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Chapter 6: Problem 16

Use the data in the table below for sitting adult males and females (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. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Find the probability that a male has a back-to-knee length between \(22.0\) in. and \(24.0\) in.

Short Answer

Expert verified
0.5867

Step by step solution

01

- Understand the Distribution

The back-to-knee length for males follows a normal distribution with a mean (\(\mu\)) of 23.5 inches and a standard deviation (\(\sigma\)) of 1.1 inches.
02

- Standardize the Data

Convert the raw scores (22.0 inches and 24.0 inches) to z-scores using the formula: \[ z = \frac{x - \mu}{\sigma} \], where \(\mu=23.5\) and \(\sigma=1.1\).
03

- Calculate the Z-Score for 22.0 inches

For 22.0 inches: \[ z_1 = \frac{22.0 - 23.5}{1.1} = -1.36 \]
04

- Calculate the Z-Score for 24.0 inches

For 24.0 inches: \[ z_2 = \frac{24.0 - 23.5}{1.1} = 0.45 \]
05

- Use the Z-Score Table

Find the probability corresponding to \( z = -1.36 \) and \( z = 0.45 \) using the standard normal distribution table or a Z-score calculator.
06

- Determine the Probabilities

From the Z-Score table, the probability for \( z_1 = -1.36 \) is approximately 0.0869, and for \( z_2 = 0.45 \) it is approximately 0.6736.
07

- Calculate the Desired Probability

The probability that a male has a back-to-knee length between 22.0 inches and 24.0 inches is: \[ P(z_1 < Z < z_2) = P(Z < 0.45) - P(Z < -1.36) = 0.6736 - 0.0869 = 0.5867 \]

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Key Concepts

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

Normal Distribution
In statistics, many real-world measurements follow a normal distribution pattern. This is also known as the bell curve because of its shape. Heights, test scores, and, in this case, the back-to-knee length of sitting adults, commonly follow this pattern. A normal distribution is characterized by two parameters: the mean (average) and the standard deviation. The mean determines the center of the distribution, while the standard deviation measures how spread out the values are. Most values cluster around the mean.
Z-Score Calculation
Z-scores help us understand where a particular value stands in a distribution. Z-scores represent how many standard deviations away a value is from the mean. The formula to calculate a z-score is: \( z = \frac{x - \text{mean}}{\text{standard deviation}} \). For instance, if you want to find the z-score for a back-to-knee length of 22.0 inches, given a mean of 23.5 inches and a standard deviation of 1.1 inches, you subtract the mean from the value and then divide by the standard deviation. This converts the measurement into a standardized form.
Probability Calculation
Understanding the probability is key in statistics. Once you have the z-scores, you can find the corresponding probabilities using a standard normal distribution table or a calculator. The z-score table shows the probability that a standard normal variable is less than or equal to a given value. For example, if you have a z-score of -1.36, you can look up this value in the table to find approximately 0.0869. This represents the probability that a randomly selected value from the normal distribution is less than 22.0 inches.
Standard Deviation
The standard deviation is a measure of how spread out numbers are in a dataset. In a normal distribution, about 68% of values will fall within one standard deviation of the mean, 95% within two, and 99.7% within three. The formula for standard deviation is \( \text{SD} = \frac{\text{Sum of (x - mean)^2}}{N} \), where \( N \) is the number of values. In our example, with a standard deviation of 1.1 inches for males' back-to-knee length, most values will fall within 22.4 to 24.6 inches (23.5卤1.1).
Mean
The mean is the average of a set of numbers and is calculated by summing all the values and dividing by the number of values. In a normal distribution, the mean is the point where the curve is highest. For our example, the mean back-to-knee length for adult males is 23.5 inches. This tells us that most men's back-to-knee lengths will be around this value. The mean provides a central value for the distribution and is essential in both calculating z-scores and understanding the overall distribution.

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Most popular questions from this chapter

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? b. If we stipulate that a baby is premature if the duration of pregnancy is in the lowest \(3 \%\), find the duration that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.

Use the population of \(\\{34,36,41,51\\}\) of the amounts of caffeine random samples of size \(n=2\) are selected with replacement. a. After identifying the 16 different possible samples, find the mean of each sample, then construct a table representing the sampling distribution of the sample mean. In the table, combine values of the sample mean that are the same. (Hint: See Table \(6-3\) in Example 2 on page 258.) b. Compare the mean of the population \(\\{34,36,41,51\\}\) to the mean of the sampling distribution of the sample mean. c. Do the sample means target the value of the population mean? In general, do sample means make good estimators of population means? Why or why not?

Use these parameters (based on Data Set 1 "Body Data" in Appendix B): \- Men's heights are normally distributed with mean \(68.6 \mathrm{in.}\) and standard deviation \(2.8 \mathrm{in} .\) \- Women's heights are normally distributed with mean 63.7 in. and standard deviation \(2.9 \mathrm{in}\). 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?

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.

Use the data in the table below for sitting adult males and females (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. (Hint: Draw a graph in each case.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Find the probability that a female has a back-to-knee length between \(22.0 \mathrm{in}\). and \(24.0 \mathrm{in}\).
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