Question

Let \(x\) be a hypergeometric random variable with \(N=15, n=3,\) and \(M=4\). Use this information to answer the questions in Exercises 14-17. What portion of the population of measurements fall into the interval \((\mu \pm 2 \sigma) ?\) Into the interval \((\mu \pm 3 \sigma) ?\) Do the results agree with Tchebysheff's Theorem?

Short answer

The portion of the population of measurements that fall into the interval \((\mu \pm 2 \sigma)\) and \((\mu \pm 3 \sigma)\) is \(\frac{198}{455}\). This is consistent with Tchebysheff's Theorem.

Step-by-step solution

Calculate the mean and standard deviation of the hypergeometric distribution

The formulas for the mean \((\mu)\) and standard deviation \((\sigma)\) of a hypergeometric distribution are: \[\mu = \frac{nM}{N}\] \[\sigma = \sqrt{\frac{nM(N-M)(N-n)}{N^2(N-1)}}\] Using the given parameters, \(N=15\), \(M=4\), and \(n=3\). Plugging these values into the formulas, we get: \[\mu = \frac{3\times4}{15} = \frac{12}{15} = \frac{4}{5}\] \[\sigma = \sqrt{\frac{3\times4\times(15-4)(15-3)}{15^2(15-1)}} = \sqrt{\frac{504}{12600}} = \frac{12}{35}\]

Calculate the probabilities for given intervals

Now we need to calculate the portion of the population of measurements that fall into the intervals \((\mu \pm 2 \sigma)\) and \((\mu \pm 3 \sigma)\). The intervals are: 1. \((\mu \pm 2 \sigma) = \left( \frac{4}{5} \pm 2 \times \frac{12}{35} \right)\) 2. \((\mu \pm 3 \sigma) = \left( \frac{4}{5} \pm 3 \times \frac{12}{35} \right)\) These intervals are continuous values and the hypergeometric distribution is discrete, so we need to include only the integers for which the hypergeometric probability formula applies. For the first interval, \(0 \le x \le 3\), so we will calculate the hypergeometric probability for each of these values. The general formula for the hypergeometric probability is: \[P(X=x) = \frac{\binom{M}{x} \binom{N-M}{n-x}}{\binom{N}{n}}\] Calculating these probabilities: 1. \(P(X=0) = \frac{\binom{4}{0} \binom{11}{3}}{\binom{15}{3}} = \frac{1 \times 165}{455} = \frac{33}{91}\) 2. \(P(X=1) = \frac{\binom{4}{1} \binom{11}{2}}{\binom{15}{3}} = \frac{4 \times 55}{455} = \frac{44}{91}\) 3. \(P(X=2) = \frac{\binom{4}{2} \binom{11}{1}}{\binom{15}{3}} = \frac{6 \times 11}{455} = \frac{6}{35}\) 4. \(P(X=3) = \frac{\binom{4}{3} \binom{11}{0}}{\binom{15}{3}} = \frac{4 \times 1}{455} = \frac{4}{455}\) Adding these probabilities gives us the portion of the population in the interval \((\mu \pm 2 \sigma)\): \[\frac{33}{91} + \frac{44}{91} + \frac{6}{35} + \frac{4}{455} = \frac{33 + 44 + 117 + 4}{455} = \frac{198}{455}\] For the second interval, all possible values of \(x\) (from 0 to 3) are included, so we don't need to calculate any new probabilities. The portion of the population in the interval \((\mu \pm 3 \sigma)\) is the same as the portion in the interval \((\mu \pm 2 \sigma)\), which is \(\frac{198}{455}\).

Check if the results agree with Tchebysheff's Theorem

Tchebysheff's Theorem states that for any non-negative number \(k\), the portion of the population with values within \(k\) standard deviations of the mean is at least \(\frac{1}{1+k^2}\). For \(k=2\), Tchebysheff's Theorem gives: \[\frac{1}{1+2^2} = \frac{1}{5}\] For \(k=3\), Tchebysheff's Theorem gives: \[\frac{1}{1+3^2} = \frac{1}{10}\] In our case, the calculated portion of the population within 2 and 3 standard deviations of the mean is \(\frac{198}{455}\). Since \(\frac{198}{455} > \frac{1}{5}\) and \(\frac{198}{455} > \frac{1}{10}\), the results are consistent with Tchebysheff's Theorem.

Key concepts

Understanding Hypergeometric Random Variables

A hypergeometric random variable emerges from a scenario where we have a finite population that consists of two types of items. We draw a sample from this population without replacement, and the variable counts the number of items of one type in the sample.

The scenario provided in the exercise is a classic example. There are fifteen items (\(N=15\) represents total number of items), and four of them are 'successes' (\(M=4\) signifies successes in the population). We draw three items (\(n=3\) is our sample size), without replacing any. The hypergeometric random variable would be the number of successes in our three-item sample.

Understanding how to calculate for specific values of a hypergeometric random variable is key. It allows you to determine probabilities for particular outcomes by using a specific mathematical formula. As the exercise instructs, you can find the probability of zero, one, two, or three successes in your sample. This kind of analysis is instrumental in fields like quality control or any scenario where 'successes' and 'failures' can be clearly defined in a sample.

Mean and Standard Deviation of Hypergeometric Distribution

The mean and the standard deviation of a hypergeometric distribution provide central measures of its behavior. They tell us what to expect on average and how much variability there might be in our outcomes.

The exercise walks through the process using the formula for the mean \[\mu = \frac{nM}{N}\] and the standard deviation \[\sigma = \sqrt{\frac{nM(N-M)(N-n)}{N^2(N-1)}}\]. By inserting the given values, you obtain a concrete sense of the 'center' of the distribution and how 'spread out' it might be. It's crucial to get that the mean represents the expected number of successes in our sample, while the standard deviation measures the level of dispersion from this mean.

To put this in context, if we sampled many groups of three items (\(n=3\)) from our population of fifteen (\(N=15\)), on average, we'd expect about \(\mu = \frac{4}{5}\) successes per sample. Moreover, the standard deviation sheds light on the typical range we'd expect most samples to fall within—quite pertinent for anticipating and interpreting real-life outcomes.

Tchebysheff's Theorem and Its Application

Tchebysheff's Theorem is a powerful tool in statistics. It provides a minimum guarantee for the proportion of results within a certain range from the mean, regardless of the distribution's shape – assuming the mean and standard deviation exist and are known. This non-specificity to the distribution makes it an incredibly versatile theorem.

The theorem states that the proportion of results within \(k\) standard deviations from the mean is at least \(1 - \frac{1}{k^2}\) for any \(k > 1\). The exercise demonstrates how the calculated proportions within \(\mu \pm 2\sigma\) and \(\mu \pm 3\sigma\) intervals indeed adhere to the theorem's bounds.

It's remarkable that without knowing the actual distribution of our hypergeometric random variable, Tchebysheff's Theorem can still estimate the minimum proportion of outcomes around the mean. This is helpful as a check on the validity of our calculated intervals. For those delving into statistics, understanding and applying Tchebysheff's Theorem can be a significant asset in your analytical toolkit.