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Many manufacturers have quality control programs that include inspection of incoming materials for defects. Suppose a computer manufacturer receives circuit boards in batches of five. Two boards are selected from each batch for inspection. We can represent possible outcomes of the selection process by pairs. For example, the pairs (1, 2) and (2, 1) represent the selection of boards 1 and 2 for inspection.

Required:

a. List the ten different possible outcomes.
b. Suppose that boards 1 and 2 are the only defective boards in a lot of five. Two boards are to be chosen at random. Define X to be the number of defective boards observed among those inspected. Find the probability distribution of X.
c. Let F(x) denote the cdf of X. First X O), F(l), and F(2); then obtain F(x) for all other x.

Respuesta :

Answer:

a) (1,2), (1,3), (1,4), (1,5), (2,3), (2,4),(2,5),(3,4), (3,5) and (4,5)

b)

PX(0) = 0.3

PX(1) = 0.6

PX(2) = 0.1

c)

  • F(x) = 0 if x<0
  • F(x) = 0.3 if 0 ≤ x < 1
  • F(x) = 0.9 if 1 ≤ x < 2
  • F(x) = 1 if x ≥2

Step-by-step explanation:

Lets call the boards 1,2,3,4 and 5

a) There are [tex]{5 \choose 2} = 10 [/tex] possible outcomes: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4),(2,5),(3,4), (3,5) and (4,5).

b) X is 0 if the selected boards doesnt contain 1 or 2. There are [tex] {3 \choose 2} = 3 [/tex] possibilities for this, in concrete, (3,4), (3,5) and (4,5).

PX(0) = 3/10 = 0.3

There is only 1 possibility so that X = 2, which is (1,2), therefore

PX(2) = 1/10 = 0.1

In any other case (there are 6 remaining), X will be equal to 1, so

PX(1) = 6/10 = 0.6

Therefore, X has the following distribution

PX(0) = 0.3

PX(1) = 0.6

PX(2) = 0.1

c) F(0) = 0.3

F(1) = 0.3+0.6 = 0.9

F(2) = 1

We have, as a consequence

  • F(x) = 0 if x<0
  • F(x) = 0.3 if 0 ≤ x < 1
  • F(x) = 0.9 if 1 ≤ x < 2
  • F(x) = 1 if x ≥2