Question

A manufacturer claims that the life span of its tires is

48,000

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

one hundred

tires at random and test them. The mean life span is

47,858

miles. Assume

sigmaσequals=900

Complete parts​ (a) through​ (c).

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

47,858

miles or ​less?

0.0573 is what I got which was correct.

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

The claim is

inaccurate/accurate?

because the sample mean

would/would not?

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

47,858

​miles? Why or why​ not?

(1)_____, because 47,858 (2) ________ within the range of a usual event, namely within (3)______ of the mean for an individual tire.

(1) no/yes

(2) does not lie/lies

(3) 1 standard deviation/2 standard deviations/3 standard deviations

#31

Answer 1

(a)

μ = 48000, x-bar = 47858, n = 100, σ = 900

z = (x-bar - μ)/(σ/√n) = (47858 - 48000)/(900/√100) = -1.5778

P(x-bar < 47858) = P(z < -1.5778) = 0.0573

(b)

Would not be considered unusual since it lies within the range of a usual​ event, namely within 2 standard deviations of the mean of the sample means.

(c)

z = (x - μ)/σ = (47858 - 48000)/900 = -0.1578

P(x < 47858) = P(z < -0.1578) = 0.4373

No, because 47,858 lies within the range of a usual event, namely within 2 standard deviations of the mean for an individual tire.

[Please give me a Thumbs Up if you are satisfied with my answer. If you are not, please comment on it, so I can edit the answer. Thanks.]