Chapter 8, Problem 1EA

Refer to the example and find the following.

(a) C(M₀) (b) C(M₂) (c) C(M₃)

(d) C(M₁₂) (e) C(M₁₃) (f) C(M₁₂₃)

To determine

The expected value.

Answer to Problem 1EA

Solution: $\underset{¯}{\$69.01}$

Explanation of Solution

Given: The tables provided in the textbook.

Explanation:

To find expected value ($C$) of $M_0$:

The required formula is

$C\left(M_0\right)=NL\left[1-\left(1-p_1\right)\left(1-p_2\right)\left(1-p_3\right)\right]$

Put the values from the provided table.

The expected value is

$\begin{array}{c}C\left(M_0\right)=3\left(54\right)\left[1-\left(1-0.9\right)\left(1-0.24\right)\left(1-0.13\right)\right]\\ =\$69.01\end{array}$

Conclusion: The expected value is $\underset{¯}{\$69.01}$.

To determine

The expected value.

Answer to Problem 1EA

Solution: $\underset{¯}{\$79.64}$

Explanation of Solution

Given: The tables provided in the textbook.

Explanation:

To find expected value ($C$) of $M_2$:

The required formula is

$C\left(M_2\right)=H_2+NL\left[1-\left(1-p_1\right)\left(1-p_3\right)\right]$

Put the values from the provided table.

The expected value is

$\begin{array}{c}C\left(M_2\right)=40+3\left(54\right)\left[1-\left(1-0.9\right)\left(1-0.13\right)\right]\\ =\$79.64\end{array}$

Conclusion: The expected value is $\underset{¯}{\$79.64}$.

To determine

The expected value.

Answer to Problem 1EA

Solution: $\$58.96$

Explanation of Solution

Given: The tables provided in the textbook.

Explanation:

To find expected value ($C$) of $M_3$:

The required formula is

$C\left(M_3\right)=H_3+NL\left[1-\left(1-p_1\right)\left(1-p_2\right)\right]$

Put the values from the provided table.

The expected value is

$\begin{array}{c}C\left(M_3\right)=9+3\left(54\right)\left[\left(1-0.9\right)\left(1-0.24\right)\right]\\ =\$58.96\end{array}$

Conclusion: The expected value is $\underset{¯}{\$58.96}$.

To determine

The expected value.

Answer to Problem 1EA

Solution: $\underset{¯}{\$82.54}$

Explanation of Solution

Given: The tables provided in the textbook.

Explanation:

To find expected value ($C$) of $M_{12}$:

The required formula is

$C\left(M_{12}\right)=H_1+H_2+NL\left[1-\left(1-p_3\right)\right]$

Put the values from the provided table.

The expected value is

$\begin{array}{c}C\left(M_{12}\right)=15+40+3\left(54\right)\left[1-\left(1-0.13\right)\right]\\ =\$82.54\end{array}$

Conclusion: The expected value is $\underset{¯}{\$82.54}$.

To determine

The expected value.

Answer to Problem 1EA

Solution: $\underset{¯}{\$62.88}$

Explanation of Solution

Given: The tables provided in the textbook.

Explanation:

To find expected value ($C$) of $M_{13}$:

The required formula is

$C\left(M_{13}\right)=H_1+H_3+NL\left[1-\left(1-p_2\right)\right]$

Put the values from the provided table.

The expected value is

$\begin{array}{c}C\left(M_{13}\right)=15+9+3\left(54\right)\left[1-\left(1-0.24\right)\right]\\ =\$62.88\end{array}$

Conclusion: The expected value is $\underset{¯}{\$62.88}$.

To determine

The expected value.

Answer to Problem 1EA

Solution: $\underset{¯}{\$64}$

Explanation of Solution

Given: The tables provided in the textbook.

Explanation:

To find expected value ($C$) of $M_{123}$:

The required formula is

$C\left(M_{123}\right)=H_1+H_2+H_3$

Put the values from the provided table.

The expected value is

$\begin{array}{c}C\left(M_{123}\right)=15+40+9\\ =\$64.00\end{array}$

Conclusion: The expected value is $\underset{¯}{\$64}$.