/*! 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 17 Use the data in the table below ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.) $$ \begin{aligned} &\text { Sitting Back-to-Knee Length (inches) }\\\ &\begin{array}{l|c|c|c} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \text { in. } & 1.1 \text { in. } & \text { Normal } \\ \hline \text { Females } & 22.7 \text { in. } & 1.0 \text { in. } & \text { Normal } \\ \hline \end{array} \end{aligned} $$ For males, find \(P_{90}\), which is the length separating the bottom \(90 \%\) from the top \(10 \%\).

Short Answer

Expert verified
The 90th percentile length for sitting back-to-knee length in males is approximately 24.908 inches.

Step by step solution

01

Understand the Given Data

The data provides the mean and standard deviation for the sitting back-to-knee length of males. The mean is 23.5 inches, and the standard deviation is 1.1 inches. The data follows a normal distribution.
02

Define the Required Percentile

We need to find the 90th percentile (\r \(P_{90}\)\r), which means we are looking for the length that separates the bottom 90% from the top 10% of the distribution.
03

Use the Z-Score for the 90th Percentile

For a normal distribution, the Z-score corresponding to the 90th percentile is approximately 1.28. This can be found using standard normal distribution tables or a calculator.
04

Calculate the Percentile Length

Use the formula to convert the Z-score into a real-world measurement: \r \r \r \r\(x = \text{mean} + (Z \times \text{standard deviation})\)\r \r \r \rInsert the known values: \r$$P_{90} = 23.5 \text{ in. } + (1.28 \times 1.1 \text{ in. }) = 24.908 \text{ in. }$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

normal distribution

Understanding normal distribution is crucial for many topics in statistics. The normal distribution is a bell-shaped curve. It's symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.

Key characteristics include:
  • Mean: The average of all data points.
  • Median: The middle value when data points are ordered.
  • Mode: The most frequently occurring value(s).
  • Standard Deviation (σ): A measure of the amount of variation in a set of values.

The area under the normal distribution curve represents probabilities. For example,
  • About 68% of data falls within one standard deviation of the mean.
  • About 95% fall within two standard deviations.
  • About 99.7% fall within three standard deviations.

This distribution helps in understanding how values are dispersed and predicting outcomes.
z-score

A Z-score, or standard score, indicates how many standard deviations an element is from the mean. If the Z-score is 0, it means the score is identical to the mean.

To calculate a Z-score, use the formula:

\ Z = \frac{x - \text{mean}}{\text{standard deviation}} \

where:
  • \( x \) is the value of the element.
  • Mean is the average of all values.
  • Standard Deviation is the measure of the spread of values.

For example, if the mean sitting back-to-knee length for males is 23.5 inches, a Z-score of 1.28 means the length is 1.28 standard deviations above the mean.
percentile rank

Percentile rank indicates the relative position of a score within a distribution. It tells you the percentage of data points that lie below a specific value.

For example, the 90th percentile (\( P_{90} \)) means that 90% of the data falls below this value.

To find a percentile using the Z-score, use the standard normal distribution table or a calculator. For the 90th percentile, the Z-score is approximately 1.28. Using this Z-score in the context of a normal distribution, we can then calculate the specific value in real-world terms.
standard deviation

Standard deviation (σ) measures the spread or dispersion of a set of data points. A low standard deviation means the data points are close to the mean, while a high standard deviation means they are spread out over a wide range.

To calculate the standard deviation, follow these steps:
  • Find the mean (average) of the data set.
  • Subtract the mean from each data point and square the result.
  • Find the average of these squared differences.
  • Take the square root of this average.

For example: With a mean sitting back-to-knee length of 23.5 inches and a standard deviation of 1.1 inches for males, this standard deviation tells you how much the length varies from the mean on average.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Identify the two requirements necessary for a normal distribution to be a standard normal distribution.

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? b. Is it fair to curve by adding 15 to each grade? Why or why not? c. If the grades are curved so that grades of \(\mathrm{B}\) are given to scores above the bottom \(70 \%\) and below the top \(10 \%\), find the numerical limits for a grade of \(\mathrm{B}\). d. Which method of curving the grades is fairer: adding 15 to each original score or using a scheme like the one given in part (c)? Explain.

In a survey of 1002 people, 701 said that they voted in a recent presidential election (based on data from ICR Research Group). Voting records showed that \(61 \%\) of eligible voters actually did vote. a. Given that \(61 \%\) of eligible voters actually did vote, find the probability that among 1002 randomly selected eligible voters, at least 701 actually did vote. b. What does the result suggest?

After 1964 , quarters were manufactured so that their weights have a mean of \(5.67 \mathrm{~g}\) and a standard deviation of \(0.06 \mathrm{~g}\). Some vending machines are designed so that you can adjust the weights of quarters that are accepted. If many counterfeit coins are found, you can narrow the range of acceptable weights with the effect that most counterfeit coins are rejected along with some legitimate quarters. a. If you adjust your vending machines to accept weights between \(5.60 \mathrm{~g}\) and \(5.74 \mathrm{~g}\), what percentage of legal quarters are rejected? Is that percentage too high? b. If you adjust vending machines to accept all legal quarters except those with weights in the top \(2.5 \%\) and the bottom \(2.5 \%\), what are the limits of the weights that are accepted?

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of \(1 .\) In each case, draw a graph, then find the probability of the given bone density test scores. If using technology instead of Table A-2, round answers to four decimal places. $$ \text { Less than }-1.23 $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.