Question

A battery for the capsule in a space flight has a lifetime, X, that is normally distributed with a mean of 100 hours and a standard deviation of 30 hours.

(a) What is the distribution of the total lifetime T of three batteries used consecutively, T = X1 + X2 + X3 ,if the lifetimes are independently distributed?

(b) What is the probability that three batteries will suffice for a mission of 200 hours?

(c) Suppose the length of the mission is independently normally distributed with a mean of 200 hours and a standard deviation of 50 hours. What is now the probability that three batteries will suffice for the mission?

Answer 1

a)

here as three batteries lifetime are normally distributed therefore there total life time should be normally distributed with mean =E(X1)+E(X2)+E(X3)=100+100+100=300

and standard deviation =sqrt(Var(X1)+Var(X2)+Var(X3))=sqrt(30²+30²+30²)=51.96

b)

probability that three batteries will suffice for a mission of 200 hours =P(X>200)

=P(Z>(200-300)/51.96)=P(Z>-1.92)=0.9726

c)

here let length of mission follow variable Y; also A=Y-T

hence mean of A =E(Y)-E(T)=200-300=-100

and std deviaiton of A =sqrt(50²+2700)=72.111

hence  probability that three batteries will suffice for the mission =P(A<0)

=P(Z<(0-(-100))/72.111)=P(Z<1.39)=0.9177