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The Wall Street Journal (February 15,1972 ) reported that General Electric was sued in Texas for sex discrimination over a minimum height requirement of \(5 \mathrm{ft}\) 7 in. 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 in. (5 ft 6 in.) and standard deviation 2 in. 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. Only approximately 69.15% of women are shorter than that. Also, approximately 69.15% of women would be excluded from employment due to the height restriction of 5 ft. 7 in.

Step by step solution

01

Analyze the Claim

First, you must calculate the Z-score to understand where a height of 5 ft. 7 in. stands within the given distribution. 5 ft. 7 in. is equal to 67 inches. The Z-score is defined by the formula: \(Z = \frac{x - \mu}{\sigma}\). Where \(x\) is the observation (67 inches in this case), \(\mu\) is the mean (66 inches), and \(\sigma\) is the standard deviation (2 inches). So the Z-score for 67 inches is \(Z = \frac{67 - 66}{2}\) which equals to 0.5.
02

Validate the Claim by finding the proportion

Now we need to verify the claim that more than 94% of adult women are shorter than 5 ft. 7 in. This can be done by finding the area to the left of the Z-score in a standard Normal Distribution. Area to the left denotes the proportion of the population that is less than the specific value. The value corresponding to Z=0.5 in standard Normal Distribution tables or using technology comes out to be 0.6915 or 69.15%. This shows that only about 69.15% women are shorter than 5 ft. 7 in. Thus the claim is incorrect.
03

Find the proportion of women excluded due to height restriction

The proportion of adult women that would be excluded from employment due to the height restriction is simply the area to the left of the Z-score (0.5) in a standard Normal Distribution, which we found to be 0.6915 or 69.15%. This means that approximately 69.15% of women would be excluded due to the height restriction of 5 ft. 7 in.

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Key Concepts

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

Z-score calculation
Calculating a Z-score is essential when you want to see how a specific value compares to the rest of the data in a normal distribution. The formula to find the Z-score is:- \(Z = \frac{x - \mu}{\sigma}\) - \(x\) is the value you are examining. - \(\mu\) is the mean of the dataset. - \(\sigma\) is the standard deviation.This formula helps you identify where your specific value lies in relation to the mean. For example, in this exercise, to determine how a height of 67 inches (5 ft 7 in) fits in the distribution of female heights, you calculate the Z-score. By substituting the values into the equation \( (x = 67, \mu = 66, \sigma = 2) \), you get a Z-score of 0.5. This means the height is half a standard deviation above the mean.Understanding the Z-score helps you analyze whether a claim about a population value, like the minimum height required, is valid based on its statistical position.
Normal distribution properties
The normal distribution, often called the bell curve, is a fundamental concept in statistics. It's symmetric, showing that most outcomes are near the mean, reflecting a balanced spread of data.Key properties include:- **Symmetric Shape**: The left and right sides of the graph are mirrored.- **Mean, Median, Mode Alignment**: The mean, median, and mode are identical and located at the center of the distribution.- **Standard Deviation Impact**: Determines the spread of the data; 68% of data lie within one standard deviation (±1\(\sigma\)), and about 95% within two standard deviations (±2\(\sigma\)).In this problem, knowing these properties helps in understanding how the heights of people spread around the average height of 66 inches. This tells us most women's heights are close to this mean, and using the standard deviation, we can gauge how common or rare certain heights are, such as 67 inches.
Proportion of a normal distribution
The proportion of a normal distribution associated with a Z-score represents the percentage of data below or above a particular point. This is also known as the area under the curve. Here's how it works: - **Finding Probability**: By referring to a Z-table or using statistical software, we can find the probability related to a specific Z-score. This reflects the proportion of the dataset expected to fall below a given point. In our exercise, we wanted to determine what proportion of women are shorter than 67 inches and thus not excluded by the height restriction. Given the calculated Z-score of 0.5, looking up this score shows about 69.15% fall below this height. Thus, only this proportion of women meet the condition, contradicting the 94% claim made in the original problem statement. Understanding this principle helps evaluate claims about cut-off points in datasets and how such restrictions might unfairly target a segment of the population.

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