/*! 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 134 If \(X\) is a hypergeometric rv,... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(X\) is a hypergeometric rv, show directly from the definition that \(E(X)=n M / N\) (consider only the case \(n

Short Answer

Expert verified
The expected value of the hypergeometric random variable is \(E(X) = \frac{nM}{N}\).

Step by step solution

01

Understand the Hypergeometric Distribution

The hypergeometric distribution is used when we are drawing samples without replacement. The formula for a hypergeometric random variable is:\[ P(X = x) = \frac{{\binom{M}{x} \binom{N-M}{n-x}}}{\binom{N}{n}} \]where \(M\) is the total number of successes in the population, \(N\) is the population size, \(n\) is the number of draws, and \(x\) is the number of observed successes.
02

Write the Expectation Formula

The expected value of a discrete random variable is calculated as:\[ E(X) = \sum_{x} x P(X = x) \]For the hypergeometric distribution, this becomes:\[ E(X) = \sum_{x} x \frac{{\binom{M}{x} \binom{N-M}{n-x}}}{\binom{N}{n}} \]
03

Factor out \(\frac{n M}{N}\)

Notice that:\[ \frac{M}{N} \cdot x = \frac{M}{N} \cdot \frac{n \binom{M-1}{x-1} \binom{N-M}{n-x}}{\binom{M}{x}} \]This allows us to factor \(\frac{M}{N}\) out of the sum for \(E(X)\), resulting in:\[ E(X) = \frac{n M}{N} \sum_{x} \frac{\binom{M-1}{x-1} \binom{N-M}{n-x}}{\binom{M}{x}} \]
04

Change the Index of the Sum

Let \(y = x - 1\), then the sum becomes over \(y\):\[ \sum_{y=0}^{n-1} \frac{\binom{M-1}{y} \binom{N-M}{n-y-1}}{\binom{N-1}{n-1}} \]This expression is the probability mass function of a hypergeometric distribution \(h(y; n-1, M-1, N-1)\), simplifying the sum to 1.
05

Simplify and Conclude

Since the sum \(\sum_{y=0}^{n-1} h(y; n-1, M-1, N-1) = 1\), we have:\[ E(X) = \frac{nM}{N} \times 1 = \frac{nM}{N} \]Thus, it is shown that the expected value of a hypergeometric random variable is \(E(X) = \frac{n M}{N}\).

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Key Concepts

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

Expected Value
The expected value, often denoted as \(E(X)\), is a fundamental concept in probability and statistics. It represents the average or mean value you would expect to get if you could repeat the random process an infinite number of times.
The expected value is key in decision-making and statistics, as it offers a single summary number that tells us the center of the distribution of the random variable. For the hypergeometric distribution, calculating the expected value involves using the distribution's specific probability function. To compute \(E(X)\), you multiply each possible outcome \(x\) by its probability and sum these products.
The formula for this calculation in a hypergeometric distribution is:
  • \(E(X) = \sum_{x} x P(X = x)\)
In the context of the problem we are working on, by simplifying the sum and factoring terms, you can show that the expected value becomes \(\frac{nM}{N}\). This formula gives us the expected number of successes when sampling \(n\) items from a set of \(N\) items, where \(M\) are successes.
Probability Mass Function
The Probability Mass Function, or PMF, describes the probability distribution of a discrete random variable. It provides the probability for each possible value of the random variable.
In other words, it maps out which values the random variable can take and the likelihood of these values.
For a hypergeometric random variable, the PMF is given by:
  • \( P(X = x) = \frac{{\binom{M}{x} \binom{N-M}{n-x}}}{\binom{N}{n}} \)
Here, \(\binom{M}{x}\) represents the number of ways to choose \(x\) successes from \(M\) possible successes, and \(\binom{N-M}{n-x}\) gives us the number of ways to choose the remaining draws from the rest of the population. The denominator, \(\binom{N}{n}\) is the total number of ways to choose \(n\) items out of \(N\) without regard to order.
This function allows us to understand how likely we are to observe a specific number of successes when drawing without replacement. It is a critical tool for describing discrete distributions like the hypergeometric distribution.
Random Variable
A random variable is a variable that can take different values based on the outcome of a random phenomenon. It encapsulates the idea of randomness in a mathematical representation.
There are two main types of random variables: discrete and continuous. Here, we focus on discrete random variables. For example, when dealing with a hypergeometric distribution, \(X\) is a discrete random variable. It represents the number of successes in a sample drawn from a finite population. In this context, the values \(X\) can take are non-negative integers, each corresponding to the different numbers of successes that can occur.
The probability that \(X\) takes a specific value is determined by the probability mass function.Understanding random variables and their distributions is crucial as they serve as the foundation for statistical inference and prediction. By modeling the random phenomenon with a random variable, you are able to use mathematical tools to analyze and interpret data.
Discrete Distribution
A discrete distribution is one where the random variable can take a finite or countably infinite number of values. Each of these values has its own associated probability. Unlike continuous distributions, where values fall within a range, discrete distributions have distinct and separate values. The hypergeometric distribution is an example of a discrete distribution. It models the number of successes in a sample drawn without replacement from a finite population. Each outcome in the distribution can be listed, and the probability of each outcome is determined using the probability mass function.
The hypergeometric distribution is particularly useful in scenarios where the lack of replacement affects the probabilities of subsequent draws, such as card games or quality control in manufacturing. Understanding discrete distributions is important as they model many real-world phenomena and help in making predictions and decisions based on data. Each scenario using a discrete distribution provides specific insights into how random variables can behave based on defined criteria and limitations.

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Most popular questions from this chapter

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

A newsstand has ondered five copies of a certain issue of a photography magazine. Let \(X=\) the number of individuals who come in to purchase this magazine. If \(X\) has a Poisson distribution with parameter \(\lambda=4\), what is the expected number of copies that are sold?

Suppose that the number of plants of a particular type found in a rectangular region (called a quadrat by ecologists) in a certain geographic area is an rv \(X\) with pmf $$ p(x)=\left\\{\begin{array}{cc} c / x^{3} & x=1,2,3, \ldots \\ 0 & \text { otherwise } \end{array}\right. $$ Is \(E(X)\) finite? Justify your answer (this is another distribution that statisticians would call heavy-tailed).

An instructor who taught two sections of statistics last term, the first with 20 students and the second with 30 , decided to assign a term project. After all projects had been tumed in, the instructor randomly ordered them before grading. Consider the first 15 graded projects. a. What is the probability that exactly 10 of these are from the second section? b. What is the probability that at least 10 of these are from the second section? c. What is the probability that at least 10 of these are from the same section? d. What are the mean value and standard deviation of the number among these 15 that are from the second section? e. What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

A company that produces fine crystal knows from experience that \(10 \%\) of its goblets have cosmetic flaws and must be classified as "seconds." a. Among six randomly selected goblets, how likely is it that only one is a second? b. Among six randomly selected goblets, what is the probability that at least two are seconds? c. If goblets are examined one by one, what is the probability that at most five must be selected to find four that are not seconds?

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