If use excel, please show the calculation steps and excel formula A satellite ha
ID: 3174559 • Letter: I
Question
If use excel, please show the calculation steps and excel formula A satellite has a life expectancy that follows a normal distribution with Au 90 3.9 months. An insurance company is going to insure the satellite for months and $500 million. If the satellite is insured for 84 months, what is the probability that it will stop functioning before the insurance coverage ends? If the insurance company charges $30 million for 84 months of insurance, how much profit does the insurance company expect to make (i.e, the expected value of the profit the insurance company will make). For how many months should the satellite be insured to be 99% confident that it will last beyond the insurance date?Explanation / Answer
Answer to question# 1)
For 84 months
Formula of Z is as follows:
Z = (x - Mean) / SD
We got x = 84
Mean = 90
SD = 3.9
.
On pluggging these values we get:
Z = (84-90)/3.9 = -1.54
From the Z table we get the probability
(Z < -1.54) = 0.06178
We can also use the excel formula to find the probability
use the following formula
=NORMSDIST(-1.54)
.
Answe to question# 2)
The payment that the insureance company need to make is $500for 84 months
If the satellite stops fucntioning before that , the loss to company would be : 500 - 30 = $470 million
The probability of loss is 0.06178
.
The profit if the satellite stops functioning after 84 months is : $30 millions
the chances of this profit are : 1- 0.06178 = 0.93822
.
Thus expected profit to company : 0.93822 * 30 - 0.06178 * 470 = $-0.89 millions
Thus the company would have loss, in this case rather than profit
.
Answer to question# 3)
For 99% confidence , we would first have to find the corresponding Z value
this can be found with the help of the excel formula
=normsinv(0.99)
We get Z = 2.3263
.
Using the Z formula with Mean = 90 and SD = 3.9 we get:
2.3263 = (X - 90) / 3.9
X = 99.07257 months
thus the insurance must be for 99.07257 months