A lumber yard is selling economy studs which are “approximately” 8 ft. long. Whe

Question

A lumber yard is selling economy studs which are “approximately” 8 ft. long. When asked about what

“approximately” means, the quality control technician explains that the length of the studs follows a

normal distribution with mean 8 ft. and standard deviation 0.5 inch. (Note: 1 ft.=12 in.)

(a) What is the probability that a randomly selected stud will be shorter than 8 ft. by more than one inch?

(b) Suppose you buy 1000 studs. What is the expected number of “short” (i.e., less than 7 ft. 11 in.) studs in your purchase?

(c) What is the probability that there will be at least 30 short studs in a batch of 1000?

Explanation / Answer

mean (mu) = 8
std. dev. (sigma) = 0.5 in (i.e. 0.0417 ft)

(A) Shorter than 1 in means less than two sigma levels.
i.e. 8 - 2*0.0417 = 7.9166

P(X < 7.9166) = P(z < (7.9166 - 8)/0.0417) = P(z < -2) = 0.0228

0.0228 is the probability that a randomly selected stud will be shorter than 8 ft. by more than one inch

(B) 7ft 11 in = 8 - 2*0.0417 = 7.9166

P(X < 7.9166) = 0.0228

Hence expected number of "short = 0.0228 * 1000 = 228

(C) p = 0.0228
n = 1000
mean = np = 228
std. devv. = sqrt(1000 * 0.0228 * (1-0.0228)) = 4.7202

Probability that there will be at least 30 short studs in a batch of 1000,
P(X >= 30) = 1 - P(X < 30)
P(X < 30) = P(z< (30 - 228)/4.7202) = P(z < -41.9473) = 0

Hence required probability = 1 - 0 = 1

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