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Chapter 6: Problem 38

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 bone density score corresponding to the 10th percentile is -1.28.

Step by step solution

01

Understand the Problem

The problem involves finding the 10th percentile of a normally distributed set of bone density scores with a mean of 0 and a standard deviation of 1. This percentile represents the score below which 10% of the scores lie.
02

Identify the Z-Score Corresponding to the 10th Percentile

Use standard normal distribution tables or a calculator to find the Z-score that corresponds to the 10th percentile. For a Z-score at the 10th percentile, we find that the Z-score is approximately \[ Z = -1.28 \]
03

Draw the Normal Distribution Curve

Draw a bell curve with the mean (\( \text{Mean} = 0 \)) at the center. Shade the left tail of the curve to represent the bottom 10% of the distribution. This shaded area corresponds to the Z-score found in the previous step.
04

Convert the Z-Score to the Bone Density Score

Use the formula for converting Z-scores to raw scores:\[ X = \text{Mean} + Z \times \text{Standard Deviation} \] Given that the mean is 0 and the standard deviation is 1, the bone density score is:\[ X = 0 + (-1.28) \times 1 \]\[ X = -1.28 \]
05

Round the Result

The bone density score has already been calculated to two decimal places in the previous step. The score at the 10th percentile is \( -1.28 \)

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Key Concepts

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

Percentile
Understanding percentiles is crucial for interpreting data in statistics, especially with normal distributions.
A percentile indicates the relative standing of a value within a dataset, showing the percentage of data below that value.
For example, the 10th percentile (P10) represents the value below which 10% of the observations fall.
This means if you are in the 10th percentile for height in your age group, 10% of individuals are shorter than you.

In our exercise, we needed to find the bone density score at the 10th percentile in a normal distribution.
Percentiles are critical in fields like healthcare to understand where patients stand compared to a standard population.
Z-Score
A Z-score quantifies the number of standard deviations a data point is from the mean of a dataset.
This is essential in standard normal distribution, where the mean is 0 and the standard deviation is 1.
It helps to understand how unusual a particular score is in the context of the entire distribution.

In our example, to find the Z-score corresponding to the 10th percentile, we used standard normal distribution tables or statistical software.
The Z-score for the 10th percentile is approximately \(-1.28\).
This negative value indicates that the score is below the mean, specifically 1.28 standard deviations below it.

Converting between Z-scores and percentiles allows us to translate abstract data into meaningful scores.
Bone Density
Bone density is a measure of the amount of minerals (mainly calcium and phosphorus) in a precise volume of bone.
It is usually measured using a DEXA scan and is crucial for diagnosing osteoporosis and other bone disorders.

In our exercise, the bone density scores follow a normal distribution with a mean of 0 and a standard deviation of 1.
This means most individuals will have bone density scores close to 0.
Using the Z-score formula, we can convert Z-scores to bone density scores: \[ X = \text{Mean} + Z \times \text{Standard Deviation} \]
Given that the mean is 0 and the standard deviation is 1:
For our 10th percentile Z-score of -1.28, the raw score is:
\(-1.28\),
indicating a bone density 1.28 standard deviations below average.

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Most popular questions from this chapter

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. If bone density scores in the bottom \(2 \%\) and the top \(2 \%\) are used as cutoff points for levels that are too low or too high, find the two readings that are cutoff values.

Use the data in the table below for sitting adult males and females (based on anthropometric survey data from Gordon, Churchill, et al.). These data are used often in the design of different seats, including aircraft seats, train seats, theater seats, and classroom seats. (Hint: Draw a graph in each case.) $$\begin{array}{|l|l|l|l|} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \mathrm{in} . & 1.1 \mathrm{in} . & \text { Normal } \\ \hline \text { Females } & 22.7 \mathrm{in} . & 1.0 \mathrm{in} . & \text { Normal } \\ \hline \end{array}$$ Find the probability that a female has a back-to-knee length between 22.0 in. and 24.0 in.

Hybridization A hybridization experiment begins with four peas having yellow pods and one pea having a green pod. Two of the peas are randomly selected with replacement from this population. a. After identifying the 25 different possible samples, find the proportion of peas with yellow pods in each of them, then construct a table to describe the sampling distribution of the proportions of peas with yellow pods. b. Find the mean of the sampling distribution. c. Is the mean of the sampling distribution [from part (b)] equal to the population proportion of peas with yellow pods? Does the mean of the sampling distribution of proportions always equal the population proportion?

Use these parameters (based on Data Set 1 "Body Data" in Appendix \(B\) ): Men's heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in. Women's heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in. Disney World requires that people employed as a Mickey Mouse character must have a height between 56 in. and 62 in. a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as Mickey Mouse characters? b. If the height requirements are changed to exclude the tallest \(50 \%\) of men and the shortest \(5 \%\) of men, what are the new height requirements?

Do the following: If the requirements of \(n p \geq 5\) and \(n q \geq 5\) are both satisfied, estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if \(n p < 5\) or n \(q < 5,\) then state that the normal approximation should not be used. With \(n=20\) births and \(p=0.512\) for a boy, find \(P\) (fewer than 8 boys).
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