5.15 After being cut by a special saw, suppose the length of aluminum alloy bill

Question

5.15 After being cut by a special saw, suppose the length of aluminum alloy billets, X, is normally distributed with mean = 96.2 inches and standard deviation = 0.2 inches. Suppose you randomly select 20 billets (assume independence) (a) Determine the probability that exactly 2 are longer than 96.5 inches. Hint: First find the probability that a single billet is longer than 96.5 inches, then use the binomial distribution. e the probability that 19 or fewer are have a length less than 96.5 inches. (c) Determine the probability that all 20 billets are between 96.0 and 96.5 inches.

Explanation / Answer

P(X > 96.5) = P ( (x - mean)/SD > (96.5 - 96.2)/0.2)

= P(Z > 1.5)

= 1 - P(Z < 1.5)

= 1 - 0.9332 = 0.0668

A) n = 20

P = 0.0668

P (X = 2) = 20C2 * (0.0668)^2 * (0.9332)^18

= 0.2443

B) P(X < 19) = 1 - P(X > 19)

= 1 - P(X = 20)

= 1 - 20C20 * (0.0668)^20 * (0.9332)^0

= 1

C) P(96 < X < 96.5) = P((96 - 96.2)/0.2 < (X - mean)/SD < (96.5 - 96.2)/0.2)

= P(-1 < Z < 1.5)

= P(Z < 1.5) - P(Z < -1)

= 0.9332 - 0.1587 = 0.7745

n = 20

P = 0.7745

P(X = 20) = 20C20 * (0.7745)^20 * (0.2255)^0

= 0.006

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