/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Let \(x\) be a hypergeometric ra... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(x\) be a hypergeometric random variable with \(N=15, n=3,\) and \(M=4\) a. Calculate \(p(0), p(1), p(2),\) and \(p(3)\). b. Construct the probability histogram for \(x\). c. Use the formulas given in Section 5.4 to calculate \(\mu=E(x)\) and \(\sigma^{2}\) d. What proportion of the population of measurements fall into the interval \((\mu \pm 2 \sigma) ?\) Into the interval \((\mu \pm 3 \sigma) ?\) Do these results agree with those given by Tchebysheff's Theorem?

Short Answer

Expert verified
Question: Calculate the value of $\mu$, $\sigma^2$, and the proportions of measurements that fall within the intervals $(\mu \pm 2\sigma)$ and $(\mu \pm 3\sigma)$ for a hypergeometric random variable with parameters N=15, n=3, and M=4. Compare with Tchebysheff's Theorem.

Step by step solution

01

Calculate probabilities

Here we have a hypergeometric random variable with the parameters N=15 (total population size), n=3 (number of objects drawn), and M=4 (number of successes in the total population). To find the probabilities p(0), p(1), p(2), and p(3), we'll use the hypergeometric probability formula: $$p(x) = \frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}$$ Calculate the probabilities using the form, substituting the appropriate values for each x: $$p(0) = \frac{\binom{4}{0}\binom{15-4}{3-0}}{\binom{15}{3}}$$ $$p(1) = \frac{\binom{4}{1}\binom{15-4}{3-1}}{\binom{15}{3}}$$ $$p(2) = \frac{\binom{4}{2}\binom{15-4}{3-2}}{\binom{15}{3}}$$ $$p(3) = \frac{\binom{4}{3}\binom{15-4}{3-3}}{\binom{15}{3}}$$ Compute these probabilities.
02

Construct the probability histogram

Now that we have the probabilities for each value of the random variable x, we can construct a probability histogram. Plot the values of x (0, 1, 2, 3) on the x-axis and the probabilities calculated in step 1 on the y-axis.
03

Calculate the mean and variance

To calculate the mean and variance, we'll use the following formulas: Mean: $$\mu = E(x) = n\frac{M}{N}$$ Variance: $$\sigma^2 = n\frac{M}{N}\left(1-\frac{M}{N}\right)\frac{N-n}{N-1}$$ Substitute the given values into the formulas: $$\mu = 3\frac{4}{15}$$ $$\sigma^2 = 3\frac{4}{15}\left(1-\frac{4}{15}\right)\frac{15-3}{15-1}$$ Compute the mean and variance.
04

Calculate proportions within intervals

To find the proportion of measurements within the interval \((\mu \pm 2\sigma)\) and \((\mu \pm 3\sigma)\), first calculate the upper and lower limits of each interval and then add the probabilities of x-values that lie within those intervals. For first interval \((\mu \pm 2\sigma)\): Find the values of \(\mu - 2\sigma\) and \(\mu + 2\sigma\), and calculate the proportion of the population measurements that fall into that interval. For second interval \((\mu \pm 3\sigma)\): Find the values of \(\mu - 3\sigma\) and \(\mu + 3\sigma\), and calculate the proportion of the population measurements that fall into that interval.
05

Compare results with Tchebysheff's Theorem

Tchebysheff's Theorem states that, for any continuous random variable with mean µ and standard deviation σ, the proportion of measurements within the interval \((\mu \pm k\sigma)\) is at least \(1-\frac{1}{k^2}\) for any k > 1. Compare the calculated proportions in step 4 to Tchebysheff's Theorem by calculating the following: For first interval \((\mu \pm 2\sigma)\): $$1-\frac{1}{2^2}$$ For second interval \((\mu \pm 3\sigma)\): $$1-\frac{1}{3^2}$$ Check whether the calculated proportions agree with these results.

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Key Concepts

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

Probability Histogram
A probability histogram is a graphical representation of the probabilities of different outcomes of a random variable. It looks similar to a bar chart, where each bar represents a specific value the random variable can take, and the height of the bar corresponds to the probability of that value. In the context of the hypergeometric distribution, this histogram helps visualize the likelihood of selecting a certain number of successes from a finite population without replacement.

To create a probability histogram for a hypergeometric random variable like in this exercise, follow these steps:
  • Identify the possible values the random variable can take (e.g., 0, 1, 2, 3 in this example).
  • Use the calculated probabilities from the hypergeometric formula to determine the height of each bar.
  • Plot these values on the x-axis, with probabilities on the y-axis.
This visual representation makes it easier to understand the distribution of probabilities and compare them intuitively.
Tchebysheff's Theorem
Tchebysheff's Theorem provides a way to understand how data is spread around the mean. It states that for any set of data, regardless of distribution shape, a minimum proportion of data will fall within a certain number of standard deviations from the mean. Specifically, the theorem guarantees that at least \(1 - \frac{1}{k^2}\) of data values lie within \(k\) standard deviations of the mean for \(k > 1\).

In this exercise, Tchebysheff's Theorem helps determine how much of the data falls within intervals \((\mu \pm 2\sigma)\) and \((\mu \pm 3\sigma)\). Here's how it's applied:
  • Calculate the proportions based on the formula \(1 - \frac{1}{2^2}\) and \(1 - \frac{1}{3^2}\).
  • Compare these theoretical results with the actual proportions found from the hypergeometric distribution probabilities.
This allows us to check whether the data align with the predictions of Tchebysheff's Theorem, providing a robust measure of spread even for non-normal distributions.
Mean and Variance Calculations
Mean and variance are fundamental concepts in probability and statistics. They provide insights into the central tendency and spread of a distribution. For the hypergeometric distribution, these values are calculated using specific formulas due to its discrete nature.

For mean \(\mu\), use the formula: \[ \mu = n\frac{M}{N} \]where \(n\) is the number of draws, \(M\) is the number of successes in the population, and \(N\) is the total population size. This gives the expected number of successes in the sample.

Variance \(\sigma^2\) measures the spread of the distribution and is calculated using: \[ \sigma^2 = n\frac{M}{N}\left(1-\frac{M}{N}\right)\frac{N-n}{N-1} \]This formula accounts for the changing probabilities as samples are drawn without replacement.

Applying these calculations to the exercise helps students grasp how expected values and variability interact in real-world scenarios, especially in hypergeometric distributions.

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Most popular questions from this chapter

Under what conditions would you use the hypergeometric probability distribution to evaluate the probability of \(x\) successes in \(n\) trials?

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