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Show that \(E(X)=n p\) when \(X\) is a binomial random variable.

Short Answer

Expert verified
The expected value of a binomial random variable is \(E(X) = n \cdot p\).

Step by step solution

01

Understand the Problem

We need to prove that the expected value of a binomial random variable is equal to the product of the number of trials and the probability of success in each trial. The binomial distribution has parameters: the number of trials \(n\) and the probability of success \(p\).
02

Define the Binomial Random Variable

A binomial random variable \(X\) can be defined as the number of successes in \(n\) independent Bernoulli trials. Each trial has only two possible outcomes (success or failure) with the probability of success \(p\).
03

Write the Expectation Formula for Discrete Variables

The expected value of a discrete random variable \(X\) is calculated as \(E(X) = \sum_{k=0}^{n} k \cdot P(X = k)\), where \(P(X = k)\) is the probability that the random variable takes the value \(k\).
04

Use the Binomial Probability Formula

For \(X\) following a binomial distribution, the probability mass function is given by \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\). Substitute this into the expectation formula: \(E(X) = \sum_{k=0}^{n} k \cdot \binom{n}{k} p^k (1-p)^{n-k}\).
05

Apply Linear Combinations of Random Variables

Notice that \(X\) can be expressed as a sum of \(n\) independent Bernoulli trials, i.e., \(X = X_1 + X_2 + \ldots + X_n\), where each \(X_i\) is a Bernoulli random variable with parameter \(p\). Since \(E(X_i) = p\), then \(E(X) = E(X_1) + E(X_2) + \ldots + E(X_n) = n \cdot p\).
06

Conclude the Derivation

We have shown that \(E(X) = n \cdot p\) directly using the properties of a linear combination of independent random variables. This confirms the expected value for a binomial random variable.

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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** is a core concept in probability and statistics that represents the average or mean value we can anticipate from a random variable over many trials. In the case of a binomial random variable, the expected value, denoted as \( E(X) \), gives us insight into the long-term average number of successes in a series of independent Bernoulli trials.

Think of the expected value as a measure of the "central tendency" of a probability distribution. For a binomial distribution with parameters \( n \) (number of trials) and \( p \) (probability of success), the expected value formula can be introduced simply as:
  • \( E(X) = n \cdot p \)
This formula highlights that the expected number of successes is directly proportional to both the number of trials and the probability of each trial resulting in a success. It's important because it helps predict outcomes for practical scenarios, such as predicting the number of heads in 100 coin tosses.For discrete probability distributions like the binomial distribution, calculating the expected value involves summing the product of each possible value of the random variable with its associated probability. This allows you to find the average expected outcome without performing endless trials.
Bernoulli Trials
**Bernoulli trials** are a building block of the binomial distribution. Each Bernoulli trial represents a single experiment or action that results in one of two outcomes: success or failure. The probability of success \( p \) remains constant in each trial.

Understanding Bernoulli trials helps us grasp why the binomial distribution is structured the way it is. When you perform several independent Bernoulli trials, where each trial's outcome does not affect others, you're essentially conducting a binomial experiment.
Examples of Bernoulli trials include:
  • Flipping a coin, where heads might be considered success
  • Checking whether a light bulb works on the first try
In any binomial distribution, every independent trial is a Bernoulli trial. The critical property here is independence; the outcome of one trial doesn't depend on or influence another, which assures each trial has the same probability \( p \).
This allows us to model real-world situations where outcomes are binary and encourages using binomial distribution principles in varied statistical analyses.
Discrete Random Variables
A **discrete random variable** is a type of random variable that takes on a countable number of possible values. These are key when analyzing scenarios with distinct outcomes.

