Answer to Question #96007 in Statistics and Probability for Anthony DiTommaso

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) (4 marks). 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.
2
(b) (3 marks). Calculate µX and σX.

Answer 1

Let the boards are numbered from 1 to 5.

a) Let X be the number of defective boards observed among the two inspected.

$(1,2), (1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5)$

If the boards 1 and 2 are the only defective boards in a lot of five, then

$(1,2),x=2;(1,3),x=1;(1,4),x=1;(1,5),x=1;$

$(2,3),x=1;(2,4),x=1;(2,5),x=1;$

$(3,4),x=0;(3,5),x=0;(4,5),x=0.$

$$P(X=0)={3 \over 10}=0.3$$ $$P(X=1)={6 \over 10}=0.6$$ $$P(X=2)={1 \over 10}=0.1$$

$$\begin{matrix} x & 0 & 1 & 2 \\ p(x) & 0.3 & 0.6 & 0.1 \end{matrix}$$

b)

$$\mu_X=\displaystyle\sum_{i=1}^3x_ip(x_i)$$ $$\mu_X=0\cdot0.3+1\cdot0.6+2\cdot0.1=0.8$$

$$\sigma_X^2=\displaystyle\sum_{i=1}^3(x_i-\mu_X)^2p(x_i)$$

$$\sigma_X^2=(0-0.8)^2\cdot0.3+(1-0.8)^2\cdot0.6+(2-0.8)^2\cdot0.1$$ $$\sigma_X^2=0.36$$

$$\sigma_X=\sqrt{\sigma_X^2}$$ $$\sigma_X=\sqrt{0.36}=0.6$$

$\mu_X=0.8,\ \sigma_X=0.6$