Grear Tire Company has produced a new tire with an estimated mean lifetime milea

Question

Grear Tire Company has produced a new tire with an estimated mean lifetime mileage of 36,500 miles. Management also believes that the standard deviation is 5000 miles and that tire mileage is normally distributed. To promote the new tire, Grear has offered to refund some money if the tire fails to reach 30,000 miles before the tire needs to be replaced. Specifically, for tires with a lifetime below 30,000 miles, Grear will refund a customer $1 per 100 miles short of 30,000.

(a) For each tire sold, what is the expected cost of the promotion? If required, round your answer to two decimal places. (b) What is the probability that Grear will refund more than $50 for a tire? If required, round your answer to three decimal places. (c) What mileage should Grear set the promotion claim if it wants the expected cost to be $2.00? If required, round your answer to the hundreds place.

Explanation / Answer

Here the Mean life mileage µ = 36500 and Standard deviation = 5000, x = 30000

It follows the normal distribution. So, Z = x - µ / = (30000-36500) / 5000 = - 6500/5000 = - 1.3

So, first we need to find the Probability P ( Z < -1.3 ) = P ( Z >1.3 ) = 0.5 - P ( 0<Z < 1.3 ) = 0.5 - 0.4032 = 0.0968

So probabilty that it will not reach 30000 is = 0.0968,

So f = N*P(Z) = 30000*0.0968 = 2904

As $1 per 100 miles, the cost of the promotion per tire = $ 2904/100 = $29.04

(b) $50 means the number of miles failed reach 30000 = 50*100 = 5000

So P( Z < (25000 - 36500)/5000 ) = P( Z <- 2.2) = P( Z > 2.2) = 0.5 - P( 0<Z < 2.2) = 0.5 - 0.4861 = 0.0139

(c) As $1 per 100 miles, it should fix the limit as 28000 for cost of promotion to $2.

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