# Answer to Question #3945 in Statistics and Probability for susan

Question #3945

A new Firestone tire is guaranteed to last for 40,000 miles. The actual mean life of the tires is 47,000 miles with a standard deviation of 4,000 miles.

a) What is the probability that a tire will not last for 40,000 miles?

b) The Ford Motor Company uses Firestone Tires and is concerned about the performance of the tires. They conduct a study and find that out of the 120 tires they tested, the mean life was only 46,500 miles. Should the Ford Motor Company be concerned about Firestone's honesty about its tires? Provide a statistical justification for your response.

a) What is the probability that a tire will not last for 40,000 miles?

b) The Ford Motor Company uses Firestone Tires and is concerned about the performance of the tires. They conduct a study and find that out of the 120 tires they tested, the mean life was only 46,500 miles. Should the Ford Motor Company be concerned about Firestone's honesty about its tires? Provide a statistical justification for your response.

Expert's answer

a) Let X is random variable that response for tire life. X is normally distributed with mean 47000 and with standard deviation 4000. Consider a random variable Y, normally distributed with mean 0 and standard deviation 1. X=4000Y+47000

We need to find probability

P(X<40000)=P(4000Y+47000<40000)=P(Y<-1.75)=(using the table)=0.04

b) Ford company tested 120 tires, and released that sum of all tire lives were 46500*120. Consider a random variable Z, that is the sum of 120 tire lives. Each of addons is normally distributed, the sum of normally distributed values is also normally distributed with parameters 120*47000 and 120*4000. We need to find probability that P(Z<120*46500)=P(120*4000Y+47000*4000<120*46500)=P(Y<-1/8)=0.45

So, the probability that average tire life is less than 46500 is 0.45. So, it is very possibly that the Ford company made mistake.

We need to find probability

P(X<40000)=P(4000Y+47000<40000)=P(Y<-1.75)=(using the table)=0.04

b) Ford company tested 120 tires, and released that sum of all tire lives were 46500*120. Consider a random variable Z, that is the sum of 120 tire lives. Each of addons is normally distributed, the sum of normally distributed values is also normally distributed with parameters 120*47000 and 120*4000. We need to find probability that P(Z<120*46500)=P(120*4000Y+47000*4000<120*46500)=P(Y<-1/8)=0.45

So, the probability that average tire life is less than 46500 is 0.45. So, it is very possibly that the Ford company made mistake.

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