Chapter 10.4, Problem 11E

Solution

To determine: The expected values is positive for the consumer who purchase laptop computer.

The expected values is negative for the consumer.

Given information:

4% of the customers require a $200 for repair, and the extended warranty costs is $79.

1% of the customers require a $300 for repair, and the extended warranty costs is $79

Formula used:

For calculating the expected value or mean use the following formula,

  $\mu ={\displaystyle \sum xP\left(x\right)}$   ...... (1)

Calculation:

4% of the customers require a $200 repair, while the extended warranty costs $79 and thus a profit of $200-$79=$121, Then find:

  $\begin{array}{c}P\left(\$121\right)=4\%\\ =0.04\end{array}$

1% of the customers require a $300 repair, while the extended warranty costs $79 and thus a profit of $300-$79=$221, Then find:

  $\begin{array}{c}P\left(\$221\right)=1\%\\ =0.01\end{array}$

The remaining 100%-4%-1% = 95% of the customers do not repair, while the extended warranty cost $79 and thus a loss of $79, Then find:

  $\begin{array}{c}P\left(-\$79\right)=95\%\\ =0.95\end{array}$

The expected value (or mean value) is the sum of the product of each possibility x with its probability P(x).

  $\begin{array}{c}\mu ={\displaystyle \sum xP\left(x\right)}\\ =\$121\times 0.04+\$221\times 0.01+\left(-\$79\right)\times 0.95\\ =-\$48.11\end{array}$

Thus, we are expected to make a loss of $48.11 when purchasing the extended coverage.