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?

Step by step solution

01

Part (a) Step 1: Given information

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

02

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.

03

Part (b) Step 1: Given information

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

04

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

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

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