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Bone Density Test A bone mineral density test is used to

Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola ISBN: 9780321836960 18

Solution for problem 1RE Chapter 6.7

Elementary Statistics | 12th Edition

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Elementary Statistics | 12th Edition | ISBN: 9780321836960 | Authors: Mario F. Triola

Elementary Statistics | 12th Edition

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Problem 1RE

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

a. For a randomly selected subject, find the probability of a bone density test score less than 2.93.

b. For a randomly selected subject, find the probability of a bone density test score greater than –1.53.

c. For a randomly selected subject, find the probability of a bone density test score between –1.07 and 2.07.

d. Find P30, the bone density test score separating the bottom 30% from the top 70%.

e. If the mean bone density test score is found for 16 randomly selected subjects, find the probability that the mean is greater than 0.27.

Step-by-Step Solution:

Answer:

Step 1 of 4</p>

The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.

a. For a randomly selected subject, the probability of a bone density test score less than 2.93.

   

We are dealing with individual value from a normally distributed population so mean is 0 and standard deviation is 1.

The Test Statistic can be expressed as

     

     

       Z  = 2.93

To find the associated probability, we have

P()

P(Z 2.93) = 0.9983 (From Area Under Normal Curve table)

Step 2 of 4</p>

b. For a randomly selected subject, the probability of a bone density test score greater than –1.53.

We are dealing with individual value from a normally distributed population so mean is 0 and standard deviation is 1.

The Test Statistic can be expressed as

     

     

       Z  = -1.53

To find the associated probability, we have

P()

P(Z -1.53) = 1 - P(Z -1.53)

                     = 1 - 0.0630 (From Area Under Normal Curve table)

                     = 0.937

Step 3 of 4

Chapter 6.7, Problem 1RE is Solved
Step 4 of 4

Textbook: Elementary Statistics
Edition: 12
Author: Mario F. Triola
ISBN: 9780321836960

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