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Chapter 6: Problem 40

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 the bone density scores that can be used as cutoff values separating the lowest \(3 \%\) and highest \(3 \%\).

Short Answer

Expert verified
-1.88 and 1.88

Step by step solution

01

- Understanding the problem

We are asked to find the bone density test scores that separate the lowest and highest 3% of normally distributed data with a mean of 0 and a standard deviation of 1.
02

- Draw the normal distribution curve

Draw a symmetric bell-shaped curve centered at the mean, which is 0. Label the mean and note that the standard deviation is 1.
03

- Identify the cutoff values

Since the scores are normally distributed, locate the cutoff values which separate the bottom 3% and the top 3% of the distribution. These are also called the z-scores corresponding to the 3rd and 97th percentiles.
04

- Lookup z-scores in the standard normal distribution table

Using the standard normal distribution table, find the z-score that corresponds to the 3rd percentile, which is approximately -1.88. Similarly, find the z-score for the 97th percentile, which is approximately 1.88.
05

- Interpretation of the z-scores

These z-scores indicate that a bone density test score of -1.88 separates the lowest 3%, and a score of 1.88 separates the highest 3% from the rest of the data.

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Key Concepts

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

Z-Scores
When dealing with normal distribution, understanding z-scores is essential. A z-score indicates how many standard deviations an element is from the mean. For example, in the given problem, the z-score of -1.88 signifies that the particular bone density score is 1.88 standard deviations below the mean. Similarly, the z-score of 1.88 is 1.88 standard deviations above the mean.

These z-scores allow us to understand where a value falls within a normal distribution. This is particularly useful for identifying cutoffs, like the lowest and highest 3% in our exercise.
Percentiles
Percentiles give us a way to understand the relative standing of a value within a dataset. A percentile rank tells us the percentage of values that fall below a certain number. For instance, the 3rd percentile means that 3% of the data values are below this point.

In our example, the bone density scores at the 3rd and 97th percentiles correspond to the z-scores of -1.88 and 1.88, respectively. This means that a score of -1.88 is higher than 3% of all other scores, and a score of 1.88 is higher than 97% of all other scores.
Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are around the mean. In a normal distribution, about 68% of values fall within one standard deviation of the mean.

Knowing that the standard deviation for our bone density scores is 1, it becomes easier to interpret the z-scores. If we have z-scores, we can quickly determine how far any given bone density score is from the mean in terms of standard deviation units.
Mean
The mean of a distribution is the average of all the data points. It serves as a central reference point in a normal distribution.

In our exercise, the mean bone density score is 0. This central point helps us understand the spread and position of other scores.

For example, z-scores of -1.88 and 1.88 tell us how far these specific scores are from the mean (i.e., 0), directly impacting our interpretation of the highest and lowest 3%.

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Most popular questions from this chapter

Sampling with Replacement The Orangetown Medical Research Center randomly selects 100 births in the United States each day, and the proportion of boys is recorded for each sample. a. Do you think the births are randomly selected with replacement or without replacement? b. Give two reasons why statistical methods tend to be based on the assumption that sampling is conducted with replacement, instead of without replacement.

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