Child Restraint Systems Use the numbers of defective child restraint systems given in Exercise 8. Find the mean, median, and standard deviation. What important characteristic of the sample data is missed if we explore the data using those statistics?

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.