Question 754501: The weekly output of a steel mill is normally distributed with the mean of 130 tons and standard deviation of 20 tons.
a.What is the probability that the mean weekly output of 9 randomly selected weeks is less than 135 tons?
b. 83% of the sample mean of 36 weeks are greater than what value?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The weekly output of a steel mill is normally distributed with the mean of 130 tons and standard deviation of 20 tons.
a.What is the probability that the mean weekly output of 9 randomly selected weeks is less than 135 tons?
Ans:
t(135) = (135-130)/[20/sqrt(9)] = 0.75
P(x-bar < 135)= P(t < 0.75 when df = 8) = tcdf(-100,0.75,8) = 0.7626
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b. 83% of the sample mean of 36 weeks are greater than what value?
The left tail is 0.17
Ans: invT(0.17,8) = -1.0146
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x-bar = z*s+u
x-bar = -1.0146*(20/3)+130 = 123.23 tons
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Cheers,
Stan H.
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