Refer to Exercise 4. The total number of assemblies that fail to meet specifications is X + Y.

a. Find μX+Y.

b. Find σX+Y.

c. Find P(X + Y = 3).

REFERENCE EXERCISE 4: In a piston assembly, the specifications for the clearance between piston rings and the cylinder wall are very tight. In a lot of assemblies, let X be the number with too little clearance and let Y be the number with too much clearance. The joint probability mass function of X and Y is given in the table below:

a. Find the marginal probability mass function of X.

b. Find the marginal probability mass function of Y.

c. Are X and Y independent? Explain.

d. Find μX and μY.

e. Find σX and σY.

f. Find Cov(X, Y).

g. Find ρ(X, Y).

Step 1 of 4 :

Here the experiment is assembly of piston. X denotes the number with too little clearance and Y denotes the number with too much clearance.

Also,the joint probability mass of X and Y is given.

Using this we have to find the required values.

Step 2 of 4:

(a)

Here we have to find the value of .This is given by,

=+

where,

=x

=y

Now we have to find the marginal probability mass functions of X and Y.

Marginal probability mass function of X is given by,

=f(x,y)

=f(0,0)+f(0,1)+f(0,2)+f(0,3) at x=0

=f(1,0)+f(1,1)+f(1,2)+f(1,3) at x=1

=f(2,0)+f(2,1)+f(2,2)+f(2,3) at x=2

=f(3,0)+f(3,1)+f(3,2)+f(3,3) at x=3

=0.15+0.12+0.11+0.10 at x=0

=0.09+0.07+0.05+0.04 at x=1

=0.06+0.05+0.04+0.02 at x=2

=0.04+0.03+0.02+0.01 at x=3

=0.48 at x=0

=0.25 at x=1

=0.17 at x=2

=0.10 at x=3

Thus,

=x

=0(0.48)+1(0.25)+2(0.17)+3(0.10)

=0.25+0.34+0.30

=0.89

Marginal probability mass function of Y is given by,

=f(0,0)+f(1,0)+f(2,0)+f(3,0) at y=0

=f(0,1)+f(1,1)+f(2,1)+f(3,1) at y=1

=f(0,2)+f(1,2)+f(2,2)+f(3,2) at y=2

=f(0,3)+f(1,3)+f(2,3)+f(3,3) at y=3

=0.15+0.09+0.06+0.04

=0.12+0.07+0.05+0.03

=0.11+0.05+0.04+0.02

=0.10+0.04+0.02+0.01

=0.34

=0.27

=0.22

=0.17

Thus,

=y

=0(0.34)+1(0.27)+2(0.22)+3(0.17)

=0.27+0.44+0.51

=1.22

Therefore,

=+

=0.89+1.22

=2.11

Hence,=2.11.