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91Ó°ÊÓ

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

Short Answer

Expert verified
\(E(X^2) = 32.5\); \(V(X) = 7.5\); Relationships: \(E[X(X-1)] = E(X^2) - E(X)\) and \(V(X) = E(X^2) - E(X)^2\).

Step by step solution

01

Understand the Given Information

We are given that \(E(X) = 5\) and \(E[X(X-1)] = 27.5\). We need to find \(E(X^2)\), \(V(X)\), and the general relationship between these quantities.
02

Identify the Relationship Formula

The provided hint tells us that \(E[X(X-1)] = E(X^2) - E(X)\). This is a formula derived from expanding \(X(X-1) = X^2 - X\).
03

Solve for \(E(X^2)\)

Substitute the known values into the relationship \(E(X^2) - E(X) = E[X(X-1)]\), which becomes \(E(X^2) - 5 = 27.5\).
04

Calculate \(E(X^2)\)

Rearranging gives \(E(X^2) = 27.5 + 5 = 32.5\). Thus, \(E(X^2) = 32.5\).
05

Use the Variance Formula

Recall that the variance of \(X\), \(V(X) = E(X^2) - (E(X))^2\).
06

Substitute to Find \(V(X)\)

We have \(E(X^2) = 32.5\) and \(E(X) = 5\). Therefore, the variance is \(V(X) = 32.5 - 5^2 = 32.5 - 25\).
07

Calculate \(V(X)\)

Thus, \(V(X) = 7.5\).
08

General Relationship

The general relationship among these quantities can be expressed as \(E[X(X-1)] = E(X^2) - E(X)\) and the variance \(V(X) = E(X^2) - (E(X))^2\).

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Key Concepts

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

Variance
Variance is a measure of how much a random variable, such as \(X\), spreads out from its expected value. It gives us an idea of how values in a data set are distributed around the mean. Calculating variance involves understanding both the expected value \(E(X)\) and the expected value of the square of \(X\), which is \(E(X^2)\).
To calculate variance, the formula we use is:
  • \(V(X) = E(X^2) - (E(X))^2\)
The variance formula tells us that we start with the expected value of the squared outcomes, \(E(X^2)\), and subtract the square of the expected value, \((E(X))^2\).
Here's a simple way to see it: You’re looking at the average of squared distances from the mean. In our example:
  • \(E(X) = 5\)
  • \(E(X^2) = 32.5\)
Substituting these into the variance formula, we get:
  • \(V(X) = 32.5 - 5^2 = 32.5 - 25 = 7.5\)
This means the variance, \(V(X)\), is 7.5, which shows the average squared deviation of the data from the mean.
Relationship Formula
The relationship formula links different statistical measures of random variables. It helps to understand how various expectations interact and can be manipulated to derive unknown quantities.
One such useful relationship presented in this exercise is:
  • \(E[X(X-1)] = E(X^2) - E(X)\)
This equation stems from an identity: multiplying \(X\) by \(X-1\) gives \(X^2 - X\). Thus, the expectation, \(E\), applies similarly like arithmetic operations:
  • \(E[X(X-1)] = E(X^2 - X) = E(X^2) - E(X)\)
Using this, we could solve for \(E(X^2)\) if given \(E(X)\) and \(E[X(X-1)]\). Plugging our values in:
  • \(27.5 = E(X^2) - 5\)
Adding 5 to both sides to find \(E(X^2)\):
  • \(E(X^2) = 32.5\)
This demonstrates how the formula allows determination of a missing expectation from known values.
Random Variables
Random variables are foundational to probability and statistics, representing numerically-valued outcomes of random phenomena. They can be thought of as variables whose possible values are numerical outcomes of a random phenomenon.
We encounter different types of random variables:
  • Discrete Random Variables: These have countable outcomes, like rolling a die or flip of a coin.
  • Continuous Random Variables: These have an infinite number of outcomes, such as measuring time or weight.
In working with random variables, concepts like the expectation and variance are crucial. The expectation, \(E(X)\), also known as the mean, tells us the average outcome if an experiment is repeated many times. Variance, as discussed, measures how spread out the values of the random variable are from the expectation.
The understanding of random variables and these measures forms the backbone for statistical inference, allowing us to make predictions and analyze patterns from data.

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Most popular questions from this chapter

Consider writing onto a computer disk and then sending it through a certifier that counts the number of missing pulses. Suppose this number \(X\) has a Poisson distribution with parameter \(\mu=.2\). (Suggested in "Average Sample Number for Semi-Curtailed Sampling Using the Poisson Distribution," \(J\). Quality Technology, 1983: 126-129.) a. What is the probability that a disk has exactly one missing pulse? b. What is the probability that a disk has at least two missing pulses? c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

The article 'Should You Report That FenderBender?" (Consumer Reports, Sept. 2013: 15) reported that 7 in 10 auto accidents involve a single vehicle (the article recommended always reporting to the insurance company an accident involving multiple vehicles). Suppose 15 accidents are randomly selected. Use Appendix Table A.l to answer each of the following questions. a. What is the probability that at most 4 involve a single vehicle? b. What is the probability that exactly 4 involve a single vehicle? c. What is the probability that exactly 6 involve multiple vehicles? d. What is the probability that between 2 and 4 , inclusive, involve a single vehicle? e. What is the probability that at least 2 involve a single vehicle? f. What is the probability that exactly 4 involve a single vehicle and the other 11 involve multiple vehicles?

Let \(X\) have a Poisson distribution with parameter \(\mu\). Show that \(E(X)=\mu\) directly from the definition of expected value. [Hint: The first term in the sum equals 0 , and then \(x\) can be canceled. Now factor out \(\mu\) and show that what is left sums to 1.]

Customers at a gas station pay with a credit card (A), debit card \((B)\), or cash \((C)\). Assume that successive customers make independent choices, with \(P(A)=.5\), \(P(B)=.2\), and \(P(C)=.3\). a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the 100 who don't pay with cash.

Give three examples of Bernoulli rv's (other than those in the text).

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