/*! 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 62 The Wall Street Journal (Februar... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The Wall Street Journal (February 15,1972 ) reported that General Electric was sued in Texas for sex discrimination over a minimum height requirement of 5 feet, 7 inches. The suit claimed that this restriction eliminated more than \(94 \%\) of adult females from consideration. Let \(x\) represent the height of a randomly selected adult woman. Suppose that \(x\) is approximately normally distributed with mean 66 inches (5 ft. 6 in.) and standard deviation 2 inches. a. Is the claim that \(94 \%\) of all women are shorter than \(5 \mathrm{ft}\). 7 in. correct? b. What proportion of adult women would be excluded from employment as a result of the height restriction?

Short Answer

Expert verified
The claim that 94% of all women are shorter than 5 ft. 7 in. is incorrect. Our calculations show that approximately 69.15% of adult women are shorter than 67 inches. Additionally, the proportion of adult women who would be excluded from employment as a result of the height restriction is also 69.15%.

Step by step solution

01

Convert the height to inches

First, we need to convert the height \(5 \mathrm{ft}. 7 \mathrm{in}\). to inches: \(5 \times 12 + 7 = 60 + 7 = 67 \mathrm{in}\)
02

Determine the z-score for the given height

To determine the percentage of women shorter than the given height, we need to calculate the z-score. A z-score represents how many standard deviations away a certain value is from the mean of a normal distribution. The z-score can be calculated as follows: \[ z = \frac{x - \mu}{\sigma} \] where: \(x = 67 \mathrm{in}\), the given height \(\mu = 66 \mathrm{in}\), the mean height of adult women \(\sigma = 2 \mathrm{in}\), the standard deviation of height of adult women
03

Calculate the z-score

Using the formula from Step 2, calculate the z-score: \[ z = \frac{67 - 66}{2} = \frac{1}{2} = 0.5 \]
04

Calculate the proportion below the given height with cumulative distribution function (CDF)

Now we can use the cumulative distribution function (CDF) to calculate the proportion of women with a height below 67 inches (5 ft. 7 in.). The CDF can be calculated using a Z-table or using a calculator with a built-in CDF function. Here, we will use a Z-table. Find the value corresponding to a z-score of 0.5 in the Z-table. The value is around 0.6915. This means that approximately \(69.15 \%\) of adult women have a height below 67 inches.
05

Compare our value with the given claim

The claim states that more than \(94 \%\) of adult women are shorter than 67 inches. However, our calculation shows that approximately \(69.15 \%\) of adult women are shorter than 67 inches. Therefore, the claim is incorrect.
06

Calculate the proportion of adult women excluded from employment

The proportion of adult women excluded from employment due to the height restriction is equivalent to the proportion of women shorter than 67 inches. Thus, approximately \(69.15 \%\) of adult women would be excluded from employment as a result of the height restriction. In conclusion, the claim that \(94 \%\) of all women are shorter than \(5 \mathrm{ft}. 7 \mathrm{in}\). is incorrect. Our calculations show that approximately \(69.15 \%\) of adult women are shorter than 67 inches. Additionally, the proportion of adult women who would be excluded from employment as a result of the height restriction is also \(69.15 \%\).

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
The Z-score is a fundamental concept in statistics that helps us understand how a particular value relates to the mean of a dataset. It tells us how many standard deviations a specific value is from the mean. For example, if a woman's height is 67 inches in a population where the mean height is 66 inches with a standard deviation of 2 inches, we can calculate the Z-score using the formula:
\[ z = \frac{x - \mu}{\sigma} \]
Here, \(x\) is the woman's height, \(\mu\) is the mean height, and \(\sigma\) is the standard deviation. In this scenario, the Z-score is:
\[ z = \frac{67 - 66}{2} = 0.5 \]
This result indicates that 67 inches is 0.5 standard deviations above the mean. A positive Z-score indicates a value above the mean, while a negative Z-score indicates a value below the mean.
Standard Deviation
Standard deviation is a measure of how spread out the values in a dataset are. In a normal distribution, it tells us the average distance between each data point and the mean.
A smaller standard deviation means that the data points tend to be closer to the mean, while a larger one indicates that they are more spread out.
In our context, where the heights of women are normally distributed with a mean of 66 inches and a standard deviation of 2 inches, the standard deviation informs us about the variability of women's heights. Knowing this helps us to understand the typical range of heights in the population.
It's crucial because it directly impacts the calculation of Z-scores. If the standard deviation were larger, the same height would have a smaller Z-score, indicating less of a deviation from the average height.
Cumulative Distribution Function (CDF)
The Cumulative Distribution Function (CDF) is a tool used in statistics to determine the probability that a random variable is less than or equal to a certain value. It is integral to probability calculations with normal distributions.
In the problem involving women's height, we used the CDF to find the proportion of women shorter than 67 inches. With a Z-score of 0.5, we looked up this value in a Z-table, which lists probabilities associated with different Z-scores.
The Z-table translates our Z-score of 0.5 into a probability of approximately 0.6915, or 69.15%. This means 69.15% of women are shorter than 67 inches, according to this normal distribution.
Using the CDF for such calculations is standard practice when analyzing normally distributed data, providing a practical way to assess probabilities and proportions.
Height Restriction Analysis
Height restriction analysis examines how specific criteria might serve as a barrier, impacting certain populations. In this case, a height restriction of 5 ft. 7 in. (or 67 inches) was imposed by General Electric. The analysis helps to evaluate the fairness and impact of this restriction.
Our calculations revealed that approximately 69.15% of adult women would be shorter than this height, contradicting General Electric's claim that 94% of women would be eliminated from consideration. This discrepancy suggests possible misestimations or misrepresentations in the original claim.
Height restrictions can inadvertently lead to discrimination if not carefully justified with valid, non-biased reasoning. Thus, it's vital for organizations to comprehensively analyze the impact of such constraints to avoid unintended exclusion and ensure equitable practices. It serves as a reminder to use statistical tools like Z-scores and the CDF wisely when setting any threshold.

