(a). Felicia is a quality- control engineer for a large electronic company. Her

Question

(a). Felicia is a quality- control engineer for a large electronic company. Her job is to thoroughly test each stereo set manufactured by the company and to classify it as acceptable or unacceptable. Due to the company’s rigorous quality standards, each set has an equally likely chance of being acceptable or unacceptable. On a Monday morning, Felicia inspects six (6) sets. Find the probability that she

(i). Rejects all the sets. (3 pts)

(ii). Accepts all the sets. (3 pts)

(iii). Accepts at least three of the sets. (7 pts)

(iv). Rejects at most four of the sets. (7 pts)

(b). If Felicia inspected 10 stereos on a Tuesday morning,

(i). What is the expected number of stereos to inspect. (2 pts)

(ii). What is the standard deviation of the number of stereos to be inspected. (4 pts)

(iii). What is the difference between the number of stereos inspected on Tuesday and on Monday. (4 pts)

Explanation / Answer

Due to the company’s rigorous quality standards, each set has an equally likely chance of being acceptable or unacceptable. On a Monday morning, Felicia inspects six (6) sets. Find the probability that she

using binomial distirbution with p =0.50 and n =6

(i). Rejects all the sets. (3 pts)

P(x=6) = 6C6 * 0.50^6 * 0.50^6-6 = 0.015625

(ii). Accepts all the sets. (3 pts)

P(x=0) = 6C0 * 0.50^0 *0.50^6 = 0.015625

(iii). Accepts at least three of the sets. (7 pts)

P(x>= 3 ) = 1 - P(x< 3)

P( x=0) = 6C0 *0.50^0 * 0.50^6-0 =0.015625

P( x=1) = 6C1 *0.50^1 * 0.50^6-1 = 0.09375

P( x=2) = 6C2 *0.50^2 * 0.50^6-2 =0.234375

P(x>= 3 ) = 1 - 0.34375 = 0.65625

(iv). Rejects at most four of the sets. (7 pts)

P ( x <= 4 )

P( x=0) = 6C0 *0.50^0 * 0.50^6-0 =0.015625

P( x=1) = 6C1 *0.50^1 * 0.50^6-1 = 0.09375

P( x=2) = 6C2 *0.50^2 * 0.50^6-2 =0.234375

P(x=3) = 6C3 *0.50^3 * 0.50^6-3 = 0.3125

P(x=4) = 6C4 *0.50^4 * 0.50^6*4 =0.234375

P ( x <= 4 )=0.890625

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