Chapter 6, Problem 4E

a)

To determine

To find the range and IQR of the weights.

Answer to Problem 4E

Range = 3.3 pounds and IQR = 0.95 pounds.

Explanation of Solution

Given:

Lowest = 4.15 pounds

Highest = 7.45 pounds

Mean = 6 pounds

First quartile = 5.6 pounds

Median = 6.2 pounds

Third quartile =6.55 pounds

Standard deviation = 0.65 pounds

Formula:

  $\begin{array}{l}Range=Highest-Lowest\\ IQR=Q_3-Q_1\end{array}$

Using formula,

  $\begin{array}{l}Range=7.45-4.15=3.3pounds\\ IQR=6.55-5.6=0.95pounds\end{array}$

b)

To determine

To explain the shape of the distribution.

Answer to Problem 4E

The distribution is right skewed.

Explanation of Solution

Given:

Lowest = 4.15 pounds

Highest = 7.45 pounds

Mean = 6 pounds

First quartile = 5.6 pounds

Median = 6.2 pounds

Third quartile =6.55 pounds

Standard deviation = 0.65 pounds

If the median is less than the mean then the distribution is left skewed. If the median is greater than mean then the distribution is right skewed. If the both equals then it is symmetric.

Here, Median > mean, the distribution is right skewed.

c)

To determine

To convert summary statistics in ounces.

Explanation of Solution

Given:

Lowest = 4.15 pounds

Highest = 7.45 pounds

Mean = 6 pounds

First quartile = 5.6 pounds

Median = 6.2 pounds

Third quartile =6.55 pounds

Standard deviation = 0.65 pounds

We need to multiply each of the value by 16. Therefore, summary statistics becomes,

Mean = 96 ounces

First quartile = 89.6 ounces

Median = 99.2 ounces

Third quartile = 104.8 ounces

Standard deviation = 10.4 ounces

Range = 52.8 ounces

IQR = 15.2 ounces

d)

To determine

To explain the change in summary statistics when added 30 ounces.

Explanation of Solution

Given:

Mean = 96 ounces

First quartile = 89.6 ounces

Median = 99.2 ounces

Third quartile = 104.8 ounces

Standard deviation = 10.4 ounces

Range = 52.8 ounces

IQR = 15.2 ounces

Only mean, quartiles and median will be increases by 30. Rest of the things will remain unchanged. Therefore, summary statistic is,

Mean = 126 ounces

First quartile = 119.6 ounces

Median = 129.2 ounces

Third quartile = 134.8 ounces

Standard deviation = 10.4 ounces

Range = 52.8 ounces

IQR = 15.2 ounces

e)

To determine

To explain which values will not change if something added to the distribution.

Explanation of Solution

Given:

Mean = 96 ounces

First quartile = 89.6 ounces

Median = 99.2 ounces

Third quartile = 104.8 ounces

Standard deviation = 10.4 ounces

Range = 52.8 ounces

IQR = 15.2 ounces

Something adding in the original distribution means creating outlier for original distribution. Therefore, standard deviations and IQR might not change as these things are unaffected by an adding the values.