/*! 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. 1.3 3. What percent of young women h... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

3. What percent of young women have heights between 62 and 72 inches? Show your work.

Short Answer

Expert verified

The percentage young women have heights between 62 and 72 inches are84%.

Step by step solution

01

Given Information

Heights of young women aged between=18to24.

Percentage of young women with heights greater than 67 inches=?

02

Explanation 

Given that,

μ=64.5

σ=2.5

P(62<x<72)=P62-64.52.5<x-μσ<72-64.52.5=P-2.52.5<z<7.52.5=P(-1<z<3)=P(z<3)-P(z<-1)

Substitute the given expression,

=0.9987-0.1587=0.84

We get,

=84%.

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 91Ó°ÊÓ!

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

T2.6. The figure shown is the density curve of a distribution. Five of the seven points marked on the density curve make up the five-number summary for this distribution. Which two points are not part of the five-number summary?

(a) Band E
(b) C and F
(c) C and E
(d) B and F
(e) A and G.

Teacher raises Refer to Exercise 20. If each teacher receives a 5% raise instead of a flat \(1000 raise, the amount of the raise will vary from \)1400 to $3000, depending on the present salary.

(a) What will this do to the mean salary? To the median salary? Explain your answers.

(b) Will a 5% raise increase the IQR? Will it increase the standard deviation? Explain your answers.

Measuring bone density Individuals with low bone density have a high risk of broken bones (fractures). Physicians who are concerned about low bone density (osteoporosis) in patients can refer them for specialized testing. Currently, the most common method for testing bone density is dual-energy X-ray absorptiometry (DEXA). A patient who undergoes a DEXA test usually gets bone density results in grams per square centimeter(g/cm2)and in standardized units. Judy, who is 25years old, has her bone density measured using DEXA. Her results indicate a bone density in the hip948g/cm2and a standardized score ofz=−1.45. In the reference population of For 25-year-old women like Judy, the mean bone density in the hip is 956g/cm2

(a) Judy has not taken a statistics class in a few years. Explain to her in simple language what the standardized score tells her about her bone density.

(b) Use the information provided to calculate the standard deviation of bone density in the reference population.

Trace the density curve onto your paper. Mark the approximate location of the median.

Questions T2.9 and T2.10 refer to the following setting. Until the scale was changed in 1995, SAT scores were based on a scale set many years ago. For Math scores, the mean under the old scale in the 1990swas 470and the standard deviation was 110. In 2009, the mean was 515and the standard deviation was 116 .
T2.10. Jane took the SAT in 1994and scored 500. Her sister Colleen took the SAT in 2009and scored 530. Who did better on the exam, and how can you tell?
(a) Colleen-she scored 30 points higher than Jane.
(b) Colleen-her standardized score is higher than Jane's.
(c) Jane-her standardized score is higher than Colleen's.
(d) Jane-the standard deviation was bigger in 2009.
(e) The two sisters did equally well-their z-scores are the same.

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.