The mean lifetime of a tire is 43 months with a standard deviation of 10 months.
If 105 tires are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 0.41 months? Round your answer to four decimal places.
Solution :
Given that,
mean = = 43
standard deviation = = 10
n = 105
= = 43
= / n = 10 / 105 = 0.9759
P(42.59 < < 43.41) = P((42.59 - 43) / 0.9759 <( - ) / < (43.41 - 43) / 0.9759 ))
= P(-0.42 < Z < 0.42)
= P(Z < 0.42) - P(Z < -0.42) Using standard normal table,
= 0.6628 - 0.3372
= 0.3256
The probability that the mean of the sample would differ from the population mean by less than 0.41 months is 0.3256.
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