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Determine the cumulative distribution function of a binomial random variable with \(n=3\) and \(p=1 / 4\).

Short Answer

Expert verified
The CDF values are: \(F(0) = 0.421875\), \(F(1) = 0.84375\), \(F(2) = 0.984375\), \(F(3) = 1.0\).

Step by step solution

01

Understand the Binomial Distribution

The binomial distribution represents the number of successes in a fixed number of independent Bernoulli trials, each with success probability \(p\). Here, \(n=3\) (the number of trials), and \(p=\frac{1}{4}\) (the probability of success on each trial).
02

Formula for Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) is the probability that the binomial random variable \(X\) takes a value less than or equal to \(x\). It is given by \( F(x) = P(X \leq x) = \sum_{k=0}^{x} \binom{n}{k} p^k (1-p)^{n-k} \).
03

Calculate Probability for Different Values of X

To find the CDF, compute \(P(X = k)\) for \(k = 0, 1, 2, 3\):- \(P(X = 0) = \binom{3}{0}(\frac{1}{4})^0(\frac{3}{4})^3 = 0.421875\)- \(P(X = 1) = \binom{3}{1}(\frac{1}{4})^1(\frac{3}{4})^2 = 0.421875\)- \(P(X = 2) = \binom{3}{2}(\frac{1}{4})^2(\frac{3}{4})^1 = 0.140625\)- \(P(X = 3) = \binom{3}{3}(\frac{1}{4})^3(\frac{3}{4})^0 = 0.015625\)
04

Compute the Cumulative Probabilities

Sum the probabilities to find the CDF:- \(F(0) = P(X \leq 0) = 0.421875\)- \(F(1) = P(X \leq 1) = 0.421875 + 0.421875 = 0.84375\)- \(F(2) = P(X \leq 2) = 0.84375 + 0.140625 = 0.984375\)- \(F(3) = P(X \leq 3) = 0.984375 + 0.015625 = 1.0\)

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Key Concepts

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

Cumulative Distribution Function (CDF)
In probability and statistics, the cumulative distribution function (CDF) is pivotal when working with random variables. It provides a way to determine the probability that a random variable takes on a value less than or equal to a certain threshold. For the binomial distribution, the CDF sums the probabilities of achieving a specific number of successes or fewer in a series of independent experiments, also known as Bernoulli trials.

To compute the CDF of a binomial distribution, you use the formula:
  • \[ F(x) = P(X \leq x) = \sum_{k=0}^{x} \binom{n}{k} p^k (1-p)^{n-k} \]
This equation showcases how you sum the probabilities of all success counts from 0 up to x. This management of probabilities is integral when you are interested in the likelihood of multiple events happening over a fixed number of trials. The CDF is an essential tool because it allows us to consider cumulative probabilities, which are often more informative in decision-making processes than individual probabilities alone.

The CDF has several practical applications, particularly in hypothesis testing and reliability engineering, where knowing the likelihood of a certain number of successes is crucial.
Probability Calculations
Calculating probabilities in the context of binomial distribution involves understanding the mathematical structure that deals with multiple trials and their results. When performing probability calculations for a binomial distribution, we focus on the probability mass function (PMF), which gives the probability of a specific number of successes in a fixed number of trials.

The PMF is calculated using the formula:
  • \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
Here, \( \binom{n}{k} \) is the binomial coefficient, which calculates the number of ways to choose k successes in n trials. The other parts of the formula, \( p^k \) and \( (1-p)^{n-k} \), represent the probability of success raised to the power of k and the probability of failure raised to the power of the remaining trials, respectively.

This formula is central to calculating probabilities for various values of k. For our example with \( n=3 \) and \( p=1/4 \), we computed probabilities for \( X = 0, 1, 2, \) and \( 3 \). By doing so, we began with the building blocks necessary for determining the CDF, as each calculated probability plays a role in determining the cumulative totals.
Bernoulli Trials
At the heart of understanding a binomial distribution is the concept of Bernoulli trials. Bernoulli trials are experiments or processes that result in a binary outcome, typically termed "success" or "failure." In any Bernoulli trial:
  • There are only two possible outcomes.
  • The probability of success is constant in each trial.
  • Each trial is independent of the others.
The primary example used is the flipping of a fair coin: getting a "heads" (success) or "tails" (failure).

The binomial distribution itself is essentially a series of Bernoulli trials. When you perform a certain number \( n \) of these trials with a constant probability of success \( p \), you are engaged in a process that can be modeled by a binomial distribution. Our task of determining the cumulative distribution function of a binomial random variable indeed relies on these foundational principles.Understanding Bernoulli trials assists in visualizing the randomness and inherent variability found in binomial experiments. Moreover, it lays the groundwork for further studies in probability and statistics, highlighting how simple binary outcomes can translate into more complex probability distributions.

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Most popular questions from this chapter

Astronomers treat the number of stars in a given volume of space as a Poisson random variable. The density in the Milky Way Galaxy in the vicinity of our solar system is one star per 16 cubic light-years. (a) What is the probability of two or more stars in 16 cubic light-years? (b) How many cubic light-years of space must be studied so that the probability of one or more stars exceeds \(0.95 ?\)

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05 flaw per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel. (a) What is the probability that there are no surface flaws in an auto's interior? (b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws? (c) If 10 cars are sold to a rental company, what is the probability that at most one car has any surface flaws?

An automated egg carton loader has a \(1 \%\) probability of cracking an egg, and a customer will complain if more than one egg per dozen is cracked. Assume each egg load is an independent event. (a) What is the distribution of cracked eggs per dozen? Include parameter values. (b) What are the probability that a carton of a dozen eggs results in a complaint? c) What are the mean and standard deviation of the number of cracked eggs in a carton of one dozen?

A shipment of chemicals arrives in 15 totes. Three of the totes are selected at random, without replacement, for an inspection of purity. If two of the totes do not conform to purity requirements, what is the probability that at least one of the nonconforming totes is selected in the sample?

Messages that arrive at a service center for an information systems manufacturer have been classified on the basis of the number of keywords (used to help route messages) and the type of message, either e-mail or voice. Also, \(70 \%\) of the messages arrive via e-mail and the rest are voice. $$ \begin{array}{llllll} \text { number of keywords } & 0 & 1 & 2 & 3 & 4 \\ \text { e-mail } & 0.1 & 0.1 & 0.2 & 0.4 & 0.2 \\ \text { voice } & 0.3 & 0.4 & 0.2 & 0.1 & 0 \end{array} $$ Determine the probability mass function of the number of keywords in a message.

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