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 $\geq \frac{75000 - 70000}{4400}$)
 = P(Z $\geq \frac{25}{22}$)
 ≈ P(Z $\geq$ 1.14)
 = 1 - P(Z < 1.14)
 ≈ 1 - 0.8729
 = 0.1271