Question

A manufacturer claims that the life span of its tires is 51,000 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 100 tires at random and test them. The mean life span is 50,844 miles. Assume σ = 800. Complete parts​ (a) through​ (c).

a.) Assuming the​ manufacturer's claim is​ correct, what is the probability that the mean of the sample is 50,844 miles or​ less?

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

The claim is (inaccurate/accurate) because the sample mean (would/would not be) considered unusual since it (does/does not) lie within the range of a usual​ event, namely within (1/2/3) 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 50,844 ​miles? Why or why​ not? (Yes/No), because 50,844 (lies/does not lie) within the range of a usual​ event, namely within (1/2/3) standard deviations of the mean for an individual tire.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	51000
std deviation   =σ=	800.0000
sample size       =n=	100
std error=σx̅=σ/√n=	80.0000

probability =

P(X<50844)

=

P(Z<-1.95)=

0.0256

b)

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

c)

No because 50,844 lies within the range of a usual​ event, namely within 2 standard deviations of the mean for an individual tire.