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Chapter 6: Problem 15

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.) $$\begin{array}{|l|l|l|l|} \hline & \text { Mean } & \text { St. Dev. } & \text { Distribution } \\ \hline \text { Males } & 23.5 \mathrm{in} . & 1.1 \mathrm{in} . & \text { Normal } \\ \hline \text { Females } & 22.7 \mathrm{in} . & 1.0 \mathrm{in} . & \text { Normal } \\ \hline \end{array}$$ Find the probability that a female has a back-to-knee length between 22.0 in. and 24.0 in.

Short Answer

Expert verified
The probability is 0.6612.

Step by step solution

01

Gathering the Data

From the table, note the mean and standard deviation for females. The mean back-to-knee length for females is 22.7 inches, and the standard deviation is 1.0 inches.
02

Understanding the Normal Distribution

The data follows a normal distribution. The probability of a female having a back-to-knee length between specific values can be found using the properties of the normal distribution.
03

Standardize the Values

Convert the given values (22.0 inches and 24.0 inches) to z-scores using the formula: \[ z = \frac{x - \text{mean}}{\text{standard deviation}} \] For 22.0 inches: \[ z_1 = \frac{22.0 - 22.7}{1.0} = -0.7 \] For 24.0 inches: \[ z_2 = \frac{24.0 - 22.7}{1.0} = 1.3 \]
04

Finding the Probability from Z-scores

Using the standard normal distribution table, find the probabilities that correspond to the z-scores. For \( z_1 = -0.7 \), the cumulative probability is approximately 0.2420. For \( z_2 = 1.3 \), the cumulative probability is approximately 0.9032.
05

Calculate the Probability Between Two Z-scores

To find the probability of a female having a back-to-knee length between 22.0 inches and 24.0 inches, subtract the cumulative probability of \( z_1 \) from the cumulative probability of \( z_2 \): \[ P(22.0 < X < 24.0) = P(z_2) - P(z_1) = 0.9032 - 0.2420 = 0.6612 \]

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Key Concepts

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

Z-Score Calculation
A z-score is a way to describe a data point's relationship to the mean of a group of data points. It shows how many standard deviations a particular value is from the mean. To calculate the z-score, you use the formula:

\[ z = \frac{x - \text{mean}}{\text{standard deviation}} \]
Here, 'x' is the value you're examining, 'mean' is the average of the data set, and 'standard deviation' is how much the values usually vary from the mean. When given values 22.0 inches and 24.0 inches, you can use the formula with the mean (22.7 inches) and the standard deviation (1.0 inches) to find the z-scores:
  • For 22.0 inches:
    \[ z_1 = \frac{22.0 - 22.7}{1.0} = -0.7 \]
  • For 24.0 inches:
    \[ z_2 = \frac{24.0 - 22.7}{1.0} = 1.3 \]
Z-scores help simplify complex data and allow you to use standard normal distribution tables for finding probabilities.
Probability Between Two Values
Finding the probability between two values involves calculating the area under the curve of the normal distribution for the given range. Once you have the z-scores, these scores map directly to probabilities in the standard normal distribution. For instance, in the exercise, the z-scores are -0.7 and 1.3.

Use a z-table, which shows the cumulative probability associated with a z-score.
  • For \( z_1 = -0.7 \), the cumulative probability is approximately 0.2420.
  • For \( z_2 = 1.3 \), the cumulative probability is approximately 0.9032.
To find the probability between these two values, subtract the probability of \( z_1 \) from \( z_2 \):
\[ P(22.0 < X < 24.0) = P(z_2) - P(z_1) = 0.9032 - 0.2420 = 0.6612 \] This means there is a 66.12% chance a randomly selected female has a back-to-knee length between 22.0 inches and 24.0 inches.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution where the mean is 0 and the standard deviation is 1. This distribution is also known as the Z-distribution. The standard normal distribution is used for easy calculation of probabilities and is crucial in statistics.

When you convert values to z-scores, you effectively standardize your data to fit this distribution. This allows you to use standard normal distribution tables, which are widely available and simple to use. These tables give cumulative probabilities for each z-score, showing the probability a value is less than a given z-score.
  • A z-score of 0 corresponds to the mean.
  • Positive z-scores indicate values above the mean.
  • Negative z-scores indicate values below the mean.
Understanding the standard normal distribution is essential for calculating probabilities and interpreting various statistical tests accurately.

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Most popular questions from this chapter

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 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.

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. Between -4.27 and 2.34

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).

Assume that a randomly selected subject is given a bone density test. Bone density 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 bone density test score corresponding to the given information. Round results to two decimal places. If bone density scores in the bottom \(2 \%\) and the top \(2 \%\) are used as cutoff points for levels that are too low or too high, find the two readings that are cutoff values.

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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