Question 133415: Hi, I'm at a lost and need some help. Could you please help me with chapter 8 problems 8.46 and 8.62 from the chapter exercise section. Also can you please help me with these problems. 1)The score on the entrance test for a well known law school has a mean score of 200 points and a standard deviation of 50 points. At value should the lowest passing score be set if the school wishes only 2.5% of those taking the entrance test to pass? and ....2)A tire manufacturer wishes to investigate the thread life of its tires. A sample of 10 tires driven 50,000 miles revealed a sample mean of 0.32 inches of thread remaining with a standard deviation of 0.09 inches. Construct a 95% confidence interval for the population mean. Would it be reasonable for the manufacturer to conclude that after 50,000 miles the population mean amount of thread remaining is 0.30 inches? I'm really confussed and is in great need of help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1)The score on the entrance test for a well known law school has a mean score of 200 points and a standard deviation of 50 points. At value should the lowest passing score be set if the school wishes only 2.5% of those taking the entrance test to pass? and ....
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You only want the upper 2.5% of the tests to pass.
Find the z-value associated with the upper 2.5% of a normal distribution.
Use your z-table or InvNorm on your TI calculator to find z = 1.9599
Now find the x-score that corresponds to that z-score where:
z(x) = (x-mu)/sigma
1.9599 = (x-200)/50
x-200 = 97.998
x = 297.998
Rounding up you get a score of 298
Only 2.5% of the scores will be above 298.
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2)A tire manufacturer wishes to investigate the thread life of its tires. A sample of 10 tires driven 50,000 miles revealed a sample mean of 0.32 inches of thread remaining with a standard deviation of 0.09 inches. Construct a 95% confidence interval for the population mean.
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E = 1.96*0.09/sqrt(10) = 0.0558
C.I. = (0.32-0.0558,0.32+0.0558)
C.I: 0.2642 < mu < 0.3758
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Would it be reasonable for the manufacturer to conclude that after 50,000 miles the population mean amount of thread remaining is 0.30 inches?
He could have 95% confidence that the mean amt. of tread is BETWEEN
0.2642 and 0.3758, not that the mean amount is any particular
value in that range.
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Cheers,
Stan H.
I'm really confussed and is in great need of help.
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