Skip to main content
Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
Chapter 6, Problem 6.r.1c

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.


c. For a randomly selected subject, find the probability of a bone density test score between -0.67 and 1.29.

Verified step by step guidance
1
Step 1: Understand the problem. The z-scores are normally distributed with a mean (μ) of 0 and a standard deviation (σ) of 1. We are tasked with finding the probability that a randomly selected z-score lies between -0.67 and 1.29.
Step 2: Recall that the probability of a z-score falling between two values can be found using the cumulative distribution function (CDF) of the standard normal distribution. The formula for this is P(a ≤ Z ≤ b) = P(Z ≤ b) - P(Z ≤ a), where Z is the standard normal variable.
Step 3: Use the standard normal table (or a statistical software) to find the cumulative probabilities for the z-scores -0.67 and 1.29. Specifically, find P(Z ≤ 1.29) and P(Z ≤ -0.67).
Step 4: Subtract the cumulative probability of the lower z-score from the cumulative probability of the higher z-score. This gives P(-0.67 ≤ Z ≤ 1.29) = P(Z ≤ 1.29) - P(Z ≤ -0.67).
Step 5: Interpret the result. The value obtained represents the probability that a randomly selected subject has a bone density test score between -0.67 and 1.29. This probability can be expressed as a percentage if needed.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Z Score

A z score indicates how many standard deviations an element is from the mean of a distribution. In the context of a bone density test, a z score of 0 represents the average bone density, while positive and negative values indicate above or below average densities, respectively. Understanding z scores is crucial for interpreting test results and assessing the likelihood of certain outcomes.
Recommended video:
Guided course
06:31
Z-Scores From Given Probability - TI-84 (CE) Calculator

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, depicting that data near the mean are more frequent in occurrence than data far from the mean. In this case, the z scores of bone density tests follow a normal distribution with a mean of 0 and a standard deviation of 1, which allows for the application of statistical methods to calculate probabilities and make inferences about the population.
Recommended video:
Guided course
09:47
Finding Standard Normal Probabilities using z-Table

Probability Calculation

Probability calculation involves determining the likelihood of a specific event occurring within a defined set of outcomes. For the bone density test, calculating the probability of a score falling between -0.67 and 1.29 requires using the properties of the normal distribution, specifically the cumulative distribution function (CDF), to find the area under the curve between these two z scores.
Recommended video:
Guided course
07:09
Probability From Given Z-Scores - TI-84 (CE) Calculator
Related Practice
Textbook Question

In Exercises 11–14, use the population of {2, 3, 5, 9} of the lengths of hospital stay (days) of mothers who gave birth, found from Data Set 6 “Births” in Appendix B. Assume that random samples of size n = 2 are selected with replacement.


Sampling Distribution of the Sample Mean


a. After identifying the 16 different possible samples, find the mean of each sample, and then construct a table representing the sampling distribution of the sample mean. In the table, combine values of the sample mean that are the same. (Hint: See Table 6-3 in Example 2.)

121
views
Textbook Question

Seat Designs. In Exercises 7–9, assume that when seated, adult males have back-to-knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. (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.


Find the probability that nine males have back-to-knee lengths with a mean greater than 23.0 in.

102
views
Textbook Question

In Exercises 7–10, use the same population of {4, 5, 9} that was used in Examples 2 and 5. As in Examples 2 and 5, assume that samples of size n = 2 are randomly selected with replacement.


Sampling Distribution of the Sample Median


c. Find the mean of the sampling distribution of the sample median.

76
views
Textbook Question

Bone Density Test A bone mineral density test is used to identify a bone disease. The result of a bone density test is commonly measured as a z score, and the population of z scores is normally distributed with a mean of 0 and a standard deviation of 1.


a. For a randomly selected subject, find the probability of a bone density test score greater than -1.37.

239
views
Textbook Question

Normal Distribution Using a larger data set than the one given for the preceding exercises, assume that cell phone radiation amounts are normally distributed with a mean of 1.17 W/kg and a standard deviation of 0.29 W/kg.

b. Find the value of Q3, the cell phone radiation amount that is the third quartile.

109
views
Textbook Question

Seat Designs. In Exercises 7–9, assume that when seated, adult males have back-to-knee lengths that are normally distributed with a mean of 23.5 in. and a standard deviation of 1.1 in. (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.


Find the probability that a male has a back-to-knee length greater than 25.0 in.

100
views