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Crank
3 years ago
6

Grear Tire Company has produced a new tire with an estimated mean lifetime mileage of 36,500 miles. Management also believes tha

t the standard deviation is 5000 miles and that tire mileage is normally distributed. To promote the new tire, Grear has offered to refund a portion of the purchase price 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?
b. What is the probability that Grear will refund more than $50 for a tire?
c. What mileage should Grear set the promotion claim if it wants the expected cost to be $2
Business
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

$0.013

0.010724

Explanation:

Given that :

Mean, m = 36500

Standard deviation, s = 5000

Refund of $1 per 100 mile short of 30,000 miles

A.) Expected cost of the promotion :

P(X < 30,000)

Using the Zscore relation :

Zscore = (x - m) / s

Zscore = (30000 - 36500) / 5000

= - 6500 / 5000

= - 1.3

100 miles = $1

1.3 / 100 = $0.013

b. What is the probability that Grear will refund more than $50 for a tire?

100 miles = $1

$50 = (100 * 50) = 5000 miles

Hence, more than $50 means x < (30000 - 5000) = x < 25000 miles

P(x < 25000) :

(25000 - 36500) / 5000

-11500 / 5000

= - 2.3

P(z < - 2.3) = 0.010724 (Z probability calculator)

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