Question

The heights (measured in inches) of men aged 20 to 29 follow
approximately the normal distribution with mean 69.5 and standard
deviation 2.9. Between what two values does the middle 91% of all
heights fall? (Please give responses to at least one decimal
place)

Answer #1

Given that,

mean = = 69.5

standard deviation = = 2.9

middle 91% of score is

P(-z < Z < z) = 0.91

P(Z < z) - P(Z < -z) = 0.91

2 P(Z < z) - 1 = 0.91

2 P(Z < z) = 1 + 0.91 = 1.91

P(Z < z) = 1.91 / 2 = 0.955

P(Z <1.70 ) =0.955

z ± 1.70 using standard normal (Z) table

Using z-score formula

x= z * +

x= - 1.70*2.9+69.5

x= 64.6

z = 1.70

Using z-score formula

x= z * +

x= 1.70*2.9+69.5

x= 74.4

answer=64.6 and 74.4

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