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According to Anthropometric Survey data, the distribution of arm spans for males is approximately Normal with a mean of \(71.4\) inches and a standard deviation of 3. 3 inches. a. What percentage of men have arm spans between 66 and 76 inches? b. Professional basketball player, Kevin Durant, has an arm span of almost 89 inches. Find the \(z\) -score for Durant's arm span. What percentage of males have an arm span at least as long as Durant's?

Short Answer

Expert verified
a) Approximately 86.72% of men have arm spans between 66 and 76 inches. b) Durant's arm span z-score is 5.33 and virtually no one (approximately 0%) has an arm span as long or longer than Durant's.

Step by step solution

01

Find Z-Scores for arm spans of 66 and 76 inches

First, we calculate the z-scores for 66 and 76 inches respectively using the formula \( z = {(x - \mu)}/{\sigma} \), where \( x \) is the data point, \( \mu \) is the mean and \( \sigma \) is the standard deviation. The z-scores, say \( z1 \) and \( z2 \) are obtained as follows: \[ z1 = (66 - 71.4) / 3.3 = -1.64 \] \[ z2 = (76 - 71.4) / 3.3 = 1.39 \]
02

Calculate percentage of men with arm span between 66 and 76 inches

The percentage of men with arm spans between 66 and 76 inches is the area under the standard normal curve between the z-scores \( z1 \) and \( z2 \), which is represented as \( P(z1 < Z < z2) \), where \( Z \) is a standard normal random variable. This can be obtained from standard normal distribution table or using statistical functions in software. It is obtained as: \[ P(-1.64 < Z < 1.39) = P(Z < 1.39) - P(Z < -1.64) = 0.9177 - 0.0505 = 0.8672 \]
03

Calculate z-score for Durant's arm span

The z-score for Kevin Durant's arm span, say \( z3 \), can be calculated using the same formula as in Step 1: \[ z3 = (89 - 71.4) / 3.3 = 5.33 \]
04

Calculate percentage of men with arm span at least as long as Durant's

The percentage of men having arm span at least as long as Durant's is the area under the standard normal curve to the right of \( z3 \), which is represented as \( P(Z > z3) \). As most standard normal tables provide cumulative probability to the left of the value, we use the property \( P(Z > z) = 1 - P(Z < z) \) to find the required probability: \[ P(Z > 5.33) = 1 - P(Z < 5.33) = 1 - 1 = 0 \] The z-score of 5.33 is so extreme that the standard normal table will not list it, implying that virtually no one has an arm span as long as Durant's.

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Key Concepts

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

Z-Score Calculation
Understanding the z-score is essential for interpreting where a certain data point lies in relation to the average in a set of data. A z-score, also known as a standard score, quantifies how many standard deviations an element is from the mean.

For any given data point, the formula for calculating the z-score is: \[ z = \frac{(x - \mu)}{\sigma} \]
where:
  • \( x \) is the value of the data point.
  • \( \mu \) is the mean of the data.
  • \( \sigma \) is the standard deviation.
In our case, to calculate the z-score for a male's arm span, we would subtract the mean arm span from the individual's arm span and then divide by the standard deviation. For example, a z-score of -1.64 for an arm span indicates that it is 1.64 standard deviations below the mean.
Standard Normal Curve
The standard normal curve, often referred to as the bell curve, represents a normal distribution with a mean of 0 and a standard deviation of 1. In the context of a normal distribution, the percentage of data that falls between two z-scores correlates to the area under the curve between those points.

For instance, to find the percentage of men with arm spans between two specific measurements, we identify the z-scores for those measurements and look at the area under the curve that lies between them. This area represents the probability of finding an arm span in that range. In statistics, areas under the standard normal curve are often found using z-tables or statistical software, simplifying the process of probability calculation.
Statistical Data Analysis
Statistical data analysis involves collecting, analyzing, interpreting, and presenting data. It allows us to understand patterns, trends, and relationships within data sets. When dealing with normally distributed data, as in our example with arm spans, we often use parameters like mean and standard deviation to summarize the data.

For in-depth analysis, knowing how to compute and interpret z-scores and the area under the standard normal curve is indispensable. This knowledge enables the comparison of individual data points to the population average, and the calculation of probabilities for certain ranges of data, which are essential for making informed decisions or predictions based on statistical data.
Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. It's a fundamental concept in statistics that includes not only the normal distribution but others like binomial, uniform, and exponential distributions. For normally distributed data, the probability distribution allows us to find the likelihood of a random variable falling within a particular range, above or below a certain value, or exactly at a specified value.

Using the properties of the normal distribution, we can predict probabilities and make sense of statistical data. For extreme values, such as Kevin Durant's arm span, the probability distribution tells us that it is an extreme outlier, which can be an exceptionally rare event within the context of the normal distribution of male arm spans.

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Most popular questions from this chapter

New York City's mean minimum daily temperature in February is \(27^{\circ} \mathrm{F}\) (http://www.ny.com). Suppose the standard deviation of the minimum temperature is \(6^{\circ} \mathrm{F}\) and the distribution of minimum temperatures in February is approximately Normal. What percentage of days in February has minimum temperatures below freezing \(\left(32^{\circ} \mathrm{F}\right) ?\)

The distribution of spring high temperatures in Los Angeles is approximately Normal, with a mean of 75 degrees and a standard deviation of \(2.5\) degrees. a. What is the probability that the high temperature is less than 70 degrees in Los Angeles on a day in spring? b. What percentage of Spring day in Los Angeles have high temperatures between 70 and 75 degrees? c. Suppose the hottest spring day in Los Angeles had a high temperature of 91 degrees. Would this be considered unusually high, given the mean and the standard deviation of the distribution? Why or why not?

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?

For each situation, identify the sample size \(n\), the probability of a success \(p\), and the number of success \(x\). When asked for the probability, state the answer in the form \(b(n, p, x)\). There is no need to give the numerical value of the probability. Assume the conditions for a binomial experiment are satisfied. Since the Surgeon General's Report on Smoking and Health in 1964 linked smoking to adverse health effects, the rate of smoking the United States have been falling. According to the Centers for Disease Control and Prevention in 2016, \(15 \%\) of U.S. adults smoked cigarettes (down from \(42 \%\) in the \(1960 \mathrm{~s}\) ). a. If 30 Americans are randomly selected, what is the probability that exactly 10 are smokers? b. If 30 Americans are randomly selected, what is the probability that exactly 25 are not smokers?

In Toronto, Canada, \(55 \%\) of people pass the drivers' road test. Suppose that every day, 100 people independently take the test. a. What is the number of people who are expected to pass? b. What is the standard deviation for the number expected to pass? c. After a great many days, according to the Empirical Rule, on about \(95 \%\) of these days, the number of people passing will be as low as and as high as (Hint: Find two standard deviations below and two standard deviations above the mean.) d. If you found that on one day, 85 out of 100 passed the test, would you consider this to be a very high number?

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