A relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution...

A relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution with a mean of 92 months and a standard deviation of 3.8 months. When this computer-relay microchip malfunctions, the entire satellite is useless. A large London insurance company is going to insure the satellite for 50 million dollars. Assume that the only part of the satellite in question is the microchip. All other components will work indefinitely.

(a) For how many months should the satellite be insured to be 98% confident that it will last beyond the insurance date? (Round your answer to the nearest month.)
months

(b) If the satellite is insured for 84 months, what is the probability that it will malfunction before the insurance coverage ends? (Round your answer to four decimal places.)

(c) If the satellite is insured for 84 months, what is the expected loss to the insurance company? (Round your answer to the nearest dollar.)
$

(d) If the insurance company charges $3 million for 84 months of insurance, how much profit does the company expect to make? (Round your answer to the nearest dollar.)

Solutions

Expert Solution

$The relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution with a mean of 92 months and a standard deviation of 3.8 months $ \\\\ $ Mean$(\mu) = 92 $ months $ \\ $ Standard deviation$(\sigma) = 3.8 $ months$

$a) $ Let the satellite be insured for "a" months to be 98\% confident that it will last beyond the insurance date $ \\\\ \Rightarrow P(x > a) = 0.98 \\\\ \Rightarrow P\Big(\ z > \frac{a - 92}{3.8}\ \Big) = 0.98 \hspace{10 mm}\Big( \ z = \frac{x - \mu}{\sigma}\ \Big) \\\\ \Rightarrow P\Big(\ z > \frac{a - 92}{3.8}\ \Big) \approx P(z < 2.05) \hspace{25 mm} ( P(z < 2.05) $ is closest to $ \hspace*{95 mm} $ 0.98 from Z table $) \\\\ \Rightarrow P\Big(\ z > \frac{a - 92}{3.8}\ \Big) \approx P(z > -2.05) \hspace{20 mm}( $ Z distribution is symmetric $ \hspace*{85 mm} $ about z = 0 $) \\\\ \Rightarrow \frac{a-92}{3.8} = -2.05 \hspace{35 mm}( $ Comparing both sides $ ) \\\\ \Rightarrow a = 92 - 2.05 * 3.8 \approx 84 \\\\ $ So, the satellite should be insured for $ \mathbf{84} $ months to be 98\% confident that it will last beyond the insurance date $$

orchestra answered 1 year ago

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