91影视

A stem-and-leaf display of the escape time data is provided.

From the provided stem-and-leaf display, the data is given as,

The number of observations is 26.

\(\begin{aligned}{l}{n_{smallest}} = 13\\{n_{{\rm{largest}}}} = 13\end{aligned}\)

The lower fourth is computed as,

\(\begin{aligned}\frac{{{n_{smallest}} + 1}}{2} &= \frac{{13 + 1}}{2}\\ &= {7^{th}}value\end{aligned}\)

The 7^th value of the smallest half is 359.

This implies that the lower fourth is 359.

The upper fourth is computed as,

\(\begin{aligned}\frac{{{n_{{\rm{largest}}}} + 1}}{2} &= \frac{{13 + 1}}{2}\\ &= {7^{th}}value\end{aligned}\)

The 11^th value of the largest half is 392.

This implies that the upper fourth is 392.

The fourth spread is computed as,

\(\begin{aligned}{f_s} &= {\rm{Upper}}\,{\rm{fourth}} - {\rm{Lower}}\;{\rm{fourth}}\\ &= 392 - 359\\ &= 33\end{aligned}\)

Therefore, the fourth spread is 33.

Referring to part a,

The lower fourth is 359.

The upper fourth is 392.

The fourth spread is 33.

Any observation farther than 1.5\({f_s}\) from the closest fourth is referred as an outlier.

An outlier is extreme if it is more than 3\({f_s}\) from the nearest fourth, otherwise, it is mild.

The calculations are as follows,

\begin{aligned} Lower\ fourth-1.5{{f}_{s}}&=359-\left( 1.5*33 \right) \\ & =359-49.5 \\ & =309.5 \end{aligned}

\begin{aligned} Upper\ fourth+1.5{{f}_{s}}&=392+\left( 1.5*33 \right) \\ & =392+49.5 \\ & =441.5 \end{aligned}

It can be observed that there are no values that are less than 309.5 and no values that exceed 441.5. This implies that there are no outliers present in the data set.

Thus, there are no extreme outliers as there are no outliers present.

Referring to part a,

The lower fourth is 359.

The upper fourth is 392.

The fourth spread is 33.

The smallest observation is 325.

The largest observation is 424.

From the provided stem-and-leaf display, the data is given as,

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

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

Thus, the median value is 369.5.

The five-number summary is,

The lower fourth is 359.

The upper fourth is 392.

The fourth spread is 33.

The smallest observation is 325.

The largest observation is 424.

The median value is 369.5.

Steps to construct a boxplot are as follows,

1. Compute the values of the five-number summary (smallest value, lower fourth, median, upper fourth, largest value).

2. Construct a line segment from the smallest value to the largest value of the dataset.

3. Construct a rectangular box from the lower fourth to the upper fourth and draw a line in the box at the median value.

The boxplot is represented as,

It can be observed from the above-represented boxplot that it is slightly positively skewed in the middle half but overall symmetric. The extent of variability appears substantial.

d.

Referring to part a,

The fourth spread is 33.

The Upper fourth is 392.

The amount by which the largest observation, 424, can be decreased without affecting the fourth spread cannot be less than the Upper fourth; that is 392.

Therefore, the amount is 32 (424-392).

Thus, the largest observation, 424, can be decreased at most 32 without affecting the fourth spread.