Question

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let x denote the number of defective boards in a batch of 25 boards, so that x has a binomial distribution with n 5 25 and 5 .05. D

Answer 1

Question:

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let x denote the number of defective boards in a batch of 25 boards, so that x has a binomial distribution with n = 25 and π = 0.05.

(a) What proportion of batches have at most 2 defective boards?

(b) What proportion of batches have at least 5 defective boards?

(c) What proportion of batches will have all 25 boards free of defects?

Answer:

(a) Proportion of batches have at most 2 defective boards 0.87291.

(b) Proportion of batches have at least 5 defective boards is 0.00714.

(c) Proportion of batches will have all 25 boards free of defects is 0.2774.

Solution:

n= 25, π = 0.05

Binomial distribution formula:

`$P(x)=(n !)/(x !(n-x) !) \pi^(x)(1-\pi)^(n-x)`

`P(x)=(25 !)/(x !(25-x) !) *(0.05)^(x)(0.95)^(25-x)`

(a) Proportion with x = 0 is given by

`P(0)=(25 !)/(0 ! 24 !)(0.05)^0 (0.95)^(25)`

P(0) = 0.2774

Proportion with x = 1 is given by

`P(1)=(25 !)/(1 ! 24 !)(0.05)^1 (0.95)^(24)`

P(1) = 0.36499

Proportion with x = 2 is given by

`P(2)=(25 !)/(2 ! 23 !)(0.05)^(2)(0.95)^(23)`

P(2) = 0.23052

Proportion with at most 2 defective = P(x ≤ 2)

= P(0) + P(1) + P(2)

= 0.2774 + 0.36499 + 0.23052

Proportion with at most 2 defective = 0.87291

(b) Proportion with x = 3 is given by

`P(3)=(25 !)/(3 ! 22 !)(0.05)^(3)(0.95)^(22)`

P(3) = 0.093

Proportion with x = 4 is given by

`P(4)=(25 !)/(4 ! 21 !)(0.05)^(4)(0.95)^(21)`

P(4) = 0.2693

Proportion with x ≥ 5 = 1 – P(x ≤ 5)

= 1 – (P(0) + P(1) + P(2) + P(3) + P(4))

= 1 – (0.2774 + 0.36499 + 0.23052 + 0.093 + 0.093)

= 0.00714

Proportion of batches have at least 5 defective boards is 0.00714.

(c) Proportion with x = 0

It is already found in (a).

Proportion of batches will have all 25 boards free of defects is 0.2774.