Question

Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose that a computer manufacturer receives computer boards in lots of five. Two boards are randomly selected without replacement from each lot for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pair (1,2) represents the selection of boards 1 and 2 for inspection regardless of the order of selection. Define r.v. X to be the number of defective boards observed among the two inspected.

(a) . By listing the ten different possible outcomes, find the probability distribution of X assuming that boards 1 and 2 are the only defective boards in a lot of five.

(b) . Calculate µX and σX.

Answer 1

Solution

Back-up Theory

If a discrete random variable, X, has probability function, p(x), x = x₁, x₂, …., x_n, then

Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x…………………………………………..…. (1)

E(X²) = Σ{x².p(x)} summed over all possible values of x……………………………………………………………….…………..(2)

Variance of X = Var(X) = σ² = E(X²) – {E(X)}²….……………………………………………………………………..……………..(3)

Standard Deviation of X = SD(X) = σ = sq.rt of Var(X) ..…………………………………………………………..………………..(4)

Now, to work out the solution,

Part (a)

Listing of outcomes and number of defectives.

Outcome #	Outcomes (Board #)	# of defectives (Board #)
1	1, 2	2 (1, 2)
2	1, 3	1 (1)
3	1, 4	1 (1)
4	1, 5	1 (1)
5	2, 3	1 (2)
6	2, 4	1 (2)
7	2, 5	1 (2)
8	3, 4	0
9	3, 5	0
10	4, 5	0

Probability Distribution of X

x	Frequency	p(x)
0	3	0.3
1	6	0.6
2	1	0.1
Total	10	1

Answer 1

Part (b)

Vide (1),

Mean (average) of X = E(X) = µ_X = 0.8 Answer 2

Vide (2),

E(X²) = 1.0

Vide (3),

Var(X²) = 1.0 – 0.8²

= 0.36

Vide (4),

σ_X = sqrt(0.36)

= 0.6 Answer 3

Calculation details

x	Frequency	p(x)	x.p(x)	x^2.p(x)
0	3	0.3	0	0
1	6	0.6	0.6	0.6
2	1	0.1	0.2	0.4
Total	10	1	0.8	1

DONE