Skip to main content
Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446당신이 사용하는 게 아니라요?교과서 변경
6장, 문제 6.4.10a

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


Designing Manholes According to the website www.torchmate.com, “manhole covers must be a minimum of 22 in. in diameter, but can be as much as 60 in. in diameter.” Assume that a manhole is constructed to have a circular opening with a diameter of 22 in. Men have shoulder widths that are normally distributed with a mean of 18.2 in. and a standard deviation of 1.0 in. (based on data from the National Health and Nutrition Examination Survey).


a. What percentage of men will fit into the manhole?

검증된 단계별 안내
1
Step 1: Understand the problem. We are tasked with finding the percentage of men who can fit into a manhole with a diameter of 22 inches. Since the opening is circular, the relevant dimension for fitting is the shoulder width of men, which is normally distributed with a mean (μ) of 18.2 inches and a standard deviation (σ) of 1.0 inch.
Step 2: Define the threshold for fitting. A man can fit into the manhole if his shoulder width is less than or equal to the diameter of the manhole, which is 22 inches. Therefore, we need to calculate the probability that a randomly selected man has a shoulder width ≤ 22 inches.
Step 3: Standardize the value of 22 inches using the z-score formula. The z-score formula is: z=x-μσ, where x is the value of interest (22 inches), μ is the mean (18.2 inches), and σ is the standard deviation (1.0 inch). Substitute these values into the formula to compute the z-score.
Step 4: Use the z-score to find the cumulative probability. Once the z-score is calculated, use a standard normal distribution table or statistical software to find the cumulative probability corresponding to that z-score. This cumulative probability represents the proportion of men with shoulder widths ≤ 22 inches.
Step 5: Interpret the result. The cumulative probability obtained in Step 4 is the percentage of men who can fit into the manhole. Multiply the probability by 100 to express it as a percentage.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
2m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Normal Distribution

Normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. In this context, men's shoulder widths are normally distributed, which allows us to use the mean and standard deviation to determine the percentage of men fitting through the manhole.
추천 영상:
가이드 코스
09:47
Finding Standard Normal Probabilities using z-Table

Z-Score

A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It is calculated by subtracting the mean from the value and then dividing by the standard deviation. In this scenario, calculating the Z-score for the manhole diameter will help determine how many men have shoulder widths that fall within the allowable range.
추천 영상:
가이드 코스
06:31
Z-Scores From Given Probability - TI-84 (CE) Calculator

Percentile

A percentile is a measure used in statistics indicating the value below which a given percentage of observations fall. In this case, once the Z-score is calculated, we can use the standard normal distribution table to find the corresponding percentile, which will tell us the percentage of men whose shoulder widths are less than or equal to the manhole diameter.
관련 실천
교과서 질문

Fatal Car Crashes There are about 15,000 car crashes each day in the United States, and the proportion of car crashes that are fatal is 0.00559 (based on data from the National Highway Traffic Safety Administration). Assume that each day, 1000 car crashes are randomly selected and the proportion of fatal car crashes is recorded.

a. What do you know about the mean of the sample proportions?

149
views
교과서 질문

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).


a. If a male passenger is randomly selected, find the probability that he can fit through the doorway without bending.

117
views
교과서 질문

College Presidents 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)?

143
views
교과서 질문

Sleepwalking Assume that 29.2% of people have sleepwalked (based on “Prevalence and Comorbidity of Nocturnal Wandering in the U.S. Adult General Population, by Ohayon et al., Neurology, Vol. 78, No. 20). Assume that in a random sample of 1480 adults, 455 have sleepwalked.


a. Assuming that the rate of 29.2% is correct, find the probability that 455 or more of the 1480 adults have sleepwalked.

106
views
교과서 질문

Curving Test Scores A professor gives a test and the scores are normally distributed with a mean of 60 and a standard deviation of 12. She plans to curve the scores.


a. If she curves by adding 15 to each grade, what is the new mean and standard deviation?

125
views
교과서 질문

Hershey Kisses Based on Data Set 38 “Candies” in Appendix B, weights of the chocolate in Hershey Kisses are normally distributed with a mean of 4.5338 g and a standard deviation of 0.1039 g.


a. What are the values of the mean and standard deviation after converting all weights of Hershey Kisses to z scores using z = (x - μ)/σ ?


b. The original weights are in grams. What are the units of the corresponding z scores?

211
views