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Chapter 6: Q2 (page 291)

Bone Density Test.

In Exercises 1–4, 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 –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~Nμ,σ2~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 –1.3 and column 0.04.

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

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

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