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91Ó°ÊÓ

Assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places. Find \(P_{10}\), the 10 th percentile. This is the bone density score separating the bottom \(10 \%\) from the top \(90 \%\).

Short Answer

Expert verified
The 10th percentile bone density test score is -1.28.

Step by step solution

01

- Understand the Problem

We need to find the bone density test score (given that the scores are normally distributed with a mean of 0 and a standard deviation of 1) which separates the bottom 10% of subjects from the top 90%.
02

- Find the Z-Score for the 10th Percentile

The 10th percentile (\(P_{10}\)) of a standard normal distribution corresponds to a Z-score that cuts off the bottom 10% from the top 90%. Use a Z-table or a calculator to find this Z-score. For the 10th percentile, the Z-score is -1.28.
03

- Map the Z-Score to the Bone Density Score

Since the bone density test scores are normally distributed with a mean (\(\mu \) = 0) and standard deviation (\(\sigma\) = 1), the Z-score itself is the bone density score. Therefore, the bone density test score corresponding to the 10th percentile is -1.28.
04

- Draw the Graph

Draw a standard normal distribution curve. Mark the mean (0) at the center. Then, locate -1.28 on the horizontal axis, which is to the left of the mean. Shade the area to the left of -1.28, which represents the bottom 10% of the distribution.

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Key Concepts

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

percentile
In statistics, a *percentile* is a measure indicating the value below which a given percentage of observations in a group falls. For example, the 10th percentile (\(P_{10}\)) is the value below which 10% of the observations may be found. Percentiles help in understanding the distribution and ranking of data.

In our problem, we're interested in finding the bone density score that separates the bottom 10% of all scores from the top 90%. To do this, we must first understand the concept of Z-scores as they are intimately related to percentiles in a normal distribution.
Z-score
A *Z-score* is a statistical measurement that describes a value's relation to the mean of a group of values. It is calculated by subtracting the mean from the value and then dividing the result by the standard deviation. The formula for Z-score is:

\[ Z = \frac{X - \mu}{\sigma} \]

In a standard normal distribution, the Z-score allows us to determine the percentile of any data point. For instance, a Z-score of -1.28 corresponds approximately to the 10th percentile. This tells us that the value is 1.28 standard deviations below the mean and separates the bottom 10% of data from the rest.

Mapping Z-scores to percentiles is key in converting abstract data points to practical, understandable terms.
standard normal distribution
The *standard normal distribution* is a normal distribution with a mean of 0 and a standard deviation of 1. It's represented by a bell-shaped curve symmetrical around the mean. This distribution is frequently used for Z-scores because it standardizes different datasets, allowing for easy comparison.

Properties of a standard normal distribution include:
  • The total area under the curve equals 1.
  • It is symmetrical around the mean (0).
  • Approximately 68% of the data fall within 1 standard deviation of the mean, 95% within 2 standard deviations, and 99.7% within 3 standard deviations.

Graphs of standard normal distributions are useful in visualizing Z-scores and percentiles, as seen in our exercise where we shaded the area to the left of -1.28 to represent the bottom 10%.
bone density
Bone density measures the amount of minerals (such as calcium) in your bones. This test is crucial for diagnosing conditions like osteoporosis, where bone density reduces, increasing fracture risk. Bone density scores are often measured using densitometry techniques like DEXA scans.

In our scenario, bone density scores follow a normal distribution pattern with a mean of 0 and a standard deviation of 1. This implies that most scores cluster around the mean, with fewer scores appearing as we move away from the mean in either direction. Calculating the 10th percentile helps in identifying individuals who may be at risk due to abnormally low bone density.

Understanding how to interpret these scores with the help of Z-scores and percentiles can facilitate early diagnosis and preventive measures.

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Most popular questions from this chapter

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Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places. $$ \text { Less than }-1.23 $$

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