Answer to Question #232832 in Statistics and Probability for qaiser

Question

 A tire manufacturer states that a certain type of tire has a mean lifetime of 60,000 miles. Suppose lifetimes are normally distributed with standard deviation σ=3,500 miles. (10 marks)

a) Find the probability that if you buy one such tire, it will last only 57,000 or fewer miles. If you had this experience, is it particularly strong evidence that the tire is not as good as claimed?

b) A consumer group buys five such tires and tests them. Find the probability that average lifetime of the five tires will be 57,000 miles or less. If the mean is so low, is that particularly strong evidence that the tire is not as good as claimed?

Answer 1

a)

If X~N($\mu , \sigma)-$ given distribution then

X₁=$\frac{X-\mu }{\sigma}=\frac{X-60000}{3500}$ ~N(0,1)-standard normal distribution

And probability P(X$\le 57000)=$$=P(X_1\le \frac{57000-60000}{3500})= P(X_1\le -0.857)=$

=pnorm(-0,857,0,1)=0.196 value of standard normal distribution at point -0.857. where Mathcad soft was applied.

This ptobability 19.6% is not enough small to reject hypothesis $\mu=60000$ if situation with X<57000 really happened for we need

p -value $\le 5$ %

b) If Y=$\frac{X^{(1)}+X^{(2)}+X^{(3)}+X^{4}+X^{(5)}}{5}$ - mean of five independent realizations of random value X, than Y also has normal distribution with the same mean M(Y)=M(X)=60000 but Y has more less standard devitation $\sigma(Y)=\frac {\sigma(X)}{\sqrt{5}}$ . Thus we can write:

Y~N$(60000,\frac{3500}{\sqrt{5}})$ =N(60000,1565) and probabily P(Y$\le$ 57000) is equal to P(N(0,1)$\le \frac{-3000}{1565}$ )=P(N(0,1)$\le$ -1.917). For calculating last probability we use program MathCad and function pnorm(p.$\mu,]sigma)$, by using it we have pnorm(-1.917,0,1)=-0.028=2.8% therefore if event Y$\le$ 57000 has happened this sitution has probability 2.8 % and with reliability 97.2 % we can say that hypothesis $\mu=60000$ is wrong and tire's life time less than 60000 miles because p-value 2.8% smaller than standard statistical level 5%.