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Assume college women's heights are approximately Normally distributed with a mean of 65 inches and a standard deviation of \(2.5\) inches. Choose the Stat- \(-\) Crunch output for finding the percentage of college women who are taller than 67 inches and report the correct percentage. Round to one decimal place. a. b.

Short Answer

Expert verified
The percentage of college women who are taller than 67 inches is 21.2%.

Step by step solution

01

Understanding Normal Distribution

Since the problem involves a normal distribution, make sure you understand that a normal distribution is a type of continuous probability distribution for a real-valued random variable. In this case, the variable is college women's heights.
02

Calculating the Z-Score

The next step is to calculate the Z-score which is a measure of how many standard deviations an element is from the mean. The formula for calculating Z-score is: \((X - \mu)/ \sigma\), where X is the height we are looking for (67 in this case), \(\mu\) is the mean (which is 65) and \(\sigma\) is the standard deviation (2.5). Thus, substituting the values gives us: \((67-65)/2.5 = 0.8\).
03

Finding the Percentage

Using the computed Z-score, find the corresponding percentage in the Z-table or using statistical software like Stat-Crunch. It is noticed that 0.8 in Stat-Crunch maps to 0.7881, which is the probability of a woman being less than 67 inches. However the problem requires finding the probability of a woman being greater than 67 inches, which can be computed as \(1-0.7881 = 0.2119\) or 21.2%

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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 helps us understand how far away a certain value is from the mean in a data set. In simpler terms, it tells us how many "steps" or standard deviations a data point is from the average. To calculate the Z-score, we use the formula:
  • Subtract the mean (\(\mu\)) from the value in question (\(X\)).
  • Then, divide by the standard deviation (\(\sigma\)).
For example, if we are looking for the Z-score of a woman's height of 67 inches, given that the average height is 65 inches with a standard deviation of 2.5 inches, we proceed as follows:
  • First, find the difference: \(67 - 65 = 2\).
  • Next, divide by the standard deviation: \(2/2.5 = 0.8\).
Thus, the Z-score is \(0.8\), indicating this height is \(0.8\) standard deviations above the mean. This Z-score becomes a powerful tool in determining how unusual or typical a certain observation is within a normal distribution.
probability distribution
A probability distribution describes how the values of a random variable are spread or distributed. For a normally distributed variable, like heights in this scenario, the probability distribution is symmetrically centered around the mean. The shape of this distribution is often referred to as a "bell curve." Think about it as a way to visualize where most observations fall.
With a normal distribution, you can find probabilities related to a specific range of outcomes. In our example, once the Z-score for a height of 67 inches is calculated, the next step is to understand what this score means in terms of probability:
  • The Z-score of \(0.8\) indicates that the height is above average.
  • Using a Z-table, or statistical software, you can find the cumulative probability up to that Z-score, which is \(0.7881\) in this case.
  • This means there's a \(78.81\%\) chance of encountering a height less than 67 inches.
To find out the probability of heights greater than 67 inches, simply subtract from 1 (since total probability sums to 1) giving: \(1 - 0.7881 = 0.2119\) or \(21.2\%\). This shows that approximately \(21.2\%\) of college women are taller than 67 inches based on this probability distribution.
StatCrunch
StatCrunch is a statistical software tool that helps you perform data analysis and view probability distributions quickly. It's like a calculator but with more capabilities, tailored to handling statistical tasks.
When dealing with normal distributions like this one, StatCrunch can be used to determine the probability or percentile related to a specific Z-score. Here’s why StatCrunch is useful:
  • It eliminates the need to manually search through Z-tables.
  • The software quickly computes cumulative probabilities, helping find the percentage of data that falls under or above a certain point.
  • StatCrunch visualizes results with graphs that help in understanding the distribution.
Using StatCrunch for calculating the height distribution in this exercise:
  • The Z-score of \(0.8\) was input into StatCrunch.
  • The software returned a probability of \(0.7881\), representing those shorter than 67 inches.
  • By subtracting from 1, the tool efficiently gave the percentage taller than 67 inches as \(21.2\%\).
This makes StatCrunch particularly handy in academic tasks, providing accuracy effortlessly, and allowing you to focus on understanding data rather than getting tangled in calculations.

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Most popular questions from this chapter

The use of drones, aircraft without onboard human pilots, is becoming more prevalent in the United States. According to a 2017 Pew Research Center report, \(59 \%\) of American had seen a drone in action. Suppose 50 Americans are randomly selected. a. What is the probability that at least 25 had seen a drone? b. What is the probability that more than 30 had seen a drone? c. What is the probability that between 30 and 35 had seen a drone? d. What is the probability that more than 30 had not seen a drone?

For each question, find the area to the right of the given \(z\) -score in a standard Normal distribution. In this question, round your answers to the nearest \(0.000\). Include an appropriately labeled sketch of the \(N(0,1)\) curve. a. \(z=-4.00\) b. \(z=-8.00\) c. \(z=-30.00\) d. If you had the exact probability for these right proportions, which would be the largest and which would be the smallest? e. Which is equal to the area in part b: the area below (to the left of) \(z=8.00\) or the area above (to the right of) \(z=8.00\) ?

The Empirical Rule applies rough approximations to probabilities for any unimodal, symmetric distribution. But for the Normal distribution we can be more precise. Use the figure and the fact that the Normal curve is symmetric to answer the questions. Do not use a Normal table or technology. According to the Empirical Rule, a. Roughly what percentage of \(z\) -scores are between \(-2\) and 2 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) b. Roughly what percentage of \(z\) -scores are between \(-3\) and 3 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) c. Roughly what percentage of \(z\) -scores are between \(-1\) and 1 . i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) d. Roughly what percentage of \(z\) -scores are greater than 0 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) e. Roughly what percentage of \(z\) -scores are between 1 and 2 ? i. almost all iii. \(50 \%\) ii. \(13.5 \%\) iv. \(2 \%\)

The Normal model \(N(69,3)\) describes the distribution of male heights in the United States. Which of the following questions asks for a probability, and which asks for a measurement? Identify the type of problem and then answer the given question. See page 316 for guidance. a. To be a member of the Tall Club of Silicon Valley a man must be at least 74 inches tall. What percentage of men would qualify for membership in this club? b. Suppose the Tall Club of Silicon Valley wanted to admit the tallest \(2 \%\) of men. What minimum height requirement should the club set for its membership criteria?

In a standard Normal distribution, if the area to the left of a \(z\) -score is about \(0.1000\), what is the approximate \(z\) -score?

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