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?
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
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