Question

19. Lifetimes of a certain brand of tires are approximately normally distributed with mean 40,000 miles and standard deviation 2,500 miles. What is the probability that the tires last less than 34,000 miles?

20. Lifetimes of a certain brand of tires are approximately normally distributed with mean 40,000 miles and standard deviation 2,500 miles. If the company making the tires did not want to replace more than 3% of the tires, what is the lowest mileage the company should make for the warranty?

Answer #1

Solution :

Given that ,

19.

mean = = 40000

standard deviation = = 2500

P(x < 34000) = P[(x - ) / < (34000 - 40000) / 2500]

= P(z < -2.4)

= 0.0082

Probability = **00082**

20.

Using standard normal table,

P(Z > z) = 3%

1 - P(Z < z) = 0.03

P(Z < z) = 1 - 0.0.03

P(Z < 1.88) = 0.97

z = 1.88

Using z-score formula,

x = z * +

x = 1.88 * 2500 + 40000 = 44700

The lowest mileage the company should make for the warranty is
**44700**

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