Question

5-7. A manufacturing company employs two devices to inspect output for quality control purposes. The first device is able to accurately detect 99.4% of the defective items it receives, whereas the second is able to do so in 99.7% of the cases. Assume that four defective items are produced and sent out for inspection. Let X and Y denote the number of items that will be identified as defective by inspecting devices 1 and 2, respectively. Assume that the devices are independent. Determine: (a) fXY (x, y) (c) E(X) (b) fx(x) (d) fyp (y)

Answer 1

(a)

Since first device detect correctly 99.4% of the defective items so X has binomial distribution with parameters as follows:

n = 4 and p = 0.994

The pdf of X is

$f_{X}(x)=P(X=x)=\binom{4}{x}(0.994)^{x}(1-0.994)^{4-x}=\binom{4}{x}(0.994)^{x}(0.006)^{4-x},x=0,1,2,3,4$

Likewise the pdf of Y is

$f_{Y}(y)=P(Y=y)=\binom{4}{y}(0.997)^{y}(1-0.997)^{4-y}=\binom{4}{y}(0.997)^{y}(0.003)^{4-y},y=0,1,2,3,4$

The joint pdf of X and Y is

$f_{XY(x,y)}=f_{X}(x)f_{Y}(y)=P(X=x)P(Y=y)=\binom{4}{x}(0.994)^{x}(0.006)^{4-x}\cdot \binom{4}{y}(0.997)^{y}(0.003)^{4-y},x=0,1,2,3,4,y=0,1,2,3,4$

Following table shows the joint and marginal pdfs:

				X
		0	1	2	3	4	P(Y=y)
	0	0	0	0	0	0	0
	1	0	0	0	0	0	0
Y	2	0	0	0	0.000001	0.000052	0.000053
	3	0	0	0.000003	0.00028	0.011609	0.011892
	4	0	0.000001	0.000211	0.023289	0.964553	0.988054
	P(X=x)	0	0.000001	0.000214	0.02357	0.976214	0.999999

(b)

The pdf of X is

$f_{X}(x)=P(X=x)=\binom{4}{x}(0.994)^{x}(1-0.994)^{4-x}=\binom{4}{x}(0.994)^{x}(0.006)^{4-x},x=0,1,2,3,4$

Following table shows the marginal pdf of X:

X	f(x)
0	1.296E-09
1	8.588E-07
2	0.0002134
3	0.0235706
4	0.9762151

(c)

The expected value of X is

$E(X)= np=4\cdot 0.994 = 3.976$

(d)

The pdf of Y is

$f_{Y}(y)=P(Y=y)=\binom{4}{y}(0.997)^{y}(1-0.997)^{4-y}=\binom{4}{y}(0.997)^{y}(0.003)^{4-y},y=0,1,2,3,4$

Following table shows the marginal pdf of Y:

Y	P(Y=y)
0	0
1	0
2	0.000053
3	0.011892
4	0.988054