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Chapter 6: Problem 23

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. $$ \text { Greater than }-2.00 $$

Short Answer

Expert verified
0.9772

Step by step solution

01

Understand the Problem

We need to find the probability that a bone density test score is greater than -2.00. The scores are normally distributed with a mean of 0 and a standard deviation of 1.
02

Draw the Normal Distribution

Draw a bell-shaped curve representing the standard normal distribution. The center of the curve (mean) is 0. Mark the point -2.00 on the horizontal axis.
03

Shade the Area of Interest

Shade the area to the right of -2.00. This represents the probability we need to find.
04

Use Z-Table or Technology

Look up the value for a z-score of -2.00 in the standard normal distribution table (Table A-2) or use technology (like a calculator or software) to find the area to the left of -2.00. For a z-score of -2.00, the table shows approximately 0.0228.
05

Find the Desired Probability

Since we need the area to the right of -2.00, subtract the table value from 1: P(Z > -2.00) = 1 - P(Z < -2.00) = 1 - 0.0228 = 0.9772.

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Key Concepts

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

Bone Density Test
A bone density test measures the density of minerals (such as calcium) in your bones. This helps assess the strength and health of bones. By determining a bone density score, medical professionals can identify conditions like osteoporosis. When the scores from these tests are collected, they often follow a normal distribution. This means most individuals will have scores close to the mean, with fewer people having very high or very low scores. In this exercise, we鈥檙e working with a standard normal distribution, where the mean score is 0.
Mean and Standard Deviation
The **mean** is the average score in the distribution. For bone density tests in this problem, the mean is 0. The **standard deviation** indicates how spread out the scores are from the mean. A standard deviation of 1 means that most scores fall within 1 unit of the mean. In a normal distribution, roughly 68% of scores are within 1 standard deviation from the mean, 95% within 2, and 99.7% within 3. This standard normal distribution helps us determine probabilities based on z-scores.
Probability Calculation
The goal in probability calculation is to find the likelihood of an event occurring. For bone density scores, we want to know the probability of scoring higher than -2.00. This involves calculating the area under the curve of our normal distribution to the right of -2.00. By using statistical tools like z-tables or technology, we can find these probabilities easily.
Z-Score
A z-score is a measure of how many standard deviations an element is from the mean. In our case, the z-score is -2.00, meaning the score is 2 standard deviations below the mean. Z-scores are crucial because they let us use the standard normal distribution to find probabilities. By looking up a z-score in a z-table, we can find the area to the left or right of that score. For a z-score of -2.00, the area to the left is 0.0228. This means 2.28% of the distribution lies below this score.
Area under Curve
In a normal distribution, the area under the curve represents probabilities. The total area of the curve is 1 (or 100%). To find the probability of a bone density score being greater than -2.00, we use the area to the left of -2.00 (0.0228). By subtracting this from 1, we get the area to the right, which is the desired probability: \( P(Z > -2.00) = 1 - P(Z < -2.00) = 1 - 0.0228 = 0.9772 \). So, there鈥檚 a 97.72% chance of scoring higher than -2.00 on this bone density test.

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Most popular questions from this chapter

There are about 4200 college presidents in the United States, and they have annual incomes with a distribution that is skewed instead of being normal. Many different samples of 40 college presidents are randomly selected, and the mean annual income is computed for each sample. a. What is the approximate shape of the distribution of the sample means (uniform, normal, skewed, other)? b. What value do the sample means target? That is, what is the mean of all such sample means?

Common tests such as the SAT, ACT, Law School Admission test (LSAT), and Medical College Admission Test (MCAT) use multiple choice test questions, each with possible answers of \(a, b, c, d, e\), and each question has only one correct answer. We want to find the probability of getting at least 25 correct answers for someone who makes random guesses for answers to a block of 100 questions. If we plan to use the methods of this section with a normal distribution used to approximate a binomial distribution, are the necessary requirements satisfied? Explain.

Data Set 4 "Births" in Appendix B includes birth weights of 400 babies. If we compute the values of sample statistics from that sample, which of the following statistics are unbiased estimators of the corresponding population parameters: sample mean; sample median; sample range; sample variance; sample standard deviation; sample proportion?

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 \(75 \%\).

When a water taxi sank in Baltimore's Inner Harbor, an investigation revealed that the safe passenger load for the water taxi was \(3500 \mathrm{lb}\). It was also noted that the mean weight of a passenger was assumed to be \(140 \mathrm{lb}\). Assume a "worst-case" scenario in which all of the passengers are adult men. Assume that weights of men are normally distributed with a mean of \(188.6 \mathrm{lb}\) and a standard deviation of \(38.9 \mathrm{lb}\) (based on Data Set 1 "Body Data" in Appendix B). a. If one man is randomly selected, find the probability that he weighs less than \(174 \mathrm{lb}\) (the new value suggested by the National Transportation and Safety Board). b. With a load limit of \(3500 \mathrm{lb}\), how many male passengers are allowed if we assume a mean weight of \(140 \mathrm{lb}\) ? c. With a load limit of \(3500 \mathrm{lb}\), how many male passengers are allowed if we assume the updated mean weight of \(188.6 \mathrm{lb}\) ? d. Why is it necessary to periodically review and revise the number of passengers that are allowed to board?
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