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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.1e

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.
e. If the mean bone density test score is found for 9 randomly selected subjects, find the probability that the mean is greater than 0.23.

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Step 1: Identify the given information. The population of z-scores is normally distributed with a mean (μ) of 0 and a standard deviation (σ) of 1. The sample size (n) is 9, and we are tasked with finding the probability that the sample mean (x̄) is greater than 0.23.
Step 2: Calculate the standard error of the mean (SE). The formula for the standard error is SE = σ / √n. Substitute the given values: σ = 1 and n = 9, so SE = 1 / √9.
Step 3: Standardize the sample mean to find the z-score corresponding to x̄ = 0.23. Use the formula z = (x̄ - μ) / SE. Substitute the values: x̄ = 0.23, μ = 0, and SE (calculated in Step 2).
Step 4: Use the z-score obtained in Step 3 to find the cumulative probability from the standard normal distribution table or a statistical software. This gives the probability that the sample mean is less than 0.23.
Step 5: Subtract the cumulative probability from 1 to find the probability that the sample mean is greater than 0.23. This is because the total probability under the normal curve is 1, and we are interested in the upper tail.

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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 means the score is exactly at the mean, while a z score of 0.23 indicates the score is 0.23 standard deviations above the mean. This standardization allows for comparison across different populations and tests.
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Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, showing 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 are normally distributed, which means that statistical methods based on this distribution can be applied to calculate probabilities and make inferences about the population.
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Central Limit Theorem

The Central Limit Theorem states that the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution, provided the sample size is sufficiently large. In this scenario, with a sample size of 9, the theorem allows us to use the normal distribution to find the probability that the mean bone density score exceeds 0.23.
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Related Practice
Textbook Question

In Exercises 8 and 9, assume that women have standing eye heights that are normally distributed with a mean of 59.7 in. and a standard deviation of 2.5 in. (based on anthropometric survey data from Gordon, Churchill, et al.).

a. If an eye recognition security system is positioned at a height that is uncomfortable for women with standing eye heights less than 54 in., what percentage of women will find that height uncomfortable?

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Textbook Question

Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.


b. If 4 randomly selected adults take the Wechsler IQ test, find the probability that their mean score is at least 131.

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Textbook Question

In Exercises 8 and 9, assume that women have standing eye heights that are normally distributed with a mean of 59.7 in. and a standard deviation of 2.5 in. (based on anthropometric survey data from Gordon, Churchill, et al.).


Significance Instead of using 0.05 for identifying significant values, use the criteria that a value x is significantly high if P(x or greater) ≤ 0.01 and a value is significantly low if P(x or less) ≤ 0.01. Find the standing eye heights of women that separate significant values from those that are not significant. Using these criteria, is a woman’s standing eye height of 67 in. significantly high?

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Textbook Question

Birth Weights Based on Data Set 6 “Births” in Appendix B, birth weights of girls are normally distributed with a mean of 3037.1 g and a standard deviation of 706.3 g.


b. What is the value of the median?

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Textbook Question

Birth Weights Based on Data Set 6 “Births” in Appendix B, birth weights of girls are normally distributed with a mean of 3037.1 g and a standard deviation of 706.3 g.


c. What is the value of the mode?

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Textbook Question

Mensa Membership in Mensa requires a score in the top 2% on a standard intelligence test. The Wechsler IQ test is designed for a mean of 100 and a standard deviation of 15, and scores are normally distributed.


a. Find the minimum Wechsler IQ test score that satisfies the Mensa requirement.

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