Chapter 7.1, Problem 25E

(a)

To determine

To calculate: The percent of healthy young adults which have osteoporosis by the WHO criterion.

Answer to Problem 25E

The percent of healthy young adults which have osteoporosis by the WHO criterion is 0.62%.

Explanation of Solution

Given information:

BMD score that is standard deviation $=2.5$

Calculation:

Here, the population distribution is roughly normal.

The given standard deviation that is, 2.5 below the mean is equivalent to the z-score which is denoted as $-2.5$ .

So, $z=-2.5$

Find the required probability by using the normal probability table.

  $\begin{array}{c}P\left(z<-2.5\right)=0.0062\\ =0.62\%\end{array}$

Hence, 0.62% of healthy young adults have osteoporosis by the WHO criterion.

(b)

To determine

To calculate: The percent of older population which has osteoporosis.

Answer to Problem 25E

The percent of older population which has osteoporosis is 30.85%.

Explanation of Solution

Given information:

BMD mean $=-2$

Formula used:

The z-score of distribution of sample x whose mean is $\mu$ and standard deviation is $\sigma$ can b e calculated by the formula, $z=\frac{x-\mu }{\sigma }$ .

Calculation:

The given population is a normal distribution; where, $\mu =-2$ and $\sigma =1$ .

To find the required percent, determine the z-score of the distribution.

The z-score can be found by using the formula $z=\frac{x-\mu }{\sigma }$ .

Substitute $-2.5$ for x, $-2$ for $\mu$ and 1 for $\sigma$ in the above formula and simplify.

  $\begin{array}{c}z=\frac{-2.5-\left(-2\right)}{1}\\ =-0.5\end{array}$

The required probability is,

  $\begin{array}{c}P\left(X<-2.5\right)=P\left(Z<-0.5\right)\\ =0.3085\\ =30.85\%\end{array}$

Hence, the percent of older population which has osteoporosis is 30.85%.