Question

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.

Short answer

The probability of a randomly selected subject having score between 0.87 and 1.78 is 0.1547.

Step-by-step solution

Given information

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

Describe the random variable

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

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

Describe the required score

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,

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

Obtain the z-score from the standard normal table

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.