Question

A manufacturer claims that the life span of its tires is 52 comma 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 51 comma 729 miles. Assume sigmaequals900. 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 729 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 ▼ 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 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 729 ​miles? Why or why​ not? ▼ Yes No ​, because 51 comma 729 ▼ 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 for an individual tire.

Answer 1

(a)

= Population mean = 52000

= Population SD = 900

n = Sample size = 100

= Sample mean = 51729

SE = /

= 900/ = 90

To find P( 51729):

Z = ( - )/SE

= (51729 - 52000/90 = - 3.0111

Table gives area = 0.4987

So,

P( < 51729) = 0.5 - 0.4987 = 0.0013

(b)

Correct option:

The claim is inaccurate because the sample mean would be considered unusual since it does not lie within the range of a usual event, namely within 3 standard cdeviations of the mean.

(c)

Z = (X - )/

= (51729 - 52000)/900 = - 0.3011

No, because 51,729 lies within the range of a usual event, namely within 3 standard deviations of the mean for an invidual tire.