The following is a joint probability distribution function f

The following is a joint probability distribution function for air conditioner maintenance.

X= service time (hr)

1 2 3 4

1 .12 .08 .07 .05

2 .08 .15 .21 .13

3 .01 .01 .02 .07

Y=#of air conditioners at location

a) show that this is a valid probability distribution function

b)what is the probability that the service time is equal to the # of air conditioner units at a location?

c)what is the probability that the service time is 2 hours?

d) Find the marginal distribution of X, P_X(x).

e) Find the marginal distribution of Y, P_Y(y).

f) Find the expected service time, E(X).

g) Find the variance in service times, V(X).

h) What is the standard deviation in service times, correct answer to minutes?

a) for it to be a valid probability distribution function we should have the sum of all probabilities to be 1

now 0.12+0.08+0.01+0.08+0.15+0.01+0.07+0.21+0.02+0.05+0.13+0.07=1

hence indeed it is a valid probability distribution function. [answer]

b) P[service time=no. of air conditioner units]=P[X=1,Y=1]+P[X=2,Y=2]+P[X=3,Y=3]=0.12+0.15+0.02=0.29 [answer]

c) P[service time is 2hours]=0.08+0.15+0.01=0.24 [answer]

d)marginal of X is

P[X=1]=0.12+0.08+0.01=0.21

P[X=2]=0.08+0.15+0.01=0.24

P[X=3]=0.21+0.07+0.02=0.30

P[X=4]=0.05+0.13+0.07=0.25

e) marginal of Y is

P[Y=1]=0.12+0.08+0.07+0.05=0.32

P[Y=2]=0.08+0.15+0.21+0.13=0.57

P[Y=3]=0.01+0.01+0.02+0.07=0.11

f) E[X]=1*0.21+2*0.24+3*0.30+4*0.25=2.59 hours [answer]

g) E[X²]=1*0.21+4*0.24+9*0.30+16*0.25=7.87

so V[X]=E[X²]-E[X]²=7.87-2.59²=1.1619 hrs² [answer]

h) standard deviation of service times= sqrt(V[X])=sqrt(1.1619)=1.077914 hours=64.6748 minutes [answer]