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Suppose the heights (in inches) of men ages 20-29) ) in the United States are normally distributed with a mean of 693 inches and a standard deviation of 2.92 inches

User Mushroom
by
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2 Answers

6 votes

Answer:

(a) 0.13567

(b) 0.68023

Complete Problem Statement:

Suppose the heights (in inches) of men ages (20-29) in the United States are normally distributed with a mean of 69.3 inches and a standard deviation of 2.92 inches. A study participant is randomly selected

Find the probability that his height is :

(a) Less than 66 inches

(b) Between 66 inches and 72 inches

Explanation:

To find the probability, we will need to calculate the Z-Score.

The formula for which is :

z = (X-μ) / σ

X= raw score

μ= mean

σ = Standard Deviation

(a) Probability that the height of participant is less than 66 inches:

First we find value of z :

z =
(66-69.3)/(2.92) = -1.1

Use this to look up at probability table, we get:

Probability (at z equals -1.1 ) = 0.13567

(b) Probability that the height of participant is between 66 inches and 72 inches

First we find value of z for X=72 inches :

z =
(72-69.3)/(2.92) = 0.9

Probability ( at z equals 0.9) = 0.8159

We already have value of z for X=66 inches , which is z = -1.1 with Probability (at z equals -1.1 ) = 0.13567

Since we are asked for probability between 66 inches and 72 inches, we subtract the two probabilities , such that:

Probability (at h=72) - Probability (at h=66) = 0.8159-0.13567=0.68023

Probability that height of participant is between 66 inches and 72 inches =0.68023

Suppose the heights (in inches) of men ages 20-29) ) in the United States are normally-example-1
Suppose the heights (in inches) of men ages 20-29) ) in the United States are normally-example-2
User Rpitting
by
8.9k points
7 votes

Answer: For a 90th percentile, X = 696.74 inches

Explanation:

X = u + Zα

where X is the required percentile

α is the standard deviation of the

variable X

u is the mean

Z is the value corresponding to X.

from the normal distributiontable

Say for the 90th percentile of men's height, using Z = 1.282 from the standard normal distribution table;

X = 693 + 1.282(2.92) = 696.74 inches

Interpretation: 90% of men (ages 20-29) in the U.S. have a height below 696.74 inches while the remaining 10% have a height above 696.74 inches.

User KlausCPH
by
8.6k points
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