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Suppose you are conducting a binomial experiment that has 15 trials and the probability of success of .02. According to the sample size requirements, you cannot use the normal distribution to approximate the binomial distribution in this situation. Use the mean and standard deviation of this binomial distribution and the empirical rule to explain why there is a problem in this situation. (Note: Drawing the graph and marking the values that correspond to the empirical rule is a good way to start.)

Short Answer

Expert verified
The mean and standard deviation of the binomial distribution of 15 trials with a success probability of 0.02 are computed. The problem arises when applying the empirical rule, which assumes a nearly symmetrical distribution (a characteristic of normal distributions but not necessarily of binomial distributions, which can be highly skewed depending on the success probability). In this instance, the probability of success is extremely low, causing the distribution to skew significantly. Therefore, the empirical rule, which approximates percentages of data falling within specific standard deviations of the mean, is not accurate in this case.

Step by step solution

01

Calculating Mean and Variance for Binomial Distribution

Calculate the mean and standard deviation of the given binomial distribution.- Mean (μ) is calculated as n*p, where n is the number of trials (15 in this case) and p is the probability of success (0.02 in this case). - Variance (σ²) is calculated as n*p*(1-p), here again use the values of n and p from above.- Standard deviation (σ) is the square root of the variance. Use a square root function on the calculated variance.
02

Application of The Empirical Rule

According to the Empirical Rule (also known as the 68-95-99.7 rule), for a normal distribution: - About 68% of the data falls within one standard deviation of the mean. - About 95% falls within two standard deviations. - Almost all (around 99.7%) falls within three standard deviations. Apply the calculated mean and standard deviation values to the empirical rule to analyze the given binomial distribution.
03

Explaining the Problem

Explain the problem that arises under these circumstances by considering the characteristics of a binomial distribution. Use the calculated mean and standard deviation to explain how the extreme skew of the distribution violates the assumptions of the Empirical Rule. Discuss how the shape of the binomial distribution (positively skewed in this case) with such a low success probability and a limited number of trials prevents the use of the Empirical Rule, as it assumes symmetry in the distribution of data.

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Key Concepts

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

Empirical Rule
The Empirical Rule, often referred to as the 68-95-99.7 rule, is a guideline for understanding the percentage of data that falls within certain standard deviations from the mean in a normal distribution. It states the following:
  • Approximately 68% of the data points lie within one standard deviation ( \(\sigma\) ) of the mean ( \(\mu\)).
  • About 95% of the data points lie within two standard deviations.
  • Nearly all, around 99.7%, are within three standard deviations.
However, for the Empirical Rule to apply, the data must follow a normal distribution, which is symmetrical. In the case of the given binomial experiment, the skewness caused by a loss of symmetry makes it difficult for the rule to be applicable.
This is especially true for distributions with low probabilities of success, like our example, where the distribution is more positively skewed. Thus, simply applying the Empirical Rule without checking the distribution's form can lead to incorrect inferences.
Mean and Variance
The mean ( \(\mu\) ) and variance ( \(\sigma^2\) ) are fundamental concepts when it comes to describing the central tendency and spread of a distribution. In a binomial distribution, the mean is calculated using the formula: \(\mu = n \cdot p\) , where \(n\) is the number of trials and \(p\) is the probability of success.
The variance is determined with: \(\sigma^2 = n \cdot p \cdot (1-p)\) . It tells us about the data's dispersion around the mean.
Then, the standard deviation ( \(\sigma\) ) is simply the square root of the variance, providing a clear distance measure from the mean for each observation.
These calculated values highlight the degree of spread and central tendency in our data. For the specific exercise, the low mean and low variance result from the small probability of success, hinting at limited variation around the mean. This reinforces the notion that the distribution is not symmetrical, impacting the applicability of the Empirical Rule.
Normal Approximation
Normal Approximation involves using a normal distribution to estimate probabilities for a binomial distribution under specific conditions. The central limit theorem provides a gateway for this approximation, primarily when the number of trials is large and \(p\) is not too close to 0 or 1. However, these conditions are not met in our exercise, as seen in the problem's requirement not to use normal approximation.
The asymmetry inherent in a binomial distribution with few trials and a low probability of success significantly distorts the results from what you would expect if the normal distribution applied.
In situations like this, the skewness due to the imbalance between the number of trials and probability directly impacts the symmetry, which is a prerequisite for applying a normal approximation. Thus, basing decisions on normal approximation under these constraints might result in significant errors.
Standard Deviation
Standard deviation ( \(\sigma\) ) is a measure of the amount of variation or dispersion in a set of values. In simpler terms, it tells us how much the individual data points deviate from the mean, on average. To compute the standard deviation of a binomial distribution, you first calculate the variance: \(\sigma^2 = n \cdot p \cdot (1-p)\) , and take its square root.
This calculation gives an insight into the expected spread of the data in a binomial distribution.
With a small probability of success and few trials, as in our exercise, the standard deviation is also small, reflecting very little variability around the mean. This compact spread, combined with a skewed distribution, suggests that approaching this with tools designed for normal distributions will yield inaccurate insights.
In practice, understanding standard deviation helps students grasp how closely or widely distributed the potential outcomes are around the mean in real-world scenarios.

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Most popular questions from this chapter

According to the National Retail Federation's recent Back to College Consumer Intentions and Actions survey, families of college students spend an average of \(\$ 616.13\) on new apparel, furniture for dorms or apartments, school supplies, and electronics (www.nrf.com/modules.php?name=News\&op= viewlive\&sp _id=966). Suppose that the expenses on such Back to College items for the current year are approximately normally distributed with a mean of \(\$ 616.13\) and a standard deviation of \(\$ 120 .\) Find the probability that the amount of money spent on such items by a randomly selected family of a college student is a. less than \(\$ 450\) b. between \(\$ 500\) and \(\$ 750\)

An office supply company conducted a survey before marketing a new paper shredder designed for home use. In the survey, \(80 \%\) of the people who tried the shredder were satisfied with it. Because of this high satisfaction rate, the company decided to market the new shredder. Assume that \(80 \%\) of all people are satisfied with this shredder. During a certain month, 100 customers bought this shredder. Find the probability that of these 100 customers, the number who are satisfied is a. exactly 75 b. 73 or fewer c. 74 to 85

Find the following probabilities for the standard normal distribution. a. \(P(z<-1.31)\) b. \(P(1.23 \leq z \leq 2.89)\) c. \(P(-2.24 \leq z \leq-1.19)\) d. \(P(z<2.02)\)

Find the value of \(z\) so that the area under the standard normal curve a. from 0 to \(z\) is \(.4772\) and \(z\) is positive b. between 0 and \(z\) is (approximately) \(.4785\) and \(z\) is negative c. in the left tail is (approximately) . 3565 d. in the right tail is (approximately) \(.1530\)

A company that has a large number of supermarket grocery stores claims that customers who pay by personal checks spend an average of \(\$ 87\) on groceries at these stores with a standard deviation of \(\$ 22\). Assume that the expenses incurred on groceries by all such customers at these stores are normally distributed. a. Find the probability that a randomly selected customer who pays by check spends more than \(\$ 114\) on groceries. b. What percentage of customers paying by check spend between \(\$ 40\) and \(\$ 60\) on groceries? c. What percentage of customers paying by check spend between \(\$ 70\) and \(\$ 105\) ? d. Is it possible for a customer paying by check to spend more than \(\$ 185\) ? Explain.

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