Let T = the distance, in miles, a tire lasts
T ~ N(70000,4400²)
P(T [tex] \geq [/tex] 75000)
= P(Z [tex] \geq \frac{75000 - 70000}{4400} [/tex])
= P(Z [tex] \geq \frac{25}{22}[/tex])
≈ P(Z [tex] \geq [/tex] 1.14)
= 1 - P(Z < 1.14)
≈ 1 - 0.8729
= 0.1271