/*! 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 27 The U.S. Air Force once used ACE... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 27

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?

Short Answer

Expert verified
Approximately 55.76% of women have weights between 140 lb and 211 lb. Thus, around 44.24% of women were excluded.

Step by step solution

01

- Understand the problem

The objective is to find the percentage of women whose weights fall between 140 lb and 211 lb. The weights are normally distributed with a mean (渭) of 171.1 lb and a standard deviation (蟽) of 46.1 lb.
02

- Calculate Z-scores

To find the percentage, first calculate the Z-scores for 140 lb and 211 lb.The formula for a Z-score is: Z = (X - 渭) / 蟽For 140 lb: Z1 = (140 - 171.1) / 46.1 鈮� -0.674For 211 lb: Z2 = (211 - 171.1) / 46.1 鈮� 0.867
03

- Use the Z-scores to find probabilities

Using the Z-score table or a standard normal distribution table, find the area to the left of Z1 and Z2. P(Z < -0.674) 鈮� 0.2502 P(Z < 0.867) 鈮� 0.8078
04

- Calculate the probability between the Z-scores

Find the difference between the probabilities to get the percentage of women within the weight limits. P(-0.674 < Z < 0.867) = P(Z < 0.867) - P(Z < -0.674) 鈮� 0.8078 - 0.2502 = 0.5576
05

- Convert to percentage

Convert the probability to a percentage by multiplying by 100. 0.5576 * 100 = 55.76%So, approximately 55.76% of women have weights within the specified limits.
06

- Evaluate the exclusion

Since 55.76% of women are within the specified weight limits, this means that around 44.24% of women were excluded with those past specifications.

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 91影视!

Key Concepts

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

Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a continuous probability distribution that is symmetric about the mean. Most of the data points cluster around the mean, and the probabilities for values furthest from the mean taper off equally on both sides.
Think of a bell-shaped curve when you hear 'normal distribution'.
  • The center of this curve is the mean \(\text{渭}\).
  • The spread of the curve is determined by the standard deviation \(\text{蟽}\).
  • For any given point \(X\), the Z-score tells us how many standard deviations \(X\) is away from the mean.
By understanding the normal distribution, we can use it to predict the probability of events within any specified range.
Z-scores
Z-scores enable us to determine the probability that a data point lies within a specific range relative to the mean of a distribution. The Z-score is calculated using the formula:
\[ Z = \frac{X - \text{渭}}{\text{蟽}} \]
Where:
  • \(X\) is the value in question.
  • \(\text{渭}\) is the mean.
  • \(\text{蟽}\) is the standard deviation.
For example, in the given exercise, to find the Z-score for a weight of 140 lb:
\[ Z = \frac{140 - 171.1}{46.1} \approx -0.674 \]
Likewise, for 211 lb:
\[ Z = \frac{211 - 171.1}{46.1} \approx 0.867 \]
These Z-scores tell us how far (in terms of standard deviations) 140 lb and 211 lb are from the mean.
Probability Calculation
Once Z-scores are determined, we can use the standard normal distribution table to find the probabilities associated with those scores.
Here鈥檚 how:
  • Find the probability that a Z-score is less than or equal to a given value (also known as the cumulative probability). For -0.674, this is approximately 0.2502. For 0.867, it is approximately 0.8078.
  • Subtract the smaller cumulative probability from the larger one to find the probability that the value falls between these two Z-scores.
For the given exercise:
\[ P(-0.674 < Z < 0.867) = P(Z < 0.867) - P(Z < -0.674) \approx 0.8078 - 0.2502 = 0.5576 \]
This means the probability that a woman's weight falls between 140 lb and 211 lb is approximately 55.76%.
Standard Deviation
Standard deviation, often represented as \(\text{蟽}\), is a measure of the amount of variation or dispersion of a set of values.
In simpler terms:
  • A low standard deviation means that most of the data points are close to the mean.
  • A high standard deviation means that the data points are spread out over a wider range.
For the given exercise:
  • The mean weight of women is 171.1 lb.
  • The standard deviation is 46.1 lb, indicating a moderate spread around the mean weight.
Understanding standard deviation is crucial because it helps us understand how spread out the weights are and how likely it is for a weight to fall within a certain range. In our problem, it allowed us to calculate the percentage of women within the specific weight limits.

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

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. $$ \text { Between }-3.00 \text { and } 3.00 $$

Because they enable efficient procedures for evaluating answers, multiple choice questions are commonly used on standardized tests, such as the SAT or ACT. Such questions typically have five choices, one of which is correct. Assume that you must make random guesses for two such questions. Assume that both questions have correct answers of " \(a\)." a. After listing the 25 different possible samples, find the proportion of correct answers in each sample, then construct a table that describes the sampling distribution of the sample proportions of correct responses. b. Find the mean of the sampling distribution of the sample proportion. c. Is the mean of the sampling distribution [from part (b)] equal to the population proportion of correct responses? Does the mean of the sampling distribution of proportions always equal the population proportion?

Assume that cans of Coke are filled so that the actual amounts are normally distributed with a mean of \(12.00 \mathrm{oz}\) and a standard deviation of \(0.11\) oz. a. Find the probability that a single can of Coke has at least \(12.19 \mathrm{oz}\). b. The 36 cans of Coke in Data Set 26 "Cola Weights and Volumes" in Appendix \(\mathrm{B}\) have a mean of \(12.19\) oz. Find the probability that 36 random cans of Coke have a mean of at least \(12.19 \mathrm{oz}\) c. Given the result from part (b), is it reasonable to believe that the cans are actually filled with a mean equal to \(12.00\) oz? If the mean is not equal to \(12.00 \mathrm{oz}\), are consumers being cheated?

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).

Assume that females have pulserates 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 鈥淏ody Data鈥� in Appendix B). a. If 1 adult female is randomly selected, find the probability that her pulse rate is greater than 70 beats per minute. b. If 25 adult females are randomly selected, find the probability that they have pulse rates with a mean greater than 70 beats per minute. c. Why can the normal distribution be used in part (b), even though the sample size does not exceed \(30 ?\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation
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.