Chapter 11.2, Problem 21E

(a)

To determine

To identify the shape of each distribution.

Answer to Problem 21E

Both the brands are skewed right.

Explanation of Solution

Given information: Box-and-whisker plot.

Calculation:

Both the brands are skewed right.

Fewer data plots are found to the right of the graph. The "tail" of the graph is pulled toward higher positive numbers, or to the right. The mean typically gets pulled toward the tail, and is greater than the median.

(b)

To determine

To find the range of upper $75\%$ data.

Answer to Problem 21E

Range of upper $75\%$ of brand A= $3.5$

Range of upper $75\%$ of brand B= $2.5$

Explanation of Solution

Given information: Box-and-whisker plot.

Formula Used: Range of the data= $max.-Q_1$

Calculation:

  

As the range of the data: $max.-Q_1$

• Brand A

Range of upper $75\%$ of brand A= $max.-Q_1$

So,

Range of upper $75\%$ of brand A= $7-3.5$

Hence,

Range of upper $75\%$ of brand A= $3.5$

• Brand B

Range of upper $75\%$ of brand B= $max.-Q_1$

So,

Range of upper $75\%$ of brand B= $5.75-3.25$

Hence,

Range of upper $75\%$ of brand B= $2.50$

(c)

To determine

To find the interquartile range of data.

Answer to Problem 21E

Interquartile Range of brand A= $1.25$

Interquartile Range of brand B= $1$

Explanation of Solution

Given information: Box-and-whisker plot.

Formula Used: Interquartile Range of the data= $Q_3-Q_1$

Calculation:

  

As the range of the data: $max.-Q_1$

• Brand A

Interquartile Range of brand A= $Q_3-Q_1$

So,

Interquartile Range of brand A= $4.75-3.5$

Hence,

Interquartile Range of brand A= $1.25$

• Brand B

Interquartile Range of brand B= $Q_3-Q_1$

So,

Interquartile Range of brand B= $4.25-3.25$

Hence,

Interquartile Range of brand B= $1$

(d)

To determine

To find the greater standard deviation.

Answer to Problem 21E

Brand A has greater standard deviation

Explanation of Solution

Given information: Box-and-whisker plot.

Formula Used: Standard deviation = $\frac{range}{4}$

Calculation:

The standard deviation is approximately equal to the  $\frac{range}{4}$

Also,

The standard deviation is approximately equal to $\frac{3}{4}\times IQR$

Both estimates work best for normal distribution, i.e. distributions that are not skewed, and the first approximation works best if they are no outliers. They will later determine additional relations between the standard deviation for normally distributed data. That reminds another useful application for the IQR is to define outliers:

Outliers are data points that fall below $Q_1-1.5\times IQR$  or above $Q_3+1.5\times IQR$

• Range A:

Range of brand A= $7-2$

Now,

Range of brand A= $5$

So,

Standard deviation = $\frac{range}{4}$

Hence,

Standard deviation of brand A= $\frac{5}{4}=1.25$

• Range B:

Range of brand B = $5.75-2.25$

Now,

Range of brand B = $3.50$

So,

Standard deviation = $\frac{range}{4}$

Hence,

Standard deviation of brand B = $\frac{3.50}{4}=0.87$

(e)

To determine

To identify the brand with more battery life.

Answer to Problem 21E

Brand A

Explanation of Solution

Given information: Box-and-whisker plot.

Calculation:

Range of brand A= $5$

and

Range of brand B = $3.50$

Hence,

Brand A will be choose because it has greater range than brand B.