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Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
All textbooksTriola 14th EditionCh. 6 - Normal Probability DistributionsProblem 6.q.4
Chapter 6, Problem 6.q.4

Bone Density Test. In Exercises 1–4, assume that scores on a bone mineral density test are normally distributed with a mean of 0 and a standard deviation of 1.


Bone Density For a randomly selected subject, find the probability of a bone density score between and 2.00.

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1
Step 1: Recognize that the problem involves a normal distribution with a mean (μ) of 0 and a standard deviation (σ) of 1. This is known as the standard normal distribution.
Step 2: Identify the range of interest for the bone density score, which is between 0 and 2.00. The goal is to find the probability that a randomly selected score falls within this range.
Step 3: Use the standard normal distribution table (Z-table) or a statistical software to find the cumulative probability for Z = 2.00. The Z-score formula is not needed here since the values are already standardized.
Step 4: Find the cumulative probability for Z = 0. Since the mean of the standard normal distribution is 0, the cumulative probability for Z = 0 is 0.5.
Step 5: Subtract the cumulative probability for Z = 0 from the cumulative probability for Z = 2.00 to find the probability of a score between 0 and 2.00. This difference represents the area under the curve between these two Z-scores.

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Key Concepts

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

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. It is characterized by its bell-shaped curve, defined by its mean and standard deviation. In this context, the bone mineral density scores are assumed to follow a normal distribution with a mean of 0 and a standard deviation of 1.
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Z-Scores

A Z-score indicates how many standard deviations an element is from the mean. It is calculated by subtracting the mean from the score and then dividing by the standard deviation. In this case, the Z-scores for the bone density scores will help determine the probability of a score falling between specific values, such as between 0 and 2.00.
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Probability and Area Under the Curve

In statistics, the probability of a score falling within a certain range in a normal distribution can be found by calculating the area under the curve between those two points. This is typically done using Z-tables or statistical software. For the bone density test, finding the probability of scores between 0 and 2.00 involves determining the area under the normal curve between these Z-scores.
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Related Practice
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.

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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.

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Textbook Question

Bone Density Test. In Exercises 1–4, assume that scores on a bone mineral density test are normally distributed with a mean of 0 and a standard deviation of 1.


Bone Density Find the bone density score that is the 90th percentile, which is the score separating the lowest 90% from the top 10%.

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Textbook Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.


Doorway Height The Boeing 757-200 ER airliner carries 200 passengers and has doors with a height of 72 in. Heights of men are normally distributed with a mean of 68.6 in. and a standard deviation of 2.8 in. (based on Data Set 1 “Body Data” in Appendix B).


d. When considering the comfort and safety of passengers, why are women ignored in this case?

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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.

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Textbook Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.


Water Taxi Safety Passengers died when a water taxi sank in Baltimore’s Inner Harbor. Men are typically heavier than women and children, so when loading a water taxi, assume a worst-case scenario in which all passengers are men. Assume that weights of men are normally distributed with a mean of 189 lb and a standard deviation of 39 lb (based on Data Set 1 “Body Data” in Appendix B). The water taxi that sank had a stated capacity of 25 passengers, and the boat was rated for a load limit of 3500 lb.


d. Is the new capacity of 20 passengers safe?

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