91影视

The data of propagation lives to reach a given crack size in fastener holes intended for use in military aircraft is provided.

The size of the sample is 16.

a.

Let x represents the propagation lives in flight hours/\({10^4}\)

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{0.736 + 0.863 + 0.865 + ... + 1.394}}{{16}}\\ &=& 1.0297\end{array}\)

Thus, the sample mean is 1.0297 flight hours /\({10^4}\).

The sample median is computed by first ordering the data in ascending order.

The data is arranged as,

For the even number of observations, the median value is computed as,

\(\begin{array}{c}\tilde x &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{n}{2}} \right)^{th}}{\rm{and}}\;{\left( {\frac{n}{2} + 1} \right)^{th}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( {\frac{{16}}{2}} \right)^{th}}{\rm{and}}\;{\left( {\frac{{16}}{2} + 1} \right)^{th}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& {\rm{average}}\;{\rm{of}}\;{\left( 8 \right)^{th}}{\rm{and}}\;{\left( 9 \right)^{th}}\;{\rm{ordered}}\;{\rm{values}}\\ &=& \frac{{1.007 + 1.011}}{2}\\ &=& 1.009\end{array}\)

Thus, the median value is 1.009flight hours/\({10^4}\).

The mean value (1.0297) is greater than the median value (1.009).

b.

The median value is 1.009.

The largest sample observation is 1.394.

The largest sample observation should be greater than the median value so that does not affect the median value.

The value is computed as,

\(\begin{array}{c}1.394 - \tilde x &=& 1.394 - 1.011\\ &=& 0.383\end{array}\)

Thus, the largest sample value can be decreased by 0.383.