An automatic machine in a manufacturing process is operating properly if the lengths of an important subcomponent are normally distributed with a mean of 118 cm and a standard deviation of 5 cm.
A. Find the probability that one selected subcomponent is longer than 120 cm. Probability =
B. Find the probability that if 3 subcomponents are randomly selected, their mean length exceeds 120 cm. Probability =
C. Find the probability that if 3 are randomly selected, all 3 have lengths that exceed 120 cm. Probability =
A) Let X be the random variable denoting the length of an important subcomponent. So, X ~ N(118, 52)
Probability that the subcomponent is longer than 120 cm =
P(X > 120)
= 1 - P(X < 120)
= 1 - P(Z < (120-118)/5)
= 1 - P( Z < 0.4)
= 1 - 0.655
= 0.345.
B) The probability that the mean length of the 3 subcomponents exceeds 120 cm will be equal to the probability that one subcomponent is longer than 120 cm, which is equal to 0.345.
C) The probability that if 3 subcomponents are randomly selected, all 3 have lengths that exceed 120 cm =
P(X > 120)3
And we have calculated that P(X > 120) = 0.345
So, P(X > 120)3 = 0.3453
= 0.0411.
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