Question

1.       The lifetime of an electrical component is modeled as an exponential random variable with

          parameter b = 2.25 years. A customer has purchased five of these components and will use one

          until the lifetime is completed, then use the second until that lifetime is completed, and so on.

          Let Y_i ~ exponential (b = 2.25 years) be a random sample of five components and consider the

          total lifetime T = Y₁ + Y₂ + Y₃ + Y₄ + Y₅ .

          A.       Derive the E(T) and Standard Deviation of T, denoted SD(T) (Hint: Find VAR(T) first).

                     You might make use of theorems from Section 5.6 (p 258-259).

          B.       Simulate 100,000 rows of 5 values per row (the Y_i’s) and calculate the sum.

                     Use the results to estimate the P( E(T) – SD(T) < T < E(T) + SD(T)), the probability the

                     random sum will be within one standard deviation of the mean of the random sum.  

Answer 1

Here Yi ,the lifetime of an electrical component ~ exp( b=2.25years )

T = Y1 + Y2 + Y3 + Y4 + Y5 =Total lifetime of all 5 bulbs

So T ~gamma (n=5 , b=2.25)

pdf of T :  

E(T) = n/b = 5/2.25 = 2.22

SD(T) = sqrt(Var(T)) = n/b² = 0.9876

B.The Simulation is shownbelow through R-code :

M=matrix(rexp(100000,rate=2.25),nrow = 100000,ncol = 5,byrow=T)
T=as.matrix(rowSums(M));T
m=mean(T)
s=sd(T)
range=c(m-s,m+s);range
count=0
for(i in 1:length(T)){
if(range[1]<T[i,] && T[i,]<range[2])
count=count+1
}
count
prob=count/length(T);prob

answered by: anonymous