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

Airlines sometimes overbook flights. Suppose that for a plane with 100 seats, an airline takes 110 reservations. Define the random variable \(x\) as \(x=\) the number of people who actually show up for a sold-out flight on this plane From past experience, the probability distribution of \(x\) is given in the following table: $$ \begin{array}{|cc|} \hline \boldsymbol{x} & \boldsymbol{p}(\boldsymbol{x}) \\ \hline 95 & 0.05 \\ 96 & 0.10 \\ 97 & 0.12 \\ 98 & 0.14 \\ 99 & 0.24 \\ 100 & 0.17 \\ 101 & 0.06 \\ 102 & 0.04 \\ 103 & 0.03 \\ 104 & 0.02 \\ 105 & 0.01 \\ 106 & 0.005 \\ 107 & 0.005 \\ 108 & 0.005 \\ 109 & 0.0037 \\ 110 & 0.0013 \\ \hline \end{array} $$ a. What is the probability that the airline can accommodate everyone who shows up for the flight? b. What is the probability that not all passengers can be accommodated? c. If you are trying to get a seat on such a flight and you are number 1 on the standby list, what is the probability that you will be able to take the flight? What if you are number 3 ?

Suppose that \(25 \%\) of the fire alarms in a large city are false alarms. Let \(x\) denote the number of false alarms in a random sample of 100 alarms. Approximate the following probabilities: a. \(P(20 \leq x \leq 30)\) b. \(P(20

The accompanying data on \(x=\) student-teacher ratio is for a random sample of 20 high schools in Maine selected from a population of 85 high schools. The data are consistent with summary values for the state of Maine that appeared in an article in the Bangor Daily News (September \(22,2016,\) bangordailynews.com/2016/09/22/mainefocus/we-discovered-a-surprise-when-we- looked-deeper-into-our-survey-of-maine-principals/?ref=morelnmidcoast, retrieved May 2, 2017). The corresponding normal scores are also shown. $$ \begin{array}{|cc|} \hline \text { Student-Teacher Ratio }(x) & \text { Normal Score } \\ \hline 9.0 & -1.868 \\ 10.0 & -1.403 \\ 11.0 & -1.128 \\ 11.2 & -0.919 \\ 11.6 & -0.744 \\ 11.7 & -0.589 \\ 11.8 & -0.448 \\ 11.9 & -0.315 \\ 12.0 & -0.187 \\ 12.1 & -0.062 \\ 12.5 & 0.062 \\ 12.6 & 0.187 \\ 13.0 & 0.315 \\ 13.2 & 0.448 \\ 13.6 & 0.589 \\ 13.7 & 0.744 \\ 14.0 & 0.919 \\ 14.5 & 1.128 \\ 14.9 & 1.403 \\ 15.0 & 1.868 \\ \hline \end{array} $$ a. Construct a normal probability plot. b. Calculate the correlation coefficient for the (normal score, \(x\) ) pairs. Compare this value to the appropriate critical \(r\) value from Table 6.2 to determine if it is reasonable to think that the distribution of student-teacher ratios for high schools in Maine is approximately normal.

Suppose that instead of three shirts, each participant was asked to choose among four shirts and that the process was performed five times. If a person can't identify her roommate by smell and is just picking a shirt at random, then \(x=\) number of correct identifications is a binomial random variable with \(n=5\) and \(p=\frac{1}{4}\). a. What are the possible values of \(x\) ? b. For each possible value of \(x\), find the associated probability \(p(x)\) and display the possible \(x\) values and \(p(x)\) values in a table. (Hint: See Example 6.27.) c. Construct a histogram displaying the probability distribution of \(x\).

Determine each of the following areas under the standard normal (z) curve: a. To the left of -1.28 b. To the right of 1.28 c. Between -1 and 2 d. To the right of 0 e. To the right of -5 f. Between -1.6 and 2.5 g. To the left of 0.23

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.