Question

The lifetime of a certain automobile tire is normally distributed with mean 40 (thousand miles) and standard deviation 5 ( thousand miles). if 8 tires are randomly selected, there would be an 83% probability that the mean tire lifetime would fall between what two values.

Answer 1

Given that,

mean = = 40

standard deviation = = 5

middle 83% of score is

P(-z < Z < z) = 0.83

P(Z < z) - P(Z < -z) = 0.83

2 P(Z < z) - 1 = 0.83

2 P(Z < z) = 1 + 0.83 = 1.83

P(Z < z) = 1.83/ 2 = 0.915

P(Z <1.37 ) = 0.915

z  ±1.37 using z table

Using z-score formula  

x= z * +

x=  ±1.37*5+40

x= 33.15 , 46.85