/*! 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} Q. 107 - Exercises BMI (2.2, 5.2, 5.3) Your body ma... [FREE SOLUTION] | 91影视

91影视

Chapter 5: Q. 107 - Exercises (page 296)

BMI (2.2, 5.2, 5.3) Your body mass index (BMI) is your weight in kilograms divided by

the square of your height in meters. Online BMI calculators allow you to enter weight in

pounds and height in inches. High BMI is a common but controversial indicator of being

overweight or obese. A study by the National Center for Health Statistics found that the

BMI of American young women (ages 20 to 29) is approximately Normally distributed

with mean 26.8 and standard deviation 7.4.

27

a. People with BMI less than 18.5 are often classed as 鈥渦nderweight.鈥 What percent of

young women are underweight by this criterion?

b. Suppose we select two American young women in this age group at random. Find the

probability that at least one of them is classified as underweight.

Short Answer

Expert verified

a.percent of young women that are underweight by this criterion is 13.101

b,.the probability that at least one of them is classifiied as underweight is0.244

Step by step solution

01

Part (a) Step 1:Given Information

We have been given

mean ()=26.8

standard deviation() =7.4

People with BMI less than18.5 are often classed as 鈥渦nderweight.鈥

02

Part (a) Step 2:Simplification

For normal distribution we will calculate z value

z=-

here,x=18.5

so,z=18.5-26.87.4

we need to find p(<18.5) as this will be considered as underweight

so,p(<18.5)=p(z<-)

p(<18.5)=p(z<18.5-26.87.4)

p(<18.5)=p(z<-1.121)

p(<18.5)=0.1310

percent of young women that are underweight by this criterion is 0.1310*100=13.10

03

Part (b) Step 1:Given Information 

Number of american young women selected randomly (n) =2

04

Part (b) Step 2:Simplification

probability that a person is underweight is (p) = 0.1310

probability that a person is not underweight is (q) = 1-p = 1-0.1310=0.869

probability that at least one of them is classified as underweight is :p(1)=1-p(<1)=1-p(0)

p(1)=1-0.8692

p(1)=0.244

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

Waiting to park Do drivers take longer to leave their parking spaces when

someone is waiting? Researchers hung out in a parking lot and collected some data. The

graphs and numerical summaries display information about how long it took drivers to

exit their spaces.

a. Write a few sentences comparing these distributions.

b. Can we conclude that having someone waiting causes drivers to leave their spaces more

slowly? Why or why not?

Smartphone addiction? A media report claims that 50%of U.S. teens with smartphones feel addicted to their devices. A skeptical researcher believes that this figure is too high. She decides to test the claim by taking a random sample of 100U.S. teens who have smartphones. Only 40of the teens in the sample feel addicted to their devices. Does this result give convincing evidence that the media report鈥檚 50%claim is too high? To find out, we want to perform a simulation to estimate the probability of getting 40or fewer teens who feel addicted to their devices in a random sample of size 100from a very large population of teens with smartphones in which 50% feel addicted to their devices.

Let 1= feels addicted and 2= doesn鈥檛 feel addicted. Use a random number generator to produce 100random integers from 1to 2. Record the number of 1鈥檚 in the simulated random sample. Repeat this process many, many times. Find the percent of trials on which the number of 1鈥檚 was40 or less.

Will Luke pass the quiz ? Luke鈥檚 teacher has assigned each student in his class an online quiz, which is made up of 10multiple-choice questions with 4options each. Luke hasn鈥檛 been paying attention in class and has to guess on each question. However, his teacher allows each student to take the quiz three times and will record the highest of the three scores. A passing score is 6or more correct out of 10. We want to perform a simulation to estimate the score that Luke will earn on the quiz if he guesses at random on all the questions.

a. Describe how to use a random number generator to perform one trial of the simulation. The dotplot shows Luke鈥檚 simulated quiz score in 50trials of the simulation.

b. Explain what the dot at 1represents.

c. Use the results of the simulation to estimate the probability that Luke passes the quiz.

d. Doug is in the same class and claims to understand some of the material. If he scored 8points on the quiz, is there convincing evidence that he understands some of the material? Explain your answer.

Teachers and college degrees Select an adult at random. Define events D: person has earned a college degree, and T: person鈥檚 career is teaching. Rank the following probabilities from smallest to largest. Justify your answer.

P(D)P(T)P(DT)P(TD)

Income tax returns Here is the distribution of the adjusted gross income (in thousands of dollars) reported on individual federal income tax returns in a recent year:

a. What is the probability that a randomly chosen return shows an adjusted gross income of \(50,000 or more?

b. Given that a return shows an income of at least \)50,000, what is the conditional

probability that the income is at least $100,000?

See all solutions

Recommended explanations on Math Textbooks

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.