machine that cuts corks for wine bottles operates in such a way that the distrib
ID: 3049020 • Letter: M
Question
machine that cuts corks for wine bottles operates in such a way that the distribution of the diameter of corks is well approximated by a Normal distribution with mean 3 cm and standard deviation 0.1 cmm The specifications call for corks with diameters between 2.9 and 3.1 cm. A cork not meeting the specifications is considered defective (a cork that 1s too small leaks an ; a cork small leaks and d causes the wine to deteriorate that is too large does not fit in the bottle) (a) What proportion of corks produced by this machine are defective? (b) If you choose 15 corks at random produced by this machine, what is the probability that 5 of them will be defective? (c) Suppose a second machine produces corks with diameters that a approximated by a Normal distribution with mean 3.05 cm and standard deviation 0.01 cm. Which machine would you recom mend? ExplainExplanation / Answer
a) P(defective) = 1 - P(non defective)
= 1 - P(2.9 < X < 3.1)
= 1 - P((2.9 - mean)/sd < (X - mean)/sd < (3.1 - mean)/sd)
= 1 - P((2.9 -3)/0.1 < Z < (3.1 - 3)/0.1)
= 1 - P(-1 < Z < 1)
= 1 - (P(Z < 1) - P(Z < -1))
= 1 - (0.8413 - 0.1587)
= 1 - 0.6826 = 0.3174
b) n = 15
p = 0.3174
P(X = x) = nCx * px * (1 - p)n - x
P( X = 5) = 15C5 * (0.3174)^5 * (1 - 0.3174)^10 = 0.2124
c) For the second machine
P(defective) = 1 - P(non defective)
= 1 - P(2.9 < X < 3.1)
= 1 - P((2.9 - mean)/sd < (X - mean)/sd < (3.1 - mean)/sd)
= 1 - P((2.9 -3.05)/0.01 < Z < (3.1 - 3.05)/0.01)
= 1 - P(-15 < Z < 5)
= 1 - (P(Z < 5) - P(Z < -15))
= 1 - (1 - 0)
= 1 - 1 = 0
As the probability of defective for the second machine is less than the forst machine, so the second machine would be recommend.
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