Question

Tires have a normal distribution with a mean of 70000 miles and a standard deviation of 4400 miles. What proportion of tires will last at least 75000 miles.

Answer 1

Let T = the distance, in miles, a tire lasts

T ~ N(70000,4400²)

P(T
`\geq` 75000)
= P(Z
`\geq (75000 - 70000)/(4400)`)
= P(Z
`\geq (25)/(22)`)
≈ P(Z
`\geq` 1.14)
= 1 - P(Z < 1.14)
≈ 1 - 0.8729
= 0.1271