One step in the manufacture of a certain metal hinge involves the drilling of fo
ID: 2960627 • Letter: O
Question
One step in the manufacture of a certain metal hinge involves the drilling of four holes. In a sample of 150 hinges, the average time needed to complete this step was 72 seconds and the standard deviation was 10 seconds.
(a) Find the probability that the average time to complete the step is between 65 and 74 seconds (inclusive).
(b) Find a 93% confidence interval for the mean time needed to complete the step.
(c) How many clamps must be sampled so that a 88% confidence interval specifies the mean to within ±1.5 seconds?
(d) Find a 97% lower confidence bound for the mean time to complete the step.
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Explanation / Answer
Given n=150, =72, s=10
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(a) Find the probability that the average time to complete the step is between 65 and 74 seconds (inclusive).
P(65<xbar<74) = P((65-72)/(10/sqrt(150)) <(xbar-)/(s/n) <(74-72)/(10/sqrt(150)))
=P( -8.57<Z< 2.45)
= 0.9929 (check standard normal table)
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(b) Find a 93% confidence interval for the mean time needed to complete the step.
Given a=0.07, Z(0.035)=1.81 (check standard normal table)
SO 93% CI is
xbar ± Z*s/n
--> 72 ± 1.81*10/sqrt(150)
--> ( 70.522, 73.478)
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(c) How many clamps must be sampled so that a 88% confidence interval specifies the mean to within ±1.5 seconds?
Given a=0.12, Z(0.06)=1.55(check standard normal table)
E=1.5
So n=(Z*s/E)^2 =(1.55*10/1.5)^2 = 106.7778
Take n=107
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(d) Find a 97% lower confidence bound for the mean time to complete the step.
Given a=0.03, Z(0.03)=1.88(check standard normal table)
So the lower confidence bound = xbar -Z*s/n = 72 - 1.88*10/sqrt(150) = 70.46499
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