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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).

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.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ For females, find the first quartile \(Q_{1}\), which is the length separating the bottom \(25 \%\) from the top \(75 \%\).

A common design requirement is that an environment must fit the range of people who fall between the 5 th percentile for women and the 95 th percentile for men. In designing an assembly work table, we must consider sitting knee height, which is the distance from the bottom of the feet to the top of the knee. Males have sitting knee heights that are normally distributed with a mean of \(21.4\) in. and a standard deviation of \(1.2\) in.; females have sitting knee heights that are normally distributed with a mean of \(19.6\) in. and a standard deviation of \(1.1\) in. (based on data from the Department of Transportation). a. What is the minimum table clearance required to satisfy the requirement of fitting \(95 \%\) of men? Why is the 95 th percentile for women ignored in this case? b. The author is writing this exercise at a table with a clearance of \(23.5\) in. above the floor. What percentage of men fit this table, and what percentage of women fit this table? Does the table appear to be made to fit almost everyone?

The heights (in inches) of men listed in Data Set 1 "Body Data" in Appendix B have a distribution that is approximately normal, so it appears that those heights are from a normally distributed population. a. If 2 inches is added to each height, are the new heights also normally distributed? b. If each height is converted from inches to centimeters, are the heights in centimeters also. normally distributed? c. Are the logarithms of normally distributed heights also normally distributed?

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. $$ \text { Greater than } 0.18 $$

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