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Chapter 3: Problem 22

A single fair die is tossed once. Let \(Y\