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Find the expected value of a random variable \(X\) having the following probability distribution: $$ \begin{array}{lllllll} \hline \boldsymbol{x} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & \frac{1}{8} & \frac{1}{4} & \frac{3}{16} & \frac{1}{4} & \frac{1}{16} & \frac{1}{8} \\ \hline \end{array} $$

Short Answer

Expert verified
The short answer for the expected value of the random variable \(X\) with the given probability distribution is: \(E[X] = \frac{9}{4}\).

Step by step solution

01

Identify the values of the random variable and their probabilities

We are given the following probability distribution: $$ \begin{array}{lllllll} \hline \boldsymbol{x} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & \frac{1}{8} & \frac{1}{4} & \frac{3}{16} & \frac{1}{4} & \frac{1}{16} & \frac{1}{8} \\ \hline \end{array} $$ From this table, we can see that the values of the random variable \(X\) are \(x=\{0,1,2,3,4,5\}\), and their corresponding probabilities are \(P(X=x)=\left\{\frac{1}{8}, \frac{1}{4}, \frac{3}{16}, \frac{1}{4}, \frac{1}{16}, \frac{1}{8}\right\}\).
02

Calculate the expected value

To calculate the expected value (\(E[X]\)) of the random variable \(X\), we sum the product of each value \(x_i\) and its corresponding probability \(P(X = x_i)\): $$ E[X] = \sum_{i=1}^n x_i P(X=x_i), $$ where \(n\) is the number of different values of \(X\). Using the values and probabilities we identified in step 1, the expected value of \(X\) is: $$ E[X] = (0)\left(\frac{1}{8}\right) + (1)\left(\frac{1}{4}\right) + (2)\left(\frac{3}{16}\right) + (3)\left(\frac{1}{4}\right) + (4)\left(\frac{1}{16}\right) + (5)\left(\frac{1}{8}\right) $$
03

Simplify the expression and find the expected value

We can simplify the expression for \(E[X]\) and calculate the final expected value: $$ E[X] = 0 + \frac{1}{4} + \frac{6}{16} + \frac{3}{4} + \frac{4}{16} + \frac{5}{8} = \frac{1}{4} + \frac{3}{8} + \frac{3}{4} + \frac{1}{4} + \frac{5}{8} $$ $$ E[X] = \frac{8+12+24+8+20}{32} = \frac{72}{32} = \frac{18}{8} $$ Finally, the expected value of the random variable \(X\) is: $$ E[X] = \frac{9}{4} $$

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Key Concepts

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

Probability Distribution
Probability distribution is a fundamental concept in the world of probability and statistics. It represents how probabilities are distributed over the various possible outcomes of a random variable. In simple terms, it tells us the likelihood of each outcome.
For a discrete random variable, a probability distribution is often displayed in a table format, showing each potential value of the variable along with its associated probability. This gives a clear picture of how the total probability (which always sums to 1) is divided among the possible outcomes.
Understanding the probability distribution is key to calculating other statistical measures, such as the expected value. It helps in making informed predictions about the outcomes based on past information or calculations.
  • It describes the likelihood of different outcomes in a random experiment.
  • The sum of all probabilities in a probability distribution equals 1.
  • It's essential for calculating expected values and variances of random variables.
Random Variable
A random variable is a mathematical function that assigns a numeric value to each outcome in a sample space of a random phenomenon. Think of it as a variable that encapsulates all potential outcomes in a probabilistic framework. Random variables can be classified mainly as discrete or continuous. In our context, we're dealing with a discrete random variable.
Discrete random variables take on a countable number of values. For instance, they might represent the roll of a die, the number of heads in a series of coin tosses, or the scores obtained from analyzing a given exercise. Each of these outcomes has a defined and distinct set of possible values.
  • They can be either discrete (countable outcomes) or continuous (infinite outcomes).
  • In this exercise, each outcome corresponds to a specific probability value.
  • Random variables serve as an important foundation for understanding probability distributions and expectations.
Discrete Probability
Discrete probability refers to probabilities associated with discrete random variables. These probabilities are calculated for finite and countable outcomes. Unlike continuous probability, which deals with probabilities of intervals, discrete probability focuses on distinct possible values, much like the example we’re analyzing.
To find the probability of a specific outcome, one simply selects the probability corresponding to that outcome from the probability distribution table. Moreover, the sum of all discrete probabilities in a given distribution will always equal 1. This property ensures that one of the potential outcomes will occur.
  • Applicable to scenarios with distinct, separate outcomes.
  • The probabilities of all possible outcomes add up to 1.
  • Ideal for modeling situations like dice rolls, coin flips, or card draws.
Mathematical Expectation
Mathematical expectation, or expected value, is an integral concept in probability and statistics, representing the average outcome one can anticipate from a random variable if a particular scenario is repeated many times. In essence, it provides a measure of the center of a probability distribution.
For a discrete random variable, the expected value is computed by multiplying each potential outcome by its corresponding probability and summing up all those values. This computation demonstrates the balance point or the "expectation" of the random variable across numerous trials.
In the exercise provided, the expected value helps understand what average result we can "expect" when this random variable (with the given probability distribution) is observed repeatedly. This forms the basis for decision-making in uncertain conditions and is a critical tool in fields ranging from finance to operations research and beyond.
  • Calculated by summing products of all outcomes and their probabilities.
  • Provides an average expected outcome for the random variable.
  • Useful for predictions and strategic decision-making.

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Most popular questions from this chapter

CoLLEGE GRADUATES At a certain university, the probability that an entering freshman will graduate within \(4 \mathrm{yr}\) is .6. From an incoming class of 2000 freshmen, find a. The expected number of students who will graduate within 4 yr. b. The standard deviation of the number of students who will graduate within \(4 \mathrm{yr}\).

FuPPING A CoIN An unbiased coin is tossed 1 million times. Show that the probability is at least. 99 that the proportion of times the coin will land heads is between \(.495\) and . 505 , inclusive. (Note: These results show that if an unbiased coin is tossed a very large number times, the proportion of times that the coin will land heads is very close to \(\frac{1}{2}\).) Hint: Use Chebychev's inequality.

To be eligible for further consideration, applicants for certain civil service positions must first pass a written qualifying examination on which a score of 70 or more must be obtained. In a recent examination it was found that the scores were normally distributed with a mean of 60 points and a standard deviation of 10 points. Determine the percentage of applicants who passed the written qualifying examination.

Let \(Z\) be the standard normal variable. Find the values of \(z\) if \(z\) satisfies a. \(P(Z>-z)=.9713\) b. \(P(Z<-z)=.9713\)

The following table gives the scores of 30 students in a mathematics examination: $$ \begin{array}{lccccc} \hline \text { Scores } & 90-99 & 80-89 & 70-79 & 60-69 & 50-59 \\ \hline \text { Students } & 4 & 8 & 12 & 4 & 2 \\ \hline \end{array} $$ Find the mean and the standard deviation of the distribution of the given data.

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