Question

A survey found that women's heights are normally distributed with mean 62.3 in. and standard deviation 2.5 in. The survey also found that men's heights are normally distributed with mean 67.4 in. and standard deviation 3.7 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 62 in. Complete parts (a) and (b) below. a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park? b. If the height requirements are changed to exclude only the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

          A survey found that women's heights are normally
distributed with mean 62.3 in. and standard deviation 2.5 in. The
survey also found that men's heights are normally distributed
with mean 67.4 in. and standard deviation 3.7 in. Most of the live
characters employed at an amusement park have height requirements
of a minimum of 57 in. and a maximum of 62 in. Complete
parts (a) and (b) below.
a. Find the percentage of men meeting the height requirement.
What does the result suggest about the genders of the people who
are employed as characters at the amusement park?
b. If the height requirements are changed to exclude only the
tallest 50% of men and the shortest 5% of men,
what are the new height requirements?

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Added by Jose D.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Joanna Quigley

03/15/2023

Step 1

and 62 in. We can use the standard normal distribution to do this: z = (x - μ) / σ where z is the standard score, x is the height requirement (either 57 in. or 62 in.), μ is the mean height of men (67.4 in.), and σ is the standard deviation of men's heights (3.7 Show more…

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A survey found that women's heights are normally distributed with mean 62.3 in. and standard deviation 2.5 in. The survey also found that men's heights are normally distributed with mean 67.4 in. and standard deviation 3.7 in. Most of the live characters employed at an amusement park have height requirements of a minimum of 57 in. and a maximum of 62 in. Complete parts (a) and (b) below. a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as characters at the amusement park? b. If the height requirements are changed to exclude only the tallest 50% of men and the shortest 5% of men, what are the new height requirements?

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00:01 In this question, we are looking at a normal distribution, so i'll start by drawing it.

00:08 The total area under this curve is 1 or 100%, and it is symmetric, so half will be below the mean, half above the mean.

00:18 For part a, we want to know what percentage of men meet height requirements to play characters at an amusement park.

00:25 So we're looking at the distribution of male heights.

00:28 M is 67 .4, sigma, a standard deviation, is 3 .6 .6 .6 .6 .6.

00:34 We're looking at between 57 and 62.

00:38 Both of those values are below the mean.

00:40 So i'm mark from on here.

00:41 We want to know what percentage of men fall into this interval.

00:46 So the first thing we do is find the corresponding z scores.

00:50 Z is x minus mu over sigma.

00:56 So we want 57 minus 67 .4, so that's minus 10 .4, divided by 3 .7, and 62 minus 67 .4.

01:06 And it's 5 .4, divided by 3 .7.

01:10 This tells us how many standard deviations away from the mean these values are.

01:15 And the first one is about minus 1 .46, the second about minus 2 .81.

01:21 But i want to use the exact values to maintain accuracy.

01:24 My next step is to turn these z scores into areas under the curve.

01:29 And i have two tools at my disposal.

01:31 They are the standard normal and the cumulative normal functions.

01:35 Standard normal gives you the area between x and mu.

01:42 So if i put in this one, it would give me this area here in red.

01:47 If i put in this one, it would give me red plus green.

01:51 Red plus green minus red leaves me my interval.

01:55 Cumulative is the area to the left.

01:59 So very similar, except the more negative score would give you red and the less negative would give you red plus green.

02:06 I'll use for standard.

02:07 So for red plus green, i'm putting this minus 10 .4 over 3 .7 into the standard normal function on my calculator.

02:23 You can also find it on excel or r.

02:25 I get 0 .4 .9753...

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