91影视

The data on time (sec) to complete the escape is provided.

The size of the sample is 26.

A stem-and-leaf display provides a visual representation of the dataset.

The steps to construct a stem-and-leaf display are as follows,

1. Select the leading digit for the stem and trailing digits for the leaves.

2 Represent the stem digits vertically and similarly the trailing digits corresponding to the stem digits.

3) Mention the units for the display.

The stem and leaf display for the provided scenario is,

Unit: 35|6=356

It can be observed that the distribution is slightly positively skewed. This implies that the mean value is greater than the median.

Let X represents the time (sec).

The sample mean is computed as,

\(\begin{array}{c}\bar x &=& \frac{{\sum {{x_i}} }}{n}\\ &=& \frac{{389 + 356 + 359 + ... + 397}}{{26}}\\ &=& \frac{{9638}}{{26}}\\ &=& 370.692\end{array}\)

Thus, the sample mean is 370.692 sec.

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

The data arranged in ascending order is as follows,

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

\(\begin{array}{c}\tilde x &=& average\;of\;{\left( {\frac{n}{2}} \right)^{th}}and\;{\left( {\frac{n}{2} + 1} \right)^{th}}\;ordered\;values\\ &=& average\;of\;{\left( {\frac{{26}}{2}} \right)^{th}}and\;{\left( {\frac{{26}}{2} + 1} \right)^{th}}\;ordered\;values\\ &=& average\;of\;{\left( {13} \right)^{th}}and\;{\left( {14} \right)^{th}}\;ordered\;values\\ &=& \frac{{369 + 370}}{2}\\ &=& 369.5\end{array}\)

Thus, the median value is 369.5 sec.

Referring to part b, the value of the sample median is 369.5.

The largest value in the dataset is 424.

The largest value can be increased by any amount as the median value is the average of 13^th and 14^th value in the provided scenario. So, the median will not be affected by any amount of increase in the largest value of the dataset.

The largest value cannot be less than the middle values so that the median value is remains unaffected.

Therefore, the largest value can be decreased by 55 (424-370) without affecting the median.

Thus, the largest value can be decreased by 55.

d.

Referring to part b,

The sample mean is 370.692 sec.

The median value is 369.5.

When the observations are re-expressed in minutes, the sample mean is computed as,

\(\begin{array}{c}\frac{{\bar x}}{{60}} &=& \frac{{370.692}}{{60}}\\ &=& 6.18\;\min \end{array}\)

The median is computed as,

\(\begin{array}{c}\frac{{\tilde x}}{{60}} &=& \frac{{369.5}}{{60}}\\ &=& 6.16\;\min \end{array}\)

Therefore, the mean and median are 6.18 min and 6.16 min respectively.