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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 31
Chapter 6, Problem 31

Aircraft Seat Width Engineers want to design seats in commercial aircraft so that they are wide enough to fit 99% of all adults. (Accommodating 100% of adults would require very wide seats that would be much too expensive.) Assume adults have hip widths that are normally distributed with a mean of 14.3 in. and a standard deviation of 0.9 in. (based on data from Applied Ergonomics). Find P99. That is, find the hip width for adults that separates the smallest 99% from the largest 1%.

Verified step by step guidance
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Step 1: Understand the problem. We are tasked with finding the hip width (P99) that separates the smallest 99% of adults from the largest 1%. This involves using the properties of the normal distribution and identifying the z-score corresponding to the 99th percentile.
Step 2: Recall the formula for converting a z-score to a value in a normal distribution: \( x = \mu + z \cdot \sigma \), where \( \mu \) is the mean, \( \sigma \) is the standard deviation, \( z \) is the z-score, and \( x \) is the value we are solving for.
Step 3: Determine the z-score for the 99th percentile. Using a z-score table or statistical software, find the z-score corresponding to a cumulative probability of 0.99. This z-score is approximately \( z = 2.33 \).
Step 4: Plug the values into the formula. Substitute \( \mu = 14.3 \), \( \sigma = 0.9 \), and \( z = 2.33 \) into the formula \( x = \mu + z \cdot \sigma \). This will give the hip width \( x \) that separates the smallest 99% of adults from the largest 1%.
Step 5: Perform the calculation. Multiply \( z \cdot \sigma \) and add the result to \( \mu \). This will yield the value of \( x \), which is the desired hip width (P99).

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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, depicting that data near the mean are more frequent in occurrence than data far from the mean. In this context, adult hip widths are assumed to follow a normal distribution, characterized by its mean (14.3 inches) and standard deviation (0.9 inches). Understanding this distribution is crucial for determining percentiles, such as P99.
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Finding Standard Normal Probabilities using z-Table

Percentiles

A percentile is a measure used in statistics to indicate the value below which a given percentage of observations fall. For example, P99 refers to the hip width below which 99% of adults fall, meaning only 1% have a hip width greater than this value. Calculating percentiles involves using the properties of the normal distribution, specifically the z-score.

Z-Score

A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, expressed in terms of standard deviations. It is calculated by subtracting the mean from the value and then dividing by the standard deviation. In this problem, the z-score corresponding to P99 will help determine the specific hip width that separates the smallest 99% from the largest 1% of adults.
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Related Practice
Textbook Question

Water Taxi Safety When a water taxi sank in Baltimore’s Inner Harbor, an investigation revealed that the safe passenger load for the water taxi was 3500 lb. It was also noted that the mean weight of a passenger was assumed to be 140 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 lb and a standard deviation of 38.9 lb (based on Data Set 1 “Body Data” in Appendix B).


c. With a load limit of 3500 lb, how many male passengers are allowed if we assume the updated mean weight of 188.6 lb?

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Textbook Question

In Exercises 25–28, use these parameters (based on Data Set 1 “Body Data” in Appendix B):


Men’s heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in.

Women’s heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.


Snow White Disney World requires that women employed as a Snow White character must have a height between 64 in. and 67 in.


a. Find the percentage of women meeting the height requirement.

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Textbook Question

Durations of Pregnancies The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.


a. In a letter to “Dear Abby,” a wife claimed to have given birth 308 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 308 days or longer. What does the result suggest?

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Textbook Question

Designing Helmets Engineers must consider the circumferences of adult heads when designing motorcycle helmets. Adult head circumferences are normally distributed with a mean of 570.0 mm and a standard deviation of 18.3 mm (based on Data Set 3 “ANSUR II 2012”). Due to financial constraints, the helmets will be designed to fit all adults except those with head circumferences that are in the smallest 5% or largest 5%. Find the minimum and maximum head circumferences that the helmets will fit.

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Textbook Question

Water Taxi Safety When a water taxi sank in Baltimore’s Inner Harbor, an investigation revealed that the safe passenger load for the water taxi was 3500 lb. It was also noted that the mean weight of a passenger was assumed to be 140 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 lb and a standard deviation of 38.9 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 lb (the new value suggested by the National Transportation and Safety Board).

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Textbook Question

In Exercises 25–28, use these parameters (based on Data Set 1 “Body Data” in Appendix B):


Men’s heights are normally distributed with mean 68.6 in. and standard deviation 2.8 in.

Women’s heights are normally distributed with mean 63.7 in. and standard deviation 2.9 in.


Mickey Mouse Disney World requires that people employed as a Mickey Mouse character must have a height between 56 in. and 62 in.


a. Find the percentage of men meeting the height requirement. What does the result suggest about the genders of the people who are employed as Mickey Mouse characters?

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