/*! 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] | 91Ó°ÊÓ

91Ó°ÊÓ

The U.S. Air Force once used ACES-II ejection seats designed for men weighing between 140 lb and 211 lb. Given that women's weights are normally distributed with a mean of 171.1 Ib and a standard deviation of 46.1 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
55.7% of women's weights are within the limits. Many women were excluded.

Step by step solution

01

- Understand the Problem

Identify the key elements of the question. We need to find the percentage of women's weights that fall between 140 lb and 211 lb, given a normal distribution with a mean of 171.1 lb and a standard deviation of 46.1 lb.
02

- Convert to Z-Scores

Convert the weight limits to Z-scores using the formula: \[ Z = \frac{X - \mu}{\sigma} \]Where \( X \) is the value, \( \mu \) is the mean (171.1 lb),and \( \sigma \) is the standard deviation (46.1 lb).
03

- Calculate Z-Scores for 140 lb

Substitute 140 lb into the Z-score formula: \[ Z = \frac{140 - 171.1}{46.1} = -0.675 \]
04

- Calculate Z-Scores for 211 lb

Substitute 211 lb into the Z-score formula: \[ Z = \frac{211 - 171.1}{46.1} = 0.866 \]
05

- Find Probabilities Using Z-Score Table

Use a Z-score table or calculator to find the probabilities for Z-scores. For a Z-score of -0.675, the probability is approximately 0.250. For a Z-score of 0.866, the probability is approximately 0.807.
06

- Calculate the Percentage

Subtract the lower probability from the upper probability to find the percentage of women within these weight limits: \[ 0.807 - 0.250 = 0.557 \]Convert this probability to a percentage by multiplying by 100: \[ 0.557 \times 100 = 55.7\% \]
07

- Interpret the Results

Approximately 55.7% of women's weights fall within the given limits (140 lb to 211 lb). Therefore, many women were excluded from using these ejection seats under the 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 91Ó°ÊÓ!

Key Concepts

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

Z-score calculation
The Z-score is a measure that describes a value's position relative to the mean of a group of values. In simpler terms, it tells us how many standard deviations a specific value (X) is from the mean (µ). The formula for calculating the Z-score is: \[ Z = \frac{X - \text{µ}}{\text{σ}} \] The mean (µ) is the average value, and the standard deviation (σ) indicates how spread out the values are. In our example, we used the weights of women to calculate the Z-scores for 140 lb and 211 lb. For 140 lb, the Z-score is calculated as follows: \[ Z = \frac{140 - 171.1}{46.1} = -0.675 \] This means that 140 lb is -0.675 standard deviations from the mean. Similarly, for 211 lb, the Z-score is: \[ Z = \frac{211 - 171.1}{46.1} = 0.866 \] This indicates that 211 lb is 0.866 standard deviations above the mean.
Percentage calculation
Calculating the percentage is essential to understand the proportion of data within a certain range. In our exercise, we needed to find out what percentage of women fall between the weights of 140 lb and 211 lb. After calculating the Z-scores, we used a Z-score table or calculator to find the corresponding probabilities. For a Z-score of -0.675, the probability is approximately 0.250. For a Z-score of 0.866, the probability is approximately 0.807. To find the percentage of women within the weight range, we subtract the lower probability from the upper probability: \[ 0.807 - 0.250 = 0.557 \] We then convert this value to a percentage by multiplying by 100: \[ 0.557 \times 100 = 55.7\text{\textpercent} \] Therefore, 55.7% of women fall within the given weight limits.
Standard deviation
Standard deviation (σ) is a measure of the amount of variation or dispersion in a set of values. It tells us how much the values in a dataset deviate from the mean. A low standard deviation indicates that the values are close to the mean, while a high standard deviation suggests that the values are spread out over a wider range. In our exercise, the standard deviation of women's weights is 46.1 lb. This means that on average, women's weights differ from the mean weight (171.1 lb) by 46.1 lb. Standard deviation helps in understanding the distribution of data, which is crucial for making accurate calculations and predictions.
Mean calculation
The mean (µ) is the average value of a dataset and is calculated as the sum of all values divided by the number of values. It serves as a measure of the central tendency of the data. In our exercise, the mean weight of women is given as 171.1 lb. This value is central to our calculations and helps in determining the Z-scores. The mean provides a reference point for understanding how individual values compare to the overall dataset. To calculate the mean manually, you add up all the weights and divide by the number of women. For instance, if the weights are 140 lb, 150 lb, and 170 lb: \[ \text{Mean} = \frac{140 + 150 + 170}{3} = 153.33 \text{ lb} \] Knowing the mean is essential for further statistical calculations, such as finding the standard deviation and Z-scores.

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

Common tests such as the SAT, ACT, LSAT, and MCAT tests use multiple choice test questions, each with possible answers of a, b, c, d, e, and each question has only one correct answer. For people who make random guesses for answers to a block of 100 questions, identify the values of \(p, q, \mu,\) and \(\sigma .\) What do \(\mu\) and \(\sigma\) measure?

Notation In general, what do the symbols \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) represent? What are the values of \(\mu_{\bar{x}}\) and \(\sigma_{\bar{x}}\) for samples of size 64 randomly selected from the population of IQ scores with population mean of 100 and standard deviation of \(15 ?\)

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.

In the year that this exercise was written, there were 879 challenges made to referee calls in professional tennis singles play. Among those challenges, 231 challenges were upheld with the call overturned. Assume that in general, \(25 \%\) of the challenges are successfully upheld with the call overturned. a. If the \(25 \%\) rate is correct, find the probability that among the 879 challenges, the number of overturned calls is exactly \(231 .\) b. If the \(25 \%\) rate is correct, find the probability that among the 879 challenges, the number of overturned calls is 231 or more. If the \(25 \%\) rate is correct, is 231 overturned calls among 879 challenges a result that is significantly high?

In a study of 420,095 cell phone users in Denmark, it was found that 135 developed cancer of the brain or nervous system. For those not using cell phones, there is a 0.000340 probability of a person developing cancer of the brain or nervous system. We therefore expect about 143 cases of such cancers in a group of 420,095 randomly selected people. a. Find the probability of 135 or fewer cases of such cancers in a group of 420,095 people. b. What do these results suggest about media reports that suggest cell phones cause cancer of the brain or nervous system?

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.