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# For an extra $49 you can buy extended warranty protection on your new HDTV for three years. If the set requires maintenance during that lime, it will be free You estimate that there is a 5% chance that you might need a$300 repair within three years and a 1% chance that you will need a \$500 repair, but the chances arc 94% that you will need no repair at all. Based on expected values, should you purchase the extended coverage?

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5% of the customers require a $\300$ repair, while the extended warranty costs $\49$ and thus a profit of $\300-\49=\251$ is then made.

$P(\251)=5\%=0.05$

1% of the customers require a $\500$ repair, while the extended warranty costs $\49$ and thus a profit of $\500-\49=\451$ is then made.

$P(\451)=1\%=0.01$

94% of the customers do not require a repair, while the extended warranty costs $\49$ and thus a loss of $\49$ is then made.

$P(-\49)=94\%=0.94$

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

\begin{align*} \mu &=\sum xP(x) \\ &=\251 \times 0.05+\451\times 0.01+(-\49)\times 0.94 \\ &=-\29 \end{align*}

Thus we are expected to make a loss of $\29$ when purchasing the extended coverage, which implies that we should not purchase the extended coverage.

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