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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. $$ \text { Between } 1.50 \text { and } 2.50 $$

Short Answer

Expert verified
The probability is approximately 0.0606.

Step by step solution

01

Understanding the Problem

The problem involves finding the probability that a bone density test score falls between 1.50 and 2.50 in a normally distributed population with a mean () of 0 and a standard deviation () of 1.
02

Standard Normal Distribution

Since the distribution is normal with mean 0 and standard deviation 1, we can use the standard normal distribution or Z-distribution for this problem.
03

Z-Scores Calculation

Given that the mean is 0 and standard deviation is 1, the Z-scores for the test scores are directly the same values: \(Z_1 = 1.50\) and \(Z_2 = 2.50\). No further calculation is necessary since the distribution parameters indicate we are already provided with Z-scores.
04

Finding the Cumulative Probabilities

Use either Table A-2 (Z-table) or technology (e.g., a statistical calculator or software) to find the cumulative probabilities corresponding to Z = 1.50 and Z = 2.50. For Z = 1.50, the cumulative probability is approximately 0.9332. For Z = 2.50, the cumulative probability is approximately 0.9938.
05

Determining the Probability Between Two Values

Subtract the cumulative probability for Z = 1.50 from the cumulative probability for Z = 2.50:
06

Calculation

The probability that a score is between 1.50 and 2.50 is given by:
07

Drawing the Graph

Draw a standard normal distribution curve (bell-shaped curve) with the mean at 0. Shade the area between Z = 1.50 and Z = 2.50 to visually represent the probability.

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Key Concepts

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

Understanding Bone Density Test Scores
Bone density tests measure the density of minerals in bones, indicating bone strength and health. These scores follow a normal distribution with a mean of 0 and a standard deviation of 1.
The normal distribution concept is like a bell curve, where most test scores are around the mean, and fewer scores are far from it. By knowing the mean and standard deviation, we can predict how likely a specific score is.
In our example, we're interested in scores between 1.50 and 2.50.
Understanding Z-Scores
Z-scores tell us how many standard deviations a data point is from the mean. When the data is normally distributed with a mean of 0 and a standard deviation of 1, the Z-score for any value is simply that value itself.
For instance, a Z-score of 1.50 means the score is 1.50 standard deviations above the mean. Similarly, a Z-score of 2.50 is 2.50 standard deviations above the mean.
Z-scores help us to compare data points from different distributions and find probabilities related to those points using standard normal distribution tables or technology.
Calculating Cumulative Probability
Cumulative probability is the probability that a random variable falls within a specified range. For normal distributions, we use the cumulative distribution function (CDF) to find this.
The CDF value at a specific Z-score tells us the total probability for scores up to that Z-score.
For example, using a Z-table or statistical software, the CDF for Z = 1.50 is 0.9332, and for Z = 2.50, it is 0.9938.
To find the probability between two Z-scores, subtract the lower cumulative probability from the higher one.
In this case, the probability of bone density scores being between 1.50 and 2.50 is: \(0.9938 - 0.9332 = 0.0606\). This means there's a 6.06% chance a score will fall in that range.

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Most popular questions from this chapter

Find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2. About \(\quad \%\) of the area is between \(z=-3.5\) and \(z=3.5\) (or within \(3.5\) standard deviations of the mean).

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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