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 For a randomly selected subject, find the probability of a score between 0.87 and 1.78.

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.

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

The probability has one-to-one correspondence with the area under the curve.

The probability of a score between 0.87 and 1.78 is expressed as,

\(\begin{aligned}{c}P\left( {0.87 < Z < 1.78} \right) = {\rm{Area}}\;{\rm{between}}\;0.87\;{\rm{and}}\;1.78\\ = {\rm{Area}}\;{\rm{to}}\;{\rm{the}}\;{\rm{left}}\;{\rm{of}}\;1.78\; - {\rm{Area}}\;{\rm{to}}\;{\rm{the}}\;{\rm{left}}\;{\rm{of}}\;0.87\\ = P\left( {Z < 1.78} \right) - P\left( {Z < 0.87} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 1 \right)\end{aligned}\)

Using the standard normal table, the cumulative probability corresponding to intersection of row 1.7 and column 0.08 is 0.9625.

Using the standard normal table, the cumulative probability corresponding to intersection of row 0.8 and column 0.07 is 0.8078.

Thus,

\(\begin{aligned}{l}P\left( {Z < 1.78} \right) = 0.9625\\P\left( {Z < 0.87} \right) = 0.8078\end{aligned}\)

Substituting in equation (1),

\(\begin{aligned}{c}P\left( {0.87 < Z < 1.78} \right) = 0.9625 - 0.8078\\ = 0.1547\end{aligned}\)

Therefore, the probability of any randomly selected subject having score between 0.87 and 1.78 is 0.1547.