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Chapter 6: Q2 (page 291)

Bone Density Test.

In Exercises 1鈥�4, assume that scores on a bone mineral density test are normally distributed with a mean of 0 and a standard deviation of 1.

Bone Density Find the score separating the lowest 9% of scores from the highest 91%.

Short Answer

Expert verified

The score separating the lowest 9% from the highest 91% is 鈥�1.34.

Step by step solution

01

Given information

The bone mineral density test scores are normally distributed with mean value of 0 and standard deviation of 1.

02

Describe the random variable

Let Z be the random variable for bone mineral density test scores.

Z~N渭,蟽2~N0,12

03

Describe the required score

Let z be the z-score such that it separates lowest 9% from the highest 91% z-scores.

Thus, PZ<z=0.09orPZ>z=0.91.

Thus, the cumulative area to the left of z is 0.09 as area and probability have one-to-one correspondence.

04

Obtain the z-score from the standard normal table

Using the standard normal table, the cumulative probability of 0.09 is closest to 0.0901, that is obtained at the intersection of row 鈥�1.3 and column 0.04.

Thus, the z-score with cumulative area of 0.09 is 鈥�1.34.

Therefore, 鈥�1.34 separated the lowest 9% scores from the highest 91% scores.

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Most popular questions from this chapter

Standard Normal DistributionIn Exercises 17鈥�36, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers

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Standard normal distribution, assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, Draw a graph, then find the probability of the given bone density test score. If using technology instead of Table A-2, round answers to four decimal places.

Greater than -3.75
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