Question

The age of a store manager follows a normal distribution with a mean age of 60 years and a standard deviation of 5 years. What values represent the middle 60% of ages of store managers?

Answer #1

Given that,

mean = = 60

standard deviation = = 5

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

Using z-score formula

x= ± z * +

x= ± 0.84 *5+60

x= 55.8 , 64.2

I'm not sure if these are able to be answered in excel, but so I
have the formula?
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