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

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

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

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

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

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

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

μ=xP(x)=$251×0.05+$451×0.01+($49)×0.94=$29\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\$29 when purchasing the extended coverage, which implies that we should not purchase the extended coverage.

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