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NFL data from the 2006 football season reported the number of yards gained by each of the league's 167 wide receivers: a) According to the Normal model, what percent of receivers would you expect to gain fewer yards than 2 standard deviations below the mean number of yards? b) For these data, what does that mean? c) Explain the problem in using a Normal model here.

Short Answer

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a) About 2.5% gain fewer yards; b) implies lower performance; c) data might not follow a normal distribution.

Step by step solution

01

Understanding the Normal Distribution

To solve the problem, we need to know that a normal distribution is a probability distribution that is symmetric about the mean. In a standard normal distribution, the mean is 0, and the standard deviation is 1. We also know that about 95% of data within a normal distribution would lie within two standard deviations (both positive and negative) from the mean.
02

Find the Percent of Receivers Gaining Fewer Yards

The percentage of receivers that gain fewer yards than 2 standard deviations below the mean corresponds to the cumulative probability for the z-score -2. In a standard normal distribution, about 2.5% of the data lies below 2 standard deviations from the mean. Thus, we expect approximately 2.5% of receivers to fall below this threshold.
03

Interpret the Results for the Data

If 2.5% of the receivers gained fewer yards than 2 standard deviations below the mean, it implies that these receivers are outliers in terms of performance, gaining significantly fewer yards compared to an average receiver of the dataset.
04

Discuss the Problem with Using the Normal Model

The issue with applying a normal model here is that real-world data, like the number of yards, might not perfectly follow a normal distribution. Factors such as team strategy, player injuries, or playtime can cause skewness, suggesting the actual data may not be symmetric or might have different peaks and tail behaviors compared to a standard normal distribution.

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Key Concepts

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

Standard Deviation
The concept of standard deviation is essential for understanding how data is distributed in a set. In any dataset, the standard deviation measures how much the data points deviate from the mean, or average, of the dataset. It's a crucial indicator of data spread. A smaller standard deviation means that the data points are closer to the mean, while a larger standard deviation signifies that the data points are spread out over a wider range.
When analyzing NFL receiver yards, the standard deviation helps determine how consistent the receivers are in gaining yards. A low standard deviation would imply most receivers gain a similar number of yards, whereas a high standard deviation would suggest a wide variance in receiver performance.
Z-score
A z-score is a statistical measurement that describes a value's position relative to the mean of a group of values. It is expressed in terms of standard deviations from the mean. Therefore, a z-score allows us to understand how far any given data point lies from the average.
For example, if an NFL receiver's yards has a z-score of -2, it means this receiver's yards are 2 standard deviations below the mean yards gained by all receivers. The z-score is vital for determining outliers or significant deviations from the average.
Cumulative Probability
Cumulative probability refers to the probability that a random variable is less than or equal to a particular value. In the context of normal distribution, cumulative probability helps identify what proportion of data falls below a certain z-score, which we use to make probabilistic predictions about the dataset.
For instance, in the NFL data example, the cumulative probability of a z-score of -2 indicates that about 2.5% of data is expected to fall below this point. Thus, it tells us directly the likelihood of receivers gaining less than a specific yard number. This information can be crucial for determining how unusual a particular performance might be.
Data Skewness
Data skewness refers to the asymmetry in a statistical distribution. In perfectly normal distributions, the data should be symmetric, meaning it has no skewness. However, in real-world scenarios, data often shows some degree of skewness either to the right or left.
When applying a normal model to data, noticing skewness is important to ensure accuracy. For the NFL data, external factors like player injuries and playing time may create skewness, making the symmetrical assumptions of a normal model less applicable. Real-world datasets often need adjustments for skewness before making precise predictions or analyses based on normal distribution.

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Most popular questions from this chapter

The first Stat exam had a mean of 80 and a standard deviation of 4 points; the second had a mean of 70 and a standard deviation of 15 points. Reginald scored an 80 on the first test and an 85 on the second. Sara scored an 88 on the first but only a 65 on the second. Although Reginald's total score is higher, Sara feels she should get the higher grade. Explain her point of view.

. Agricultural scientists are working on developing an improved variety of Roma tomatoes. Marketing research indicates that customers are likely to bypass Romas that weigh less than 70 grams. The current variety of Roma plants produces fruit that averages 74 grams, but \(11 \%\) of the tomatoes are too small. It is reasonable to assume that a Normal model applies. a) What is the standard deviation of the weights of Romas now being grown? b) Scientists hope to reduce the frequency of undersized tomatoes to no more than \(4 \%\). One way to accomplish this is to raise the average size of the fruit. If the standard deviation remains the same, what target mean should they have as a goal? c) The researchers produce a new variety with a mean weight of 75 grams, which meets the \(4 \%\) goal. What is the standard deviation of the weights of these new Romas? d) Based on their standard deviations, compare the tomatoes produced by the two varieties.

The mean of the 100 car speeds in Exercise 20 was \(23.84 \mathrm{mph}\), with a standard deviation of \(3.56 \mathrm{mph}\). a) Using a Normal model, what values should border the middle \(95 \%\) of all car speeds? b) Here are some summary statistics. $$ \begin{array}{lll} \hline \text { Percentile } & & \text { Speed } \\ \hline 100 \% & \text { Max } & 34.060 \\ 97.5 \% & & 30.976 \\ 90.0 \% & & 28.978 \\ 75.0 \% & \text { Q3 } & 25.785 \\ 50.0 \% & \text { Median } & 23.525 \\ 25.0 \% & \text { Q1 } & 21.547 \\ 10.0 \% & & 19.163 \\ 2.5 \% & & 16.638 \\ 0.0 \% & \text { Min } & 16.270 \\ \hline \end{array} $$

Consider the IQ model \(N(100,16)\) one last time. a) What IQ represents the 15 th percentile? b) What IQ represents the 98 th percentile? c) What's the IQR of the IQs?

Companies that design furniture for elementary school classrooms produce a variety of sizes for kids of different ages. Suppose the heights of kindergarten children can be described by a Normal model with a mean of \(38.2\) inches and standard deviation of \(1.8\) inches. a) What fraction of kindergarten kids should the company expect to be less than 3 feet tall? b) In what height interval should the company expect to find the middle \(80 \%\) of kindergarteners? c) At least how tall are the biggest \(10 \%\) of kindergarteners?

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