Question

For a normal distribution with µ = 500 and  = 100, find the following values.

What X values form the boundaries for the middle 60% of the distribution?

Answer 1

Given that,

mean = = 500

standard deviation = =100

middle 60% of score is

P(-z < Z < z) = 0.60

P(Z < z) - P(Z < -z) = 0.60

2 P(Z < z) - 1 = 0.60

2 P(Z < z) = 1 + 0.60= 1.60

P(Z < z) = 1.60 / 2 = 0.8

P(Z <0.84 ) = 0.8

z  ± 0.84 (see the probability 0.80 in standard normal (Z) table corresponding value is 0.84 )

Using z-score formula  

x= z * +

x= -0.84*100+500

x= 416

z = 0.84

Using z-score formula  

x= z * +

x=0.84 *

x= 584