Question

Dem bones ($2.2$) Osteoporosis is a condition in which the bones become brittle due to the loss of minerals. To diagnose osteoporosis, an elaborate apparatus measures bone mineral density (BMD). BMD is usually reported in a 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?

Short answer

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

(b) The $30.85\%$of this older population has osteoporosis.

Step-by-step solution

Part (a) Step 1: Given information

We need to find the percentage of healthy young adults who have osteoporosis by the WHO criterion.

Part (a) Step 2: Explanation

When the standard deviation is $2.5$less than the mean, The $z$-score is equal to $-2.5$

$z=-2.5$

Calculate the probability using the normal probability

$P(z<-2.5)=0.0062=0.62\%$

According to the WHO definition, $0.62$percent of healthy young individuals have osteoporosis.

Part (b) Step 1: Given information

We need to find out the percentage of this older population has osteoporosis.

Part (b) Step 2: Explanation

We know that

Mean is, $\mu =-2$

And from part (a)

The standard deviation is, $\sigma =1$

So, $z$-score is,

$z\text{'}=\frac{z-\mu }{\sigma }=-0.5$

Therefore, the probability is

$P(X<-2.5)=P(z\text{'}<-0.5)=0.3085=30.85\%$