Question

A manufacturer claims that the life span of its tires is

52 comma 00052,000

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

51 comma 79951,799

miles. Assume

sigma?equals=700700.

Complete parts? (a) through? (c).

?(a) Assuming the? manufacturer's claim is? correct, what is the probability that the mean of the sample is

51 comma 79951,799

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

?

because the sample mean

?

be considered unusual since it

?

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

51 comma 79951,799

?miles? Why or why? not?

?

No

Yes

?, because

51 comma 79951,799

?

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.

Answer 1

a)probability that the mean of the sample is less than 51799=P(Xbar<51799)=P(Z<(51799-52000)/(700/sqrt(100))

=P(Z<-2.87)=0.0021 ( please try 0.0020 if this comes wrong)

b)The claim is wrong;because the sample mean can be be considered unusual since it

does not lie within the range of a usualnamely within 2 standard deviationsof the mean of the sample means.

c)

No because 51,799 lie within the range of a usual vent, namely within 2 standard deviations of the mean for an individual tire.