/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 Use the data in the table below ... [FREE SOLUTION] | 91Ó°ÊÓ

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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.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Instead of using \(0.05\) for identifying significant values, use the criteria th?t a value \(x\) is significantly high if \(P(x\) or greater \() \leq 0.01\) and a value is significantly low if \(P(x\) or less \() \leq 0.01\). Find the back-to- knee lengths for males, separating significant values from those that are not significant. Using these criteria, is a male back-to-knee length of 26 in. significantly high?

Short Answer

Expert verified
No, a male back-to-knee length of 26 inches is not significantly high.

Step by step solution

01

Understand the Normal Distribution

The sitting back-to-knee lengths for both males and females are normally distributed. The mean length (μ) for males is 23.5 inches, and the standard deviation (σ) is 1.1 inches.
02

Setup the Criteria

The criteria given state that a value is significantly high if the probability of a value x or greater is less than or equal to 0.01, and significantly low if the probability of a value x or less is less than or equal to 0.01.
03

Convert Length to Z-Score

The Z-score formula is given by \( Z = \frac{x - \text{mean}}{\text{standard deviation}} \). For a back-to-knee length of 26 inches, calculate the Z-score.
04

Calculate the Z-Score

For males, mean (μ) = 23.5 and standard deviation (σ) = 1.1. Using these: \[ Z = \frac{26 - 23.5}{1.1} = \frac{2.5}{1.1} \approx 2.27 \]
05

Find the Probability from Z-Score

Using the Z-score table, find the probability corresponding to the Z-score of 2.27.
06

Determine Significant Values

From the Z-score table, the probability for Z = 2.27 is about 0.9884. Therefore, \( P(X \geq 26) = 1 - 0.9884 = 0.0116 \). Since 0.0116 is greater than 0.01, 26 inches is not significantly high.

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Key Concepts

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

Z-score calculation
The Z-score is a way to measure how far a data point is from the mean, expressed in terms of standard deviations. It is calculated using the formula: \( Z = \frac{x - \text{mean}}{\text{standard deviation}} \). In the context of the exercise, if you want to determine whether a back-to-knee length of 26 inches is significant, you first calculate its Z-score. With a mean (μ) of 23.5 inches and a standard deviation (σ) of 1.1 inches, the Z-score for 26 inches can be computed as follows: \( Z = \frac{26 - 23.5}{1.1} \ = \frac{2.5}{1.1} \ \roughly\thinspace 2.27 \). This Z-score helps us understand how extreme or unusual a measurement is.
significance level
A significance level is a critical value that helps decide whether to reject a null hypothesis. It is the threshold used in hypothesis testing. Typically, a significance level of 0.05 is common, but in this exercise, the criteria are adjusted. A value is significantly high if the probability associated with it is \( \text{P}(x \text{ or greater}) \ \thinspace \thinspace\thinspace \leq 0.01 \), and significantly low if \( \text{P}(x \text{ or less}) \ \ \thinspace\thinspace\thinspace\leq 0.01 \). This more stringent criterion allows for higher confidence when determining the extremity of values.
probability distribution
A probability distribution describes how probabilities are distributed over the possible values of a random variable. In this problem, the distribution is normal, meaning it follows a bell-shaped curve. Most values cluster around the mean, with fewer values appearing as you move away. For normally distributed data, the probabilities decrease symmetrically as you move away from the mean. Understanding the probability distribution allows us to determine the likelihood of obtaining values within a specific range.
standard deviation
Standard deviation is a measure of how dispersed the values in a dataset are relative to the mean. A lower standard deviation indicates that the values tend to be close to the mean, whereas a higher standard deviation means they are more spread out. In this exercise, the standard deviation for males' sitting back-to-knee length is 1.1 inches, which tells us about the variability in the measurements. Knowing the standard deviation is crucial for calculating the Z-score and assessing the significance of values.
mean value
The mean value, also known as the average, is a central point around which the data points are distributed. For the measurement of males' sitting back-to-knee length, the mean is 23.5 inches. The mean provides a reference point for understanding whether a particular measurement is above, below, or exactly at the expected average. Calculating the mean involves summing all values and dividing by the number of values. In normally distributed data, the mean is located at the peak of the bell curve.
normal distribution curve
The normal distribution curve, often referred to as the bell curve, is defined by a symmetrical shape with the highest point at the mean. Most data points lie within one standard deviation from the mean, creating a smooth, continuous line. The tails of the curve extend infinitely in both directions, but almost all data points lie within three standard deviations. This characteristic shape allows us to use Z-scores and standard deviations to find probabilities and make predictions about data points. Graphing the data in a normal distribution helps visualize the spread and central tendency of the dataset.
anthropometric data analysis
Anthropometric data analysis involves measuring and analyzing the dimensions of the human body. This data is critical in fields like ergonomics, biomechanics, and design engineering. For example, designing seats for airplanes, trains, and classrooms uses anthropometric data to ensure comfort and functionality for the average user. In this problem, sitting back-to-knee length data helps inform how to design seats that fit a wide range of users. Using statistical methods, such as Z-scores and significance levels, we can interpret the data and make informed decisions about design criteria.

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Most popular questions from this chapter

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.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ Instead of using \(0.05\) for identifying significant values, use the criteria that a value \(x\) is significantly high if \(P(x\) or greater \() \leq 0.025\) and a value is significantly low if \(P(x\) or less \() \leq 0.025 .\) Find the female back-to-knee length, separating significant values from those that are not significant. Using these criteria, is a female back-to-knee length of 20 in. significantly low?

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 breadths 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? b. Assume that the Connecticut's Eversource company employs 36 men who work in manholes. If 36 men are randomly selected, what is the probability that their mean shoulder breadth is less than \(18.5\) in.? Does this result suggest that money can be saved by making smaller manholes with a diameter of \(18.5 \mathrm{in} . ?\) Why or why not?

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 \(6200 \mathrm{lb}\). 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 \(189 \mathrm{lb}\) and a standard deviation of \(39 \mathrm{lb}\) (based on Data Set 1 "Body Data" in Appendix B).

Do the following: If the requirements of \(n p \geq 5\) and \(n q \geq 5\) are both satisfied, estimate the indicated probability by using the normal distribution as an approximation to the binomial distribution; if \(n p<5\) or \(n q<5\), then state that the normal approximation should not be used. With \(n=20\) births and \(p=0.512\) for a boy, find \(P(\) fewer than 8 boys).

The U.S. Air Force once used ACES-II ejection seats designed for men weighing between \(140 \mathrm{lb}\) and \(211 \mathrm{lb}\). Given that women's weights are normally distributed with a mean of \(171.1 \mathrm{lb}\) and a standard deviation of \(46.1 \mathrm{lb}\) (based on data from the National Health Survey), what percentage of women have weights that are within those limits? Were many women excluded with those past specifications?

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