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Osteoporosis. 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 \(2.5\) standard deviations below the mean for healthy young adults. BMD measurements in a population of people similar in age and sex roughly follow a Normal distribution. (a) What percent of healthy young adults have osteoporosis by the WHO criterion? (b) Women aged \(70-79\) are, of course, not young adults. The mean BMD in this age is about \(-2\) on the standard scale for young adults. Suppose the standard deviation is the same as for young adults. What percent of this older population have osteoporosis?

Short Answer

Expert verified
(a) 0.62% of young adults; (b) 0.62% of older women have osteoporosis by the WHO criterion.

Step by step solution

01

Understanding the WHO Criterion

According to the WHO criterion, osteoporosis is diagnosed if the BMD is more than 2.5 standard deviations below the mean. This means we need to find the percentage of individuals who have a BMD that is at or below \(-2.5\).
02

Finding the Percent for Young Adults

Since BMD follows a Normal distribution with mean 0 and standard deviation 1 for healthy young adults, we need to determine the proportion of the distribution at or below \(-2.5\). The standard normal table (Z-table) shows that about 0.62\% lie at or below this value.
03

Mean BMD for Older Adults

For women aged 70-79, the mean BMD is given as \(-2\). The osteoporosis criterion remains the mean minus 2.5 standard deviations, which translates to a BMD value of \(-4.5\).
04

Calculating Percent for Older Adults

For the older population, calculate the Z-score for \(-4.5\) using the mean of \(-2\) and the standard deviation of 1. So, the Z-score is \((-4.5 - (-2)) / 1 = -2.5\). The percentage of older women that have osteoporosis is again about 0.62\%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal distribution
The concept of a Normal distribution is fundamental in statistics and health research. It's a way of describing how data is spread out. When something, like bone mineral density (BMD), follows a Normal distribution, it means that most of the data points lie around the mean, or the average, and fewer points lie further away from the mean.
This distribution is symmetrical, and its shape resembles a bell curve where:
  • Approximately 68% of data points fall within one standard deviation of the mean.
  • About 95% fall within two standard deviations.
  • Nearly 99.7% fall within three standard deviations.
In health research, such as studying osteoporosis, using the Normal distribution helps researchers easily calculate probabilities. For example, in our exercise, knowing that BMD follows a Normal distribution allows us to find out what percentage of the population has osteoporosis by understanding how many fall below a specific threshold. This distribution makes it straightforward to use Z-tables to find these probabilities.
Bone mineral density (BMD)
Bone mineral density (BMD) is a measure used to diagnose conditions like osteoporosis. It quantifies the amount of minerals, mainly calcium, found in a specific volume of bone. The higher the BMD, the denser the bone, which usually means it's stronger and less likely to fracture.
In the context of diagnosing osteoporosis, BMD values are reported in standardized form. This standardization allows for comparisons across ages and populations. For healthy young adults, the BMD is set to a mean of 0 with a standard deviation of 1. So, any BMD measurement gets compared against this reference model.
The standardized score tells whether an individual's BMD is above or below the average. In our exercise, a young adult's BMD being 2.5 standard deviations below the mean indicates osteoporosis according to the World Health Organization (WHO). Thus, understanding BMD is crucial for health professionals in assessing bone health and implementing preventive measures or treatments.
World Health Organization (WHO) criterion
The World Health Organization (WHO) criterion provides clear guidelines for diagnosing osteoporosis. According to the WHO, an individual is said to have osteoporosis if their BMD is 2.5 or more standard deviations below the mean of a healthy young adult population.
This criterion creates a standard for health practitioners worldwide, ensuring consistency in diagnoses. It relies heavily on statistical concepts like the Normal distribution to quantify how much a given BMD value deviates from the norm. By setting this threshold, WHO allows clinicians to identify individuals at high risk of fractures due to low bone density.
In the exercise, we saw this criterion in action to estimate the percentage of people with osteoporosis. For young adults, the calculation showed a small percentage below this threshold, highlighting the method's precision and reliability in distinguishing between healthy and at-risk individuals.

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Most popular questions from this chapter

Heights of men and women. The heights of women aged 20-29 follow approximately the \(N(64.2,2.8)\) distribution. Men the same age have heights distributed as \(N(69.4,3.0)\). What percent of men aged \(20-29\) are taller than the mean height of women aged 20-29?

Grading managers. Some companies "grade on a bell curve" to compare the performance of their managers and professional workers. This forces the use of some low performance ratings so that not all workers are listed as "above average." Ford Motor Company's "performance management process" for this year assigned \(10 \%\) A grades, \(80 \%\) B grades, and \(10 \% \mathrm{C}\) grades to the company's managers. Suppose Ford's performance scores really are Normally distributed. This year, managers with scores less than 25 received \(\mathrm{C}\) grades and those with scores above 475 received A grades. What are the mean and standard deviation of the scores?

Understanding density curves. Remember that it is areas under a density curve, not the height of the curve, that give proportions in a distribution. To illustrate this, sketch a density curve that has a tall, thin peak at 0 on the horizontal axis but has most of its area close to 1 on the horizontal axis without a high peak at \(1 .\)

To completely specify the shape of a Normal distribution, you must give (a) the mean and the standard deviation. (b) the five-number summary. (c) the median and the quartiles.

Weights aren't normal. The heights of people of the same sex and similar ages follow a Normal distribution reasonably closely. Weights, on the other hand, are not Normally distributed. The weights of women aged 20-29 in the United States have mean \(161.9\) pounds and median \(149.4\) pounds. The first and third quartiles are \(126.3\) pounds and \(181.2\) pounds, respectively. What can you say about the shape of the weight distribution? Why?

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