.) is computed as shown below:

\(\begin{aligned}{c}S.D. = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {3 - 7.3} \right)}^2} + {{\left( {2 - 7.3} \right)}^2} + ..... + {{\left( {15 - 7.3} \right)}^2}}}{{10 - 1}}} \\ = 4.2\end{aligned}\)

Thus, the standard deviation is equal to 4.2 defective frames.

The computed statistics do not reveal anything about the changing pattern of the data over time.