A relay microchip in a telecommunications satellite has a life expectancy that f

Question

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 3.7 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 indefinitey (a) For how many months should the satellite be insured to be 91% confident that it will last beyond the insurance date? Round your answer to the nearest month.) months 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.) (e) If the satellite is insured for 84 months, what is the expected loss to the insurance company? (Round your answer to the nearest dollar. oen (a) 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 doilar,) (d) If the insurance company charges $3 million for 84 m Noed Help? L Rosdn My Notes As Your Teacher 12. 0.87 points Beunderstat126. 3038 + distributed, with pairs at a large restaurant was observed over a long period of time to be appeoximately normally The amount of money spent weekly on cleaning, maintenance, and re mean = 595 and standard deviation -S42. decimal places.) asa ed for next week, what is the probability that the actual costs will exceed the budgeted amount? (Round your answer to four given week is only o d be budgeted for weekly repairs, cleaning, and maintenance so that the probability that the budgeted amount will be exceeded in a (b) How much shoul (Round your answer to the nearest dollar.)

Explanation / Answer

(a)

z = (x - 92)/3.7)

P(z < (x - 92)/3.7 ) = 0.91

P( z < 1.341 ) = 0.91

(x - 92)/3.7 = 2.0537
or, x = 2.0537*3.1 + 90 = 96.96 = 97 months

c.

P(X<84)=P(Z<84-92/3.7)=P(Z<-2.16)=0.0153

Expected loss=0.0153*84=1.285

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