Question

Statistics: The mean height of adult males is 72 inches, with a standard deviation of 2.5 inches. Consider the heights to follow a normal distribution.

A. If a man is 71 inches tall, in what percentile does he lie?

B. What percentage of men will be between 71 and 75 inches tall?

C. How tall must a man be to be in the 98th percentile?


Some help would be sincerely appreciated, this is my 3rd all nighter in a row, posting again because I got no real answers.

Answer 1

Using the normal distribution principle, the results obtained for each of the questions are :

• 34.5th percentile
• 54%
• 77.135 inches

Given the Parameters :

• Mean, μ= 72
• Standard deviation, σ= 2.5

Zscore = (X - μ) ÷ σ

A.)

X = 71 inches

Zscore = (71 - 72) ÷ 2.5 = - 0.4

P(Z < - 0.4) = 0.3445 = (0.3445 × 100%) = 34.5th percentile

B.)

Between 71 and 75 inches tall :

For X = 75 Inches :

Zscore = (75 - 72) ÷ 2.5 = 1.2

P(Z < 1.2) = 0.8849 = (0.8849 × 100%) = 88.5th percentile

P(Z < 1.2) - P(Z < - 0.4) = 88.5 - 34.5 = 54%

C)

Using the normal distribution table :

98% has a Zscore of 2.054

Using the Zscore formula :

2.054 = (X - 72) / 2.5

2.054 × 2.5 = X - 72

5.135 = X - 72

X = 5.135 + 72

X = 77.135

Therefore, the man must be 77.135 inches tall to be in the 98 percentile

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