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Chapter 3: Problem 42

Suppose \(E(X)=5\) and \(E[X(X-1)]=27.5\). What is a. \(E\left(X^{2}\right)\) ? [Hint: \(E[X(X-1)]=E\left[X^{2}-X\right]=E\left(X^{2}\right)-\) \(E(X)] ?\) b. \(V(X)\) ? c. The general relationship among the quantities \(E(X)\), \(E[X(X-1)]\), and \(V(X)\) ?

Short Answer

Expert verified
a. \(E[X^2] = 32.5\); b. \(V(X) = 7.5\); c. Relationships use identities involving expectations and variance.

Step by step solution

01

Understand the Given Information

We are given two expected values: \(E(X)=5\) and \(E[X(X-1)]=27.5\). Also, the hint provides the formula \(E[X(X-1)]=E[X^2]-E(X)\). Our task is to find \(E(X^2)\) using this equation.
02

Calculate E(X^2)

As given, \(E[X(X-1)]=E[X^2]-E(X)\). Substituting the known values:\[27.5 = E[X^2] - 5.\]To find \(E[X^2]\), rearrange the equation:\[E[X^2] = 27.5 + 5 = 32.5.\]
03

Calculate Variance V(X)

Variance \(V(X)\) is calculated using the formula \(V(X) = E[X^2] - (E[X])^2\). We already found \(E[X^2] = 32.5\) and \(E(X) = 5\). Now substitute these into the variance formula:\[V(X) = 32.5 - 5^2 = 32.5 - 25 = 7.5.\]
04

Describe General Relationship

The relationship among the given quantities is based on the identity used in variance calculation and the problem hint. From the hint, we have \(E[X(X-1)]=E[X^2]-E(X)\). Variance \(V(X)\) is related as \(V(X) = E[X^2] - (E[X])^2\). Hence the relationships between these expectations and variance involve manipulation of expectations and the definitions of variance.

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Key Concepts

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

Expected Value
Expected value, often denoted as \(E(X)\), is a fundamental concept in probability theory. It represents the average or mean value of a random variable over a large number of trials. In simpler terms, it's what we "expect" the outcome to be when we perform a random experiment multiple times. For instance, if we expect something to happen five times out of ten trials, then the expected value would be 5.
  • The concept provides a summary measure that captures the central tendency of a distribution.
  • Expected value is calculated by multiplying each possible outcome by the probability of that outcome, then summing all those values.
In the context of our problem, we are given \(E(X) = 5\). This means, on average, the random variable \(X\) takes the value of 5. This information is pivotal for further calculations like finding variance.
Variance Calculation
Variance is a measure that describes the spread or dispersion of a set of values. It tells us how much the values deviate from the expected value.
  • More specifically, variance \(V(X)\) quantifies how far each outcome of a random variable \(X\) is from the expected value \(E(X)\).
  • To calculate variance, we use the formula \(V(X) = E[X^2] - (E[X])^2\).
In our case, we already determined \(E(X) = 5\) and \(E[X^2] = 32.5\). Now substituting these values into the variance formula gives us:
\[V(X) = 32.5 - 5^2 = 32.5 - 25 = 7.5.\]
This simple calculation allows us to understand the spread of our variable around its expected value, and a variance of 7.5 indicates the degree of this spread.
Expected Product
The concept of the expected product comes into play when we deal with the expectation involving products of variables, like \(E[X(X - 1)]\). This concept is vital for extending the traditional idea of expectation to cases involving products of expressions.
  • The function \(X(X - 1)\) represents a slightly different calculation than \(X^2\), accounting for outcomes in succession.
  • Utilizing the relation \(E[X(X-1)] = E[X^2] - E(X)\) helps reformulate and simplify complex expressions.
For our problem, knowing \(E[X(X-1)] = 27.5\) allows us to calculate \(E[X^2]\) by rearranging the formula, giving \(E[X^2] = 32.5\). It's a neat trick to bridge between different forms of expectations, aiding in calculating variances as well.

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Most popular questions from this chapter

The negative binomial rv \(X\) was defined as the number of \(F\) 's preceding the \(r\) th \(S\). Let \(Y=\) the number of trials necessary to obtain the \(r\) th \(S\). In the same manner in which the pmf of \(X\) was derived, derive the pmf of \(Y\).

A \(k\)-out-of-n system is one that will function if and only if at least \(k\) of the \(n\) individual components in the system function. If individual components function independently of one another, each with probability \(.9\), what is the probability that a 3 -out-of-5 system functions?

A chemical supply company currently has in stock \(100 \mathrm{lb}\) of a certain chemical, which it sells to customers in 5 -lb lots. Let \(X=\) the number of lots ordered by a randomly chosen customer, and suppose that \(X\) has pmf $$ \begin{array}{l|llll} x & 1 & 2 & 3 & 4 \\ \hline p(x) & .2 & .4 & .3 & .1 \end{array} $$ Compute \(E(X)\) and \(V(X)\). Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left.

An educational consulting firm is trying to decide whether high school students who have never before used a handheld calculator can solve a certain type of problem more easily with a calculator that uses reverse Polish logic or one that does not use this logic. A sample of 25 students is selected and allowed to practice on both calculators. Then each student is asked to work one problem on the reverse Polish calculator and a similar problem on the other. Let \(p=P(S)\), where \(S\) indicates that a student worked the problem more quickly using reverse Polish logic than without, and let \(X=\) number of \(S\) 's. a. If \(p=.5\), what is \(P(7 \leq X \leq 18)\) ? b. If \(p=.8\), what is \(P(7 \leq X \leq 18)\) ? c. If the claim that \(p=.5\) is to be rejected when either \(X \leq\) 7 or \(X \geq 18\), what is the probability of rejecting the claim when it is actually correct? d. If the decision to reject the claim \(p=.5\) is made as in part (c), what is the probability that the claim is not rejected when \(p=.6\) ? When \(p=.8\) ? e. What decision rule would you choose for rejecting the claim \(p=.5\) if you wanted the probability in part (c) to be at most .01?

Consider a collection \(A_{1}, \ldots, A_{k}\) of mutually exclusive and exhaustive events, and a random variable \(X\) whose distribution depends on which of the \(A_{i}\) 's occurs (e.g., a commuter might select one of three possible routes from home to work, with \(X\) representing the commute time). Let \(E\left(X \mid A_{i}\right)\) denote the expected value of \(X\) given that the event \(A_{i}\) occurs. Then it can be shown that \(E(X)=\) \(\Sigma E\left(X \mid A_{i}\right) \cdot P\left(A_{i}\right)\), the weighted average of the individual "conditional expectations" where the weights are the probabilities of the partitioning events. a. The expected duration of a voice call to a particular telephone number is 3 minutes, whereas the expected duration of a data call to that same number is 1 minute. If \(75 \%\) of all calls are voice calls, what is the expected duration of the next call? b. A deli sells three different types of chocolate chip cookies. The number of chocolate chips in a type \(i\) cookie has a Poisson distribution with parameter \(\lambda_{i}=i+1\) \((i=1,2,3)\). If \(20 \%\) of all customers purchasing a chocolate chip cookie select the first type, \(50 \%\) choose the second type, and the remaining \(30 \%\) opt for the third type, what is the expected number of chips in a cookie purchased by the next customer?
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