A manufacturer claims that the life span of its tires is 50 comma 00050,000 mile

Question

A manufacturer claims that the life span of its tires is 50 comma 00050,000 miles. You work for a consumer protection agency and you are testing these tires. Assume the life spans of the tires are normally distributed. You select 100100 tires at random and test them. The mean life span is 49 comma 83149,831 miles. Assume sigmaequals=800800. Complete parts (a) through (c). (a) Assuming the manufacturer's claim is correct, what is the probability that the mean of the sample is 49,831 miles or less? nothing (Round to four decimal places as needed.) (b) Using your answer from part (a), what do you think of the manufacturer's claim? The claim is accurate inaccurate because the sample mean would not would be considered unusual since it does not lie lies within the range of a usual event, namely within 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means. (c) Assuming the manufacturer's claim is true, would it be unusual to have an individual tire with a life span of 49 comma 83149,831 miles? Why or why not? Yes No , because 49,831 lies does not lie within the range of a usual event, namely within 1 standard deviation 2 standard deviations 3 standard deviations of the mean for an individual tire.

Explanation / Answer

A manufacturer claims that the life span of its tires is 50 comma 00050,000 miles. You work for a consumer protection agency and you are testing these tires. Assume the life spans of the tires are normally distributed. You select 100100 tires at random and test them. The mean life span is 49 comma 83149,831 miles. Assume sigmaequals=800800. Complete parts (a) through (c).

(a) Assuming the manufacturer's claim is correct, what is the probability that the mean of the sample is 49,831 miles or less? nothing (Round to four decimal places as needed.)

Standard error = sd /sqrt(n) =800/sqrt(100) =80

Z value for 49831, z=(49831-50000)/80 = -2.11

P( mean x < 49831) = P( z < -2.11)

=0.0174

(b) Using your answer from part (a), what do you think of the manufacturer's claim?

The claim is accurate inaccurate because the sample mean would be considered unusual since it does not lie lies within the range of a usual event, namely within 1 standard deviation 2 standard deviations 3 standard deviations of the mean of the sample means.

(c) Assuming the manufacturer's claim is true, would it be unusual to have an individual tire with a life span of 49 comma 83149,831 miles? Why or why not?

Yes , because 49,831 does not lie within the range of a usual event, namely within 1 standard deviation 2 standard deviations 3 standard deviations of the mean for an individual tire.

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