Chapter 6.2, Problem 39E

a.

To determine

To find: The mean of G

Answer to Problem 39E

  ${\mu }_G=76$

Explanation of Solution

Given:

The output is:

  

Calculation:

Consider,

  $G=10X$

The mean is computed as:

  $\begin{array}{c}{\mu }_G={\mu }_{10X}\\ =10\left(7.6\right)\\ =76\end{array}$

Thus, the mean is 76.

b.

To determine

To find: The ${\sigma }_G$

Answer to Problem 39E

  ${\sigma }_G=2.38$

Explanation of Solution

Calculation:

The value of the ${\sigma }_G$ can be calculated as:

  $\begin{array}{c}{\sigma }_G={\sigma }_X\\ =10\times {\sigma }_X\\ =10\left(1.32\right)\\ =13.2\end{array}$

The average number of tickets are 2.38.

c.

To determine

To find: The variance of G and X .

Explanation of Solution

Calculation:

Here,

  $G=10X$

  $\begin{array}{l}{\sigma }_G={\sigma }_{10X}=10{\sigma }_X\\ \text{Variance}\\ {\sigma }^2{}_G={\sigma }^2{}_{10X}=10{\sigma }^2{}_X\end{array}$

Here, it could be seen that variance of G is 100 multiples of variance of X.