Question

The following is the probability mass function for the number of times a certain computer program will malfunction: 0 1 2 3 4 5 p(x) | 0.05 0.28 0.42 0.15 0.08 0.02 (a) What is the probability that the computer program will malfunction more than 3 times? (b) Compute E(X), E(X2), and V(X). E(X) = E(X2) = V(X) = (C) Suppose Y, the time in minutes to fix malfunctions, relates to the number of times the program malfunctions, by the function: Y = 9*X . What is the expected time in minutes needed to fix malfunctions in the program?d) What is the variance of the time in minutes to fix the malfunctions?

Answer 1

a)P(X>3)=0.1

b)

E(X)=1.99

E(X²)=5.09

V(X)=1.13

c) E(Y)=17.91

d) V(Y)=81*V(X)=91.5

Explanation:

a)

P(X>3)=P(X=4)+P(X=5)=0.08+0.02=0.1

b)

E(X)=∑x*p(x)=0*0.05+1*0.28+2*0.42+3*0.15+4*0.08+5*0.02=1.99

E(X²)=∑x²*p(x)=0*0.05+1*0.28+4*0.42+9*0.15+16*0.08+25*0.02=5.09

V(X)=E(X²)-(E(X))²=5.09-3.96=1.13

c)

E(Y)=9*E(X)=9*1.99=17.91

d)

V(Y)=9²*V(X)=81*V(X)=81*1.13=91.5