Homework SourseA relay microchip in a telecommunications satellite has a li
A relay microchip in a telecommunications satellite has a life expectancy that follows a normal distribution with a mean of 90 months and a standard deviation of 3.3 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.)
$
Solution
A)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.02
Then, using table or technology,
z = -2.053748911
As x = u + z * s,
where
u = mean = 90
z = the critical z score = -2.053748911
s = standard deviation = 3.3
Then
x = critical value = 83.22262859 = 83 months [ANSWER]
*******************
b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 84
u = mean = 90
s = standard deviation = 3.3
Thus,
z = (x - u) / s = -1.818181818
Thus, using a table/technology, the left tailed area of this is
P(z < -1.818181818 ) = 0.034518174 [ANSWER]
*********************
c)
Expected loss = 0.034518174*50000000 = $1725908.7 [ANSWER]
**********************
d)
Expected profit = 3000000 - 1725908.7 = $ 1274091.3 [ANSWER]
