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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.

Sagot :

superkoopasirf
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
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