Question

A driver’s reaction time to brake lights on a decelerating vehicle can be modeled with a normal distribution where the mean is 1.25 seconds and the standard deviation is 0.4 seconds. (e) About 68% of the reaction times lie between what two values (use the original parameters)?

Answer 1

solution

Given that,

mean = = 1.25

standard deviation = =0.4

middle 68% of score is

P(-z < Z < z) = 0.68

P(Z < z) - P(Z < -z) = 0.68

2 P(Z < z) - 1 = 0.68

2 P(Z < z) = 1 + 0.68 = 1.68

P(Z < z) = 1.68 / 2 = 0.84

P(Z < 0.99) = 0.84

z  ± 0.99 using z table

Using z-score formula  

x= z * +

x= ± 0.99 *0.4+1.25

x= 0.854 ,1.646

68% of the reaction times lie between 0.854 to 1.646