/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Heights of Men in the US Heights... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 35

Heights of Men in the US Heights of adult males in the US are approximately normally distributed with mean 70 inches \((5 \mathrm{ft} 10 \mathrm{in})\) and standard deviation 3 inches. (a) What proportion of US men are between \(5 \mathrm{ft}\) 8 in and \(6 \mathrm{ft}\) tall \((68\) and 72 inches, respectively)? (b) If a man is at the 10 th percentile in height, how tall is he?

Short Answer

Expert verified
(a) Approximately 49.72% of US men are between 5 ft 8 in and 6 ft tall. (b) A man at the 10th percentile in height is approximately 5 ft 6.16 in tall.

Step by step solution

01

Convert Heights into Z-Scores for Part (a)

First, convert the heights to Z-scores using the formula \(Z = (X - 渭) / 蟽\), where \(X\) is the height, \(渭\) is the mean and \(蟽\) is the standard deviation. This transforms the height into standard units of deviation from the mean. Thus, for the lower bound (68 in), \( Z_1 = (68 - 70) / 3 = -0.67\), and for the upper bound (72 in), \(Z_2 = (72 - 70) / 3 = 0.67\).
02

Calculate Proportion for Part (a)

Next, look up the Z-scores in a standard normal distribution table (or use a calculator function) to find the area to the left of each point under the curve, which represents the proportion of men. For \(Z_1 = -0.67\), the table gives 0.2514 and for \(Z_2 = 0.67\), the table yields 0.7486. To find the proportion between these two Z-scores, subtract the area of \(Z_1\) from that of \(Z_2\). Hence, the proportion is 0.7486 - 0.2514 = 0.4972. So approximately 49.72% of men are between 5 ft 8 in and 6 ft tall.
03

Convert the Percentile into a Z-Score for Part (b)

The Z-score for the 10th percentile is -1.28 (from a standard normal distribution table or calculator function that produces percentile ranks). This follows from the fact that 10% of the area under the standard normal curve lies to the left of this Z-score.
04

Calculate Height for the 10th Percentile

Finally, convert the Z-score back to height using our original mean and standard deviation, with the formula \(X = Z * 蟽 + 渭\). Hence, the height at the 10th percentile is -1.28 * 3 + 70 = 66.16 inches or approximately 5 ft 6.16 in.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Exercises 5.44 to 5.49 , find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(95 \%\) confidence interval for a difference in proportions \(p_{1}-p_{2}\) if the samples have \(n_{1}=50\) with \(\hat{p}_{1}=0.68\) and \(n_{2}=80\) with \(\hat{p}_{2}=0.61,\) and the standard error is \(S E=0.085\)

In Exercises 5.56 and \(5.57,\) find the p-value based on a standard normal distribution for each of the following standardized test statistics. (a) \(z=0.84\) for an upper tail test for a difference in two proportions (b) \(z=-2.38\) for a lower tail test for a difference in two means (c) \(z=2.25\) for a two-tailed test for a correlation

(a) The area to the left of the endpoint on a \(N(5,2)\) curve is about 0.10 . (b) The area to the right of the endpoint on a \(N(500,25)\) curve is about 0.05

Smoke-Free Legislation and Asthma Hospital admissions for asthma in children younger than 15 years was studied \(^{11}\) in Scotland both before and after comprehensive smoke-free legislation was passed in March \(2006 .\) Monthly records were kept of the annualized percent change in asthma admissions, both before and after the legislation was passed. For the sample studied, before the legislation, admissions for asthma were increasing at a mean rate of 5.2\% per year. The standard error for this estimate is \(0.7 \%\) per year. After the legislation, admissions were decreasing at a mean rate of \(18.2 \%\) per year, with a standard error for this mean of \(1.79 \% . \mathrm{In}\) both cases, the sample size is large enough to use a normal distribution. (a) Find and interpret a \(95 \%\) confidence interval for the mean annual percent rate of change in childhood asthma hospital admissions in Scotland before the smoke-free legislation. (b) Find a \(95 \%\) confidence interval for the same quantity after the legislation. (c) Is this an experiment or an observational study? (d) The evidence is quite compelling. Can we conclude cause and effect?

Exercises 5.21 to 5.28 ask you to convert an area from one normal distribution to an equivalent area for a different normal distribution. Draw sketches of both normal distributions, find and label the endpoints, and shade the regions on both curves. The area above 13.4 for a \(N(10,2)\) distribution converted to a standard normal distribution,
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.