The distribution of adult men’s heights is approximately normally distributed with mean 175.0 cm and standard deviation 6.40 cm. State answers rounded to one place of decimal. [4] a) Approximately what percentage of men are taller than 183 cm? b) What height is such that approximately 90% of men are taller than this height?
Given,
= 175, = 6.4
We convert this to standard normal as
P(X < x) = P( Z < x - / )
a)
P( X > 183) = P( Z > 183 - 175 / 6.4)
= P( Z > 1.25)
= 1 - P( Z < 1.25)
= 1 - 0.8944
= 0.1056
= 10.6%
b)
We have to calculate x such that
P( X > x) = 0.90
That is P( X < x) = 0.10
P( Z < x - / ) = 0.10
From the Z table, z-score for the probability of 0.10 is -1.2816
x - / = -1.2816
Put the values of and in above equation and solve for x
x - 175 / 6.4 = -1.2816
x = 166.8
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