Chapter 7, Problem 8CT

(a)

To determine

To graph: the representation of the data and make the double box.

Answer to Problem 8CT

Least value: $7$ and greatest value: $12.5$ .

Explanation of Solution

Given:

  

Concept used:

Arithmetic mean:

Mean: $\stackrel{-}{x}=\frac{{\displaystyle \sum xi}}{N}$ .

Median $=\left\{\begin{array}{c}\frac{N}{2}\text{th term; }N\text{ is odd}\\ \frac{N}{2}\text{th term; }N\text{ is even}\end{array}\right\}$ .

Mode= the value in the data set that occurs most frequently.

Calculation:

Steps in making a box and whisker plot:

Step $1$ : order the data: find the median and the quartiles.

Step $2$ : draw a number line that includes the least and greatest values.

Graph points above the number line for the $5$ number summary.

Step $3$ : draw a box using $Q_1$ and $Q_3$ . Draw a line through the median.

Draw whiskers from the box to the least and greatest values.

Brand $A$ :

  $8.5,11.5,13.5,13.5,14.5,15.5,16.25,16.75,18.5,20.75$ .

Least value: $8.5$ .

First quartile: median of lower half:

  $\left(8.5,11.5,13.5,13.5,14.5\right)\to 13.5$ .

Median: $\frac{14.5+15.5}{2}=15.$

Third quartile:

Median of upper half:

  $\left(15.5,16.25,16.75,18.5,20.75\right)\to 16.75$ .

Greatest value: $20.75$ .

Brand $B$ :

  $7,8.5,8.5,9,9,9.5,9.75,10.25,10.5,\mathrm{12.5.}$

Least value: $7$ .

First quartile: median of lower half:

  $\left(7,8.5,8.5,9,9\right)\to 8.5$ .

median $=\frac{9+9.5}{2}=9.25$ .

Third quartile: median of upper half:

  $\left(9.5,9.75,10.25,10.5,12.5\right)\to 10.25$ .

Greatest value: $12.5$ .

The double box:

  

Hence, Least value: $7$ and greatest value: $12.5$ .

(b)

To determine

To find: The Identification of the shape of each distribution.

Answer to Problem 8CT

The brand $A$ is skewed left and brand $B$ is skewed right.

Explanation of Solution

Given:

  

Concept used:

For the symmetric distributions, the mean is approximately equal to the median and right tails are equally balanced meaning that they have about the same length.

If the right whisker is longer than the left whisker, the distribution is positively skewed.

Calculation:

For brand $A$ ,the left whisker is longer than the right whisker and most of the data are on the right side of the plot. Therefore, the distribution is skewed left.

For brand $B$ , the right whisker is longer than the left whisker and most of the data are on the left side of the plot. Therefore, the distribution is skewed right.

Since both distributions are skewed, use the median to describe the centre and the five numbers summary to describe the variation.

Hence, the brand $A$ is skewed left and brand $B$ is skewed right.

(c)

To determine

To find: the battery lives of two brands that are most spread out and explain.

Answer to Problem 8CT

Brand $A$ battery lives are more spread out.

Explanation of Solution

Given:

  

Concept used:

For the symmetric distributions, the mean is approximately equal to the median and right tails are equally balanced meaning that they have about the same length.

If the right whisker is longer than the left whisker, the distribution is positively skewed.

Calculation:

Brand $A$ battery lives are more spread out since the plots $A$ is longer than brand $B$ as shown in the graph.

Hence, Brand $A$ battery lives are more spread out.

(d)

To determine

To find: the comparison of the distribution using there shapes and appropriate measurement of centre and variation.

Answer to Problem 8CT

The brand $A$ is skewed left and brand $B$ is skewed right.

Explanation of Solution

Given:

  

Concept used:

For the symmetric distributions, the mean is approximately equal to the median and right tails are equally balanced meaning that they have about the same length.

If the right whisker is longer than the left whisker, the distribution is positively skewed.

Calculation:

For brand $A$ ,the left whisker is longer than the right whisker and most of the data are on the right side of the plot. Therefore, the distribution is skewed left.

For brand $B$ , the right whisker is longer than the left whisker and most of the data are on the left side of the plot. Therefore, the distribution is skewed right.

Since both distributions are skewed, use the median to describe the centre and the five numbers summary to describe the variation.

Hence, the brand $A$ is skewed left and brand $B$ is skewed right.