Chapter 6.2, Problem 42E

(a)

To determine

To Explain: the shape of the distribution T

Explanation of Solution

Given:

  $\begin{array}{c}{\mu }_X=23.8\\ {\sigma }_X=12.63\end{array}$

X = score on a randomly selected roll of the ball

T= number of tickets gets on a arbitrary selected roll

For every 10 points score one ticket is received it means that

  $\text{Total number of tickets }\left(\text{T}\right)\text{= }\frac{\text{number of points}}{10}$

So if every data is divided by the constant, then the shape of the distribution remains unchanged. The distribution is skewed to the right, because the lowest bar is to the right in the histogram and head of higher bars to the right thus T has the same shape as the distribution, skewed to the right.

(b)

To determine

To Calculate: the mean of T

Answer to Problem 42E

  ${\mu }_M=2.38$

Explanation of Solution

Given:

  $\begin{array}{c}{\mu }_X=23.8\\ {\sigma }_X=12.63\end{array}$

X = score on a randomly selected roll of the ball

T= number of tickets gets on a arbitrary selected roll

Calculation:

  $\begin{array}{c}{\mu }_M=\frac{{\mu }_M}{10}\\ =\frac{23.8}{10}\\ =2.38\end{array}$

For every 10 points score one ticket is received it means that

  $\text{Total number of tickets }\left(\text{T}\right)\text{= }\frac{\text{number of points}}{10}$

So if every data is divided by the constant, then the center of the distribution will be affected in the same way, so the mean will be divided by that same constant 10. The number of tickets receives on a arbitrary selected roll is average 2.38

(c)

To determine

To Calculate: the standard deviation of T

Answer to Problem 42E

  ${\sigma }_M=1.263$

Explanation of Solution

  $\begin{array}{c}{\mu }_X=23.8\\ {\sigma }_X=12.63\end{array}$

X = score on a randomly selected roll of the ball

T= number of tickets gets on a arbitrary selected roll

Calculation:

  $\begin{array}{c}{\sigma }_M=\frac{{\sigma }_X}{10}\\ =\frac{12.63}{10}\\ =1.263\end{array}$

For every 10 points score one ticket is received it means that

  $\text{Total number of tickets }\left(\text{T}\right)\text{= }\frac{\text{number of points}}{10}$

So if every data is divided by the constant, and then the spread of the distribution T will be affected in the same way, so the standard deviation $\sigma$ is also dived by that same constant 10 so the number of tickets received on an arbitrary selected roll varies on average by 1.236 from the mean number of tickets