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 greater than -2.93.

Short answer

The probability of a randomly selected score being greater than –2.93 is 0.9983.

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 of a randomly selected score being greater than –2.93 is expressed as,

\(P\left( {Z > - 2.93} \right)\)

Since the probability has one-to-one correspondence with the area under the curve, the expression infers to the right tailed area of –2.93.

Thus, the area to the right of –2.93 is 1 minus the area to the left of –2.93.

Mathematically,

\(\begin{aligned}{c}P\left( {Z > - 2.93} \right) = 1 - {\rm{Area}}\;{\rm{to}}\;{\rm{the}}\;{\rm{left}}\;{\rm{of}}\; - 2.93\\ = 1 - P\left( {Z < - 2.93} \right)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 1 \right)\end{aligned}\)

Obtain the z-score from the standard normal table

Using the standard normal table, the cumulative probability of corresponding to intersection of row –2.9 and column 0.03 is 0.0017.

Thus, \(P\left( {Z < - 2.93} \right) = 0.0017\).

Substituting in equation (1),

\(\begin{aligned}{c}P\left( {Z > - 2.93} \right) = 1 - 0.0017\\ = 0.9983\end{aligned}\)

Therefore, the probability of any randomly selected score being greater than –2.93 is 0.9983.