/*! 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 17. Measuring bone density Individua... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Measuring bone density Individuals with low bone density (osteoporosis) have a high risk of broken bones (fractures). Physicians who are concerned about low bone density in patients can refer them for specialized testing. Currently, the most common method for testing bone density is dual-energy X-ray absorptiometry (DEXA). The bone density results for a patient who undergoes a DEXA test usually are reported 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 bone density in the hip of 948 g/cm2 and a standardized score of z=−1.45The mean bone density in the hip is 956 g/cm2in the reference population of 25-year-old women like Judy.

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

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

Short Answer

Expert verified

Part (a) Judy's bone density of 948g/cm2 is 1.45 standard deviations below the mean bone density of 25-year-old women.

Part (b) Standard deviation,σ=5.52

Step by step solution

01

Part (a) Step 1: Given information

Value, x=948

Mean,μ=956

z − score, z=−1.45

02

Part (a) Step 2: Concept

The formula used:z=(x-μ)σ

03

Part (a) Step 3: Explanation

The amount of standard deviations a value deviates from the mean is described by the z−score.

The value is below the mean when the z−score is negative.

Whereas,

Positive z−implies that the value is greater than the mean.

Thus,

Judy's bone density of 948g/cm2 is 1.45 standard deviations below the mean bone density of 25-year-old women, according to the standardized z- score.

04

Part (b) Step 1: Calculation

Calculate the z− score:

z=x−μσ

Substitute values,

−1.45=948−956σ

That becomes

σ=−8−1.45

That further becomes

σ=81.45≈5.52

Because the standard deviation and data values have the same units.

Thus,

The standard deviation is approx. 5.52g/cm2

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

Potato chips The weights of 9-ounce bags of a particular brand of potato

chips can be modeled by a Normal distribution with mean μ=9.12ounces and standard deviation σ=0.05ounce. Sketch the Normal density curve. Label the mean and the points that are 1,2, and 3 standard deviations from the mean.

Making money (2.1) The parallel dot plots show the total family income of randomly chosen individuals from Indiana (38 individuals) and New Jersey (44 individuals). Means and standard deviations are given below the dot plots.

Consider individuals in each state with total family incomes of $95,000 Which individual has a higher income, relative to others in his or her state? Use percentiles and z-scores to support your answer.

Put a lid on it! Refer to Exercise 56. The supplier is considering two changes to reduce the percent of its large-cup lids that are too small to less than 1%. One strategy is to adjust the mean diameter of its lids. Another option is to alter the production process, thereby decreasing the standard deviation of the lid diameters.

(a) If the standard deviation remains at σ=0.02inches, at what value should the supplier set the mean diameter of its large-cup lids to ensure that less than 1%are too small to fit? Show your method.

(b) If the mean diameter stays at μ=3.98inches, what value of the standard deviation will result in less than 1%of lids that are too small to fit? Show your method.

(c) Which of the two options in (a) and (b) do you think is preferable? Justify your answer. (Be sure to consider the effect of these changes on the percent of lids that are too large to fit.)

Flight times An airline flies the same route at the same time each day. The flight time varies according to a Normal distribution with unknown mean and standard deviation. On 15% of days, the flight takes more than an hour. On 3% of days, the flight lasts 75 minutes or more. Use this information to determine the mean and standard deviation of the flight time distribution.

Long jump A member of a track team was practicing the long jump and

recorded the distances (in centimeters) shown in the dot plot. Some numerical summaries of the data are also provided.

After chatting with a teammate, the jumper realized that he measured his jumps from the back of the board instead of the front. Thus, he had to subtract 20 centimeters from each of his jumps to get the correct measurement for each jump.

a. What shape would the distribution of corrected long jump distance have?

b. Find the mean and median of the distribution of corrected long-jump distance.

c. Find the standard deviation and interquartile range (IQR) of the distribution of corrected long-jump distance.

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.