Question

Men’s heights in the USA are **normally
distributed** with a mean of 69 inches and a standard
deviation of 2.7 inches.

(a) What is the probability that a randomly selected man has a height of at least 68 inches?

(b) What height represents the 96^{th} percentile?

Answer #1

Part a)

X ~ N ( µ = 69 , σ = 2.7 )

P ( X ≥ 68 ) = 1 - P ( X < 68 )

Standardizing the value

Z = ( X - µ ) / σ

Z = ( 68 - 69 ) / 2.7

Z = -0.3704

P ( ( X - µ ) / σ ) > ( 68 - 69 ) / 2.7 )

P ( Z > -0.3704 )

P ( X ≥ 68 ) = 1 - P ( Z < -0.3704 )

P ( X ≥ 68 ) = 1 - 0.3555

P ( X ≥ 68 ) = 0.6445

Part b)

X ~ N ( µ = 69 , σ = 2.7 )

P ( X < x ) = 96% = 0.96

To find the value of x

Looking for the probability 0.96 in standard normal table to
calculate Z score = 1.7507

Z = ( X - µ ) / σ

1.7507 = ( X - 69 ) / 2.7

**X = 73.7269 ≈ 74 inches**

P ( X < 73.7269 ) = 0.96

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