Question 1164888: The average number of miles (in thousands) that a car's tire will function before needing replacement is 66 and the standard deviation is 11. Suppose that 47 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.
What is the distribution of
X? X~ N(________,________)
What is the distribution of
¯x?¯x~ N(_______,__________)
If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 67.3 and 69.2.
For the 47 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 67.3 and 69.2.
For part d), is the assumption that the distribution is normal necessary? Yes or No
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! X ~ N(66, 121)
x bar is ~ N (66, 121/47) that can be converted to a decimal 2.57
for the first, the probability is
z=(67.3-66)/11 and z=(69.2-66)/11 or z between 0.118 and 0.291
=0.0675 probability (0.0674 not rounded on the calculator)
for a sample n=47, the sd is 11/sqrt (47) or 1.605, z between 0.810 and 2.00
probability is 0.1859
With a sample size of 47 and a distribution that is likely not skewed, normal assumption can be used. The sample size >30 is used by many, but tire wear is likely not greatly skewed.
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