Chapter 6.2, Problem 59E

a.

To determine

To find: The mean and standard deviation of the time required for the entire operation of positioning.

Answer to Problem 59E

The mean and standard deviation are 31 and 4.4721.

Explanation of Solution

Given:

For $X_1$ :

Population mean $\left(\mu \right)=11$

Population standard deviation $\left(\sigma \right)=2$

For $X_2$ :

Population mean $\left(\mu \right)=20$

Population standard deviation $\left(\sigma \right)=4$

Calculation:

The mean value and standard deviation can be computed as:

  $\begin{array}{c}{\mu }_{X_1+X_2}=11+20\\ =31\end{array}$

  $\begin{array}{c}{\sigma }_{X_1+X_2}=\sqrt{{\sigma }^2{}_{X_1}+{\sigma }^2{}_{X_2}}\\ =\sqrt{2^2+4^2}\\ =\sqrt{20}\\ =4.4721\end{array}$

Thus, mean and standard deviation are 31 and 4.4721.

b.

To determine

To find: The probability that goal is met.

Answer to Problem 59E

The probability is 0.4129.

Explanation of Solution

The probability that goal is met can be calculated as:

  $\begin{array}{c}P\left(X_1+X_2<30\right)=P\left(Z<\frac{30-31}{4.4721}\right)\\ =P\left(Z<-0.22\right)\\ =0.4129\end{array}$

Thus, the required probability is 0.4129.