Q.25 - The Practice of Statistics for AP

Question

Dem bones (2.2) Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD score that is $2.5$ standard deviations below the mean for young adults. BMD measurements in a population of people similar in age and gender roughly follow a Normal distribution.

(a) What percent of healthy young adults have osteoporosis by the WHO criterion? 

(b) Women aged $70$ to $79$ are, of course, not young adults. The mean BMD in this age group is about 2 on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis? 

Step-by-Step Solution

Answer

a). $0.62\%$ of the healthy young adults have osteoporosis by the WHO criterion.

b). $30.85\%$ of women aged $70$ to $79$ have osteoporosis.

Part (a) Step 1: Given Information

The WHO criterion is a BMD score that is $2.5$ standard deviations below the mean.

The population distribution is roughly normal.

Part (a) Step 2: Explanation

$2.5$ standard deviations below the mean is equivalent with the $z$-score being equal to $-2.5$.

$z=-2.5$

Using the normal probability table in the appendix, calculate the corresponding probability:

$$\begin{gathered} P(z<-2.5)=0.0062 \\ =0.62\% \end{gathered}$$

Part (b) Step 1: Given Information

The population is a normal distribution with:

$\mu =-2$

$\sigma =1$

Part (b) Step 2: Explanation

The z-score is the difference between the mean and the standard deviation:

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{-2.5-(-2)}{1} \\ =-0.5 \end{gathered}$$

Using table A, calculate the corresponding probability:

$$\begin{gathered} P(X<-2.5)=P(Z<-0.5) \\ =0.3085 \\ =30.85\% \end{gathered}$$