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%.

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

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

\(\begin{aligned}{c}Z \sim N\left( {\mu ,{\sigma ^2}} \right)\\ \sim N\left( {0,{1^2}} \right)\end{aligned}\)

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

Thus, \(P\left( {Z < z} \right) = 0.09\)or\(P\left( {Z > z} \right) = 0.91\).

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

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.