Question

1.   The lifetime of a tire can be represented by a normal distribution with an average of 35,000 kilometers and a standard deviation of 4,000 kilometers.

1. Find the probability that the time of a tire is more than 30,000 kilometers.
2. Find the value of K such that the probability that the life time of a tire is less than K is 0.95.
3. A sample of 80,000 was taken from these tires. Approximately how many have a life time greater than 38,000 kilometers?
4. A sample of 5 of their tires was taken. What is the (approximate) probability between 1 (inclusive) and 4 (not inclusive) of them having a life time of less than 40,000 kilometers?

Answer 1

1. z = (x - µ)/σ = (30000 - 35000)/4000 = -1.25

P (z > -1.25) = 0.89435

2. z = 1.645

1.645 = (x- 35000)/4000

x = 1.645*4000 + 35000 = 41580

3. z = (x - µ)/σ = (38000 - 35000)/4000 = 0.75

P (z > 0.75) = 0.226627

n = 0.226627*80000 = 18131

4. z = (x - µ)/σ = (40000 - 35000)/4000 = 1.25

P (z < 1.25) = 0.89435

z = (x - µ)/σ/√n = (40000 - 35000)/4000/√5 = 2.80

P (z < 2.80) = 0.997406

Probability = 0.997406 - 0.89435 = 0.103055

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