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Bernoulli, the mean. Use the probability rules from Section 3.5 to derive the mean of a Bernoulli random variable, i.e. a random variable \(X\) that takes value 1 with probability \(p\) and value 0 with probability \(1-p .\) That is, compute the expected value of a generic Bernoulli random variable.

Short Answer

Expert verified
The expected value of a Bernoulli random variable is \(p\).

Step by step solution

01

Understanding the Bernoulli Random Variable

A Bernoulli random variable, denoted by \(X\), can take two values: 1 with probability \(p\), and 0 with probability \(1-p\). This describes a simple experiment with only two possible outcomes: success (1) or failure (0).
02

Formula for Expectation of a Random Variable

The expected value or mean of a discrete random variable \(X\) is given by the formula: \(E(X) = \sum_{i=1}^{n} x_i P(X = x_i)\), where \(x_i\) are the possible values \(X\) can take and \(P(X = x_i)\) is the probability of each value.
03

Apply the Expectation Formula to Bernoulli

For the Bernoulli random variable \(X\), the possible values \(x_i\) are 0 and 1. Thus, we apply the formula: \(E(X) = 0 \times P(X=0) + 1 \times P(X=1)\).
04

Substitute the Probabilities

Substituting the probabilities into the formula: \(P(X = 0) = 1 - p\) and \(P(X = 1) = p\), we get: \(E(X) = 0 \times (1 - p) + 1 \times p\).
05

Calculate the Expected Value

Calculate \(E(X) = 0 + p = p\). This shows the expected value of a Bernoulli random variable is simply its success probability \(p\).

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Key Concepts

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

Expected Value
The concept of expected value provides a way to summarize and predict outcomes of random experiments. Imagine a random variable, like the number of goals a team might score in a game. The expected value helps us understand the average or most likely outcome over multiple trials.

To find the expected value of a discrete random variable, we use the formula:
  • \(E(X) = \sum_{i=1}^{n} x_i P(X = x_i)\)
This equation tells us to multiply each possible value the variable can take by its likelihood (probability), and then sum up all these products.

For instance, with a Bernoulli random variable, which takes values 0 and 1, the expected value would mean calculating \(E(X) = 0 \times (1-p) + 1 \times p\), which simplifies to \(E(X) = p\).

The expected value \(p\) here represents the probability of a 'success,' offering a straightforward measure of central tendency for such binary outcomes.
Discrete Random Variable
Discrete random variables are quite intriguing as they can only take specific, countable values. When dealing with probability, it's important to differentiate between discrete and continuous random variables.

Discrete random variables are defined by a set of exhaustive possible outcomes which could be finite or infinite but countable. For example, the number of students in a classroom is a discrete random variable because you can count the number of students without any in-between values.

In the case of a Bernoulli random variable, the scenario is even simpler: you only have two options, typically coded as 0 and 1.
  • 0 represents failure or absence of a specific outcome.
  • 1 represents success or occurrence of a particular event.
When solving problems involving discrete random variables, you'll often employ simple algebraic techniques to calculate probabilities and outcomes, making them a cornerstone in understanding the broader topic of probability theory.
Probability Rules
Probability rules are fundamental principles that aid in the computation and understanding of random events. These rules help you work through problems involving random variables like the Bernoulli variable seamlessly.

Basic Probability Rules:

  • The sum of probabilities for all possible outcomes of a random variable is always 1, i.e., \( P(A) + P(eg A) = 1 \).
  • If two events cannot occur simultaneously (mutually exclusive), the probability of one or the other occurring is the sum of their individual probabilities.

Bernoulli Distribution:

When working with a Bernoulli random variable, important rules include:
  • \(P(X=0) = 1 - p\), which is the probability of a failure.
  • \(P(X=1) = p\), which is the probability of success.
Using these rules helps simplify the task of finding the expected value, as you know precisely how to break down the probabilities according to the outcomes defined by the Bernoulli variable.

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Most popular questions from this chapter

Spray paint, Part II. As described in Exercise 4.14 , the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribution with a mean of 25 square feet and a standard deviation of 3 square feet. (a) What is the probability that the area covered by a can of spray paint is more than 27 square feet? (b) Suppose you want to spray paint an area of 540 square feet using 20 cans of spray paint. On average, how many square foet must each can be able to cover to spray paint all 540 square feet? (c) What is the probability that you can cover a 540 square feet area using 20 cans of spray paint? (d) If the area covered by a can of spray paint had a slightly skewed distribution, could you still calculate the probabilities in parts (a) and (c) using the normal distribution?

Area under the curve, Part I. What percent of a standard normal distribution \(N(\mu=0, \sigma=1)\) is found in each region? Be sure to draw a graph. (a) \(Z<-1.35\) (b) \(Z>1.48\) (c) \(-0.42\)

Defective rate. A machine that produces a special type of transistor (a component of computers) has a \(2 \%\) defective rate. The production is considered a random process where each transistor is independent of the others. (a) What is the probability that the \(10^{t h}\) transistor produced is the first. with a defect? (b) What is the probability that the machine produces no defective transistors in a batch of \(100 ?\) (c) On average, how many transistors would you expect to be produced before the first with a defect? What is the standard deviation? (d) Another machine that also produces transistors has a \(5 \%\) defective rate where each transistor is produced independent of the others. On average how many transistors would you expect to be produced with this machine before the first with a defect? What is the standard deviation? (e) Based on your answers to parts (c) and (d), how does increasing the probability of an event affect the mean and standard deviation of the wait time until success?

Stats final scores. Each year about 1500 students take the introductory statistics course at a large university. This year scores on the final exam are distributed with a median of 74 points, a mean of 70 points, and a standard deviation of 10 points. There are no students who scored above 100 (the maximum score attainable on the final) but a few students scored below 20 points. (a) Is the distribution of scores on this final exam symmetric, right skewed, or left skewed? (b) Would you expect most students to have scored above or below 70 points? (c) Can we calculate the probability that a randomly chosen student scored abave 75 using the normal distribution? (d) What is the probability that the average score for a random sample of 40 students is above \(75 ?\) (e) How would eutting the sample size in half affect the standard deviation of the mean?

Eye color, Part I. A husband and wife both have brown eyes but carry genes that make it possible for their children to have brown eyes (probability 0.75\()\), blue eyes \((0.125),\) or green eyes (0.125) . (a) What is the probability the first blue-eyed child they have is their third child? Assume that the eye colors of the children are independent of each other. (b) On average, how many children would such a pair of parents have before having a blue-eyed child? What is the standard deviation of the number of children they would expect to have until the first blue-eyed child?

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