Question

A manufacturer of golf balls uses a production process that produces 10 percent defective balls. A quality inspector takes samples of a week's output with replacement. Using the cumulative binomial probability table available in your text, the inspector can determine which of the following probabilities?

	a.	If 15 units are inspected, the probability of at least 10 of these units being defective is .547.
	b.	If 10 units are inspected, the probability of 5 or 6 of these units being defective is .002.
	c.	If 20 units are inspected, the probability of at least 19 of these units being defective is .0009.
	d.	If 5 units are inspected, the probability of at most 3 of these units being defective is .984.

Answer 1

Correct option:

b. If 10 units are inspected, the probability of 5 or 6 of these units being defective is .002

EXPLANATION:

(a)

n= 15

p = 0.1

q = 1 -p = 0.9

P(At least 10 are defective) = P(X $\geq$ 10) = 1 - P(X $\leq$ 10) = 1 - 0.999999 = 0.00000019

(b)

n= 10

p = 0.1

q = 1 -p = 0.9

P(5 or 6 are defective) = P(X $\leq$ 6) - P(X $\leq$ 4)= 0.999991 - 0.998365 = 0.002

(c)

n= 20

p = 0.1

q = 1 -p = 0.9

P(At least 19 are defective) = P(X $\geq$ 19) = 1 - P(X $\leq$ 19) = 1 - 0.999999 = 0.0000001

(d)

n= 5

p = 0.1

q = 1 -p = 0.9

P(At most 5 are defective) = P(X $\leq$ 5) = 0.99985