Question

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?

Answer #1

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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