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Chapter 6: Problem 23

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. 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? b. If the height requirements are changed to exclude the tallest \(50 \%\) of men and the shortest \(5 \%\) of men, what are the new height requirements?

Short Answer

Expert verified
a. The percentage of men meeting the height requirement is almost 0\text%. b. The new height requirements are between 63 in. and 68.6 in.

Step by step solution

01

Title - Convert Heights to Z-Scores for Men

We first convert the height limits (56 in. and 62 in.) to z-scores using the formula: \[ z = \frac{x - \mu}{\sigma} \] Where \( x \) is the height, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. For men, \( \mu = 68.6 \) and \( \sigma = 2.8 \).
02

Title - Calculate Z-Score for Lower Limit (56 in)

For \( x = 56 \) in, the z-score is: \[ z = \frac{56 - 68.6}{2.8} \approx -4.5 \]
03

Title - Calculate Z-Score for Upper Limit (62 in)

For \( x = 62 \) in, the z-score is: \[ z = \frac{62 - 68.6}{2.8} \approx -2.36 \]
04

Title - Look Up Z-Scores in Z-Table

Using the z-table, find the percentage corresponding to \( z = -4.5 \) and \( z = -2.36 \). These z-scores correspond to almost 0\text% and 0.009\text% respectively. The percentage of men meeting this height requirement is the difference between these two values.
05

Title - Calculate the Percentage of Men Meeting Height Requirement

Subtract the cumulative probability of \( z = -4.5 \) from \( z = -2.36 \). The difference is negligible because both of these values are almost 0\text%. Therefore, the percentage of men meeting the height requirement is essentially 0\text%.
06

Title - Conclusion for Part A

The percentage of men who meet the height requirement is extremely low, suggesting that Mickey Mouse characters are predominantly not men.
07

Title - Determine the New Height Limits to Exclude Tallest 50% of Men

To exclude the tallest 50% of men, find the z-score that corresponds to the 50th percentile, which is \( z = 0 \). Use the z-score formula to find x: \[ x = z \cdot \sigma + \mu \]. For men, \( x = 0 \cdot 2.8 + 68.6 = 68.6 \) inches.
08

Title - Determine the New Height Limit to Exclude Shortest 5% of Men

To exclude the shortest 5% of men, find the z-score that corresponds to the 5th percentile, which is about \( z = -1.645 \). Use the z-score formula to find x: \[ x = -1.645 \cdot 2.8 + 68.6 \approx 63.0 \] inches.
09

Title - Conclusion for Part B

The new height requirements to exclude the tallest 50% and the shortest 5% of men are between 63 in. and 68.6 in.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

z-scores in statistics education
In statistics, z-scores are a measure of how many standard deviations an element is from the mean. This concept is crucial for comparing different data points within a distribution. The z-score formula is:

\( z = \frac{x - \mu}{\sigma} \)

Here, \( x \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation of the data set.

Z-scores allow us to normalize scores on different scales, providing a way to compare them directly. For instance, to determine how many men meet the height requirement for a job, we convert the height limits into z-scores.

Converting heights into z-scores standardizes the measure, which helps in determining the probability of specific heights within the population. For example, men's heights usually follow a normal distribution with a mean (\( \mu \)) of 68.6 inches and a standard deviation (\( \sigma \)) of 2.8 inches.
height distribution
Understanding height distribution involves analyzing how heights are spread out within a given population. It is common to assume a normal distribution (bell curve) for heights, as most people's heights are around the average, with fewer people being extremely tall or short. In this scenario, we are considering the height distributions for men and women:
  • Men's heights: mean (\( \mu \)) = 68.6 inches, std deviation (\( \sigma \)) = 2.8 inches
  • Women's heights: mean (\( \mu \)) = 63.7 inches, std deviation (\( \sigma \)) = 2.9 inches

To find the percentage of individuals within a specific height range (e.g., 56 inches to 62 inches for a job role), we use z-scores. These help us determine the proportion of the population within those limits by referencing the standard normal distribution table, simplifying the calculation of probabilities.
percentiles in statistics
Percentiles indicate the relative standing of a value within a data set. A percentile represents the value below which a given percentage of observations in a group of observations falls. For example, if you are in the 90th percentile for height, it means you are taller than 90% of the population.

In our task, we use percentiles to redefine the height requirements for a job to exclude the tallest 50% and the shortest 5% of men. This involves looking up corresponding z-scores in a z-table:
  • 50th percentile corresponds to a z-score of \( 0 \)
  • 5th percentile corresponds to a z-score of around \( -1.645 \)

These z-scores help in determining new height limits when applied to the normal distribution's mean and standard deviation. Calculating new limits ensures that only the selected percentage from a population meets the criteria. For example, the height at the 5th percentile allows us to determine the cutoff height for the shortest 5% of men.

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Most popular questions from this chapter

Common tests such as the SAT, ACT, Law School Admission test (LSAT), and Medical College Admission Test (MCAT) use multiple choice test questions, each with possible answers of a, b, c, d, e, and each question has only one correct answer. We want to find the probability of getting at least 25 correct answers for someone who makes random guesses for answers to a block of 100 questions. If we plan to use the methods of this section with a normal distribution used to approximate a binomial distribution, are the necessary requirements satisfied? Explain.

Elevator Safety Example 2 referred to an elevator with a maximum capacity of 4000 ib. When rating elevators, it is common to use a \(25 \%\) safety factor, so the elevator should actually be able to carry a load that is \(25 \%\) greater than the stated limit. The maximum capacity of 4000 ib becomes 5000 ib after it is increased by \(25 \%,\) so 27 adult male passengers can have a mean weight of up to 185 ib. If the elevator is loaded with 27 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than 185 ib. (As in Example \(2,\) assume that weights of males are normally distributed with a mean of 189 ib and a standard deviation of 39 Ib.) Does this elevator appear to be safe?

Loading Aircraft Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The Bombardier Dash 8 aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6200 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than \(6200 \mathrm{lb} / 37=167.6 \mathrm{lb}\). What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? 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 I "Body Data" in Appendix B).

A common design requirement is that an environment must fit the range of people who fall between the Sth percentile for women and the 95 th percentile for men. In designing an assembly work table, we must consider sitting knee height, which is the distance from the bottom of the feet to the top of the knee. Males have sitting knee heights that are normally distributed with a mean of 21.4 in. and a standard deviation of 1.2 in.; females have sitting knee heights that are normally distributed with a mean of 19.6 in. and a standard deviation of 1.1 in. (based on data from the Department of Transportation). a. What is the minimum table clearance required to satisfy the requirement of fitting \(95 \%\) of men? Why is the 95 th percentile for women ignored in this case? b. The author is writing this exercise at a table with a clearance of 23.5 in. above the floor. What percentage of men fit this table, and what percentage of women fit this table? Does the table appear to be made to fit almost everyone?

Using the Central Limit Theorem assume that females have pulse rates that are normally distributed with a mean of 74.0 beats per minute and a standard deviation of 12.5 beats per minute (based on Data Set 1 "Body Data" in Appendix \(B\) ). a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 80 beats per minute. b. If 16 adult females are randomly selected, find the probability that they have pulse rates with a mean less than 80 beats per minute. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed \(30 ?\)
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