In probability, a random variable acts as a function assigning numerical values to outcomes of a random phenomenon. For discrete random variables, this function maps sample space outcomes to integers or whole numbers.
Some examples of discrete random variables:
  • Number of successful free throws out of ten attempts
  • Total count of defective items in a shipment
In a binomial distribution, the random variable \( X \) includes values that represent the number of successes in a fixed number of independent Bernoulli trials. Because these can only take integer values, like 0, 1, 2, up to \( n \), they are precisely defined as discrete.
Given this setup, understanding discrete random variables is essential for comprehending how probabilities are assigned to these outcomes in a structured, logical manner.
Probability Mass Function
A **probability mass function (PMF)** is an essential tool used to specify the probability that a discrete random variable is exactly equal to a certain value. The PMF is crucial in the context of discrete distributions, such as the binomial distribution.

For a random variable \( X \), the PMF \( P(X = k) \) tells us the probability that \( X \) will take the value \( k \). In a binomial distribution, this is expressed as:
  • \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \)
Where \( \binom{n}{k} \) represents the binomial coefficient, often read as "n choose k," calculating combinations of successes. Each part of the function plays a role:
✨ \( p^k \) is the probability of having \( k \) successes
✨ \( (1-p)^{n-k} \) represents the probability of having the rest as failures
The PMF provides a clear structure to visualize how probability is distributed across possible outcomes. Authors and statisticians use PMFs to work through, summarize, and dynamically calculate probabilities for events dictated by binomial random variables. These functions allow for easy computation and analysis in various real-world scenarios like quality control, insurance prediction, etc.

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Most popular questions from this chapter

The number of pumps in use at both a six-pump station and a four-pump station will be determined. Give the possible values for each of the following random variables: a. \(T=\) the total number of pumps in use b. \(X=\) the difference between the numbers in use at stations 1 and 2 c. \(U=\) the maximum number of pumps in use at either station d. \(Z=\) the number of stations having exactly two pumps in use

Of all customers purchasing automatic garage-door openers, \(75 \%\) purchase a chain-driven model. Let \(X=\) the number among the next 15 purchasers who select the chain-driven model. a. What is the pmf of \(X\) ? b. Compute \(P(X>10)\). c. Compute \(P(6 \leq X \leq 10)\). d. Compute \(\mu\) and \(\sigma^{2}\). e. If the store currently has in stock 10 chaindriven models and 8 shaft- driven models, what is the probability that the requests of these 15 customers can all be met from existing stock?

A mail-order computer business has six telephone lines. Let \(X\) denote the number of lines in use at a specified time. Suppose the pmf of \(X\) is as given in the accompanying table. \begin{tabular}{c|ccccccc} \(x\) & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(.10\) & \(.15\) & \(.20\) & \(.25\) & \(.20\) & \(.06\) & \(.04\) \end{tabular} Calculate the probability of each of the following events. a. \(\\{\) at most three lines are in use \(\\}\) b. \\{fewer than three lines are in use \(\\}\) c. \\{at least three lines are in use \(\\}\) d. \\{between two and five lines, inclusive, are in use \\} e. \\{between two and four lines, inclusive, are not in use \(\\}\) f. \\{at least four lines are not in use \(\\}\)

Suppose that you read through this year's issues of the \(N e w\) York Times and record each number that appears in a news article-the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be \(1,2, \ldots, 8\), or 9 . Your first thought might be that the leading digit \(X\) of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford's law: \(p(x)=P(1\) st digit is \(x)=\log _{10}(1+1 / x) x=1,2, \ldots, 9\) a. Compute the individual probabilities and compare to the corresponding discrete uniform distribution. b. Obtain the cdf of \(X\). c. Using the cdf, what is the probability that the leading digit is at most 3 ? At least 5 ? [Note: Benford's law is the basis for some auditing procedures used to detect fraud in financial reporting-for example, by the Internal Revenue Service.]

A test for the presence of a certain disease has probability \(.20\) of giving a false-positive reading (indicating that an individual has the disease when this is not the case) and probability . 10 of giving a false-negative result. Suppose that ten individuals are tested, five of whom have the disease and five of whom do not. Let \(X=\) the number of positive readings that result. a. Does \(X\) have a binomial distribution? Explain your reasoning. b. What is the probability that exactly three of the ten test results are positive?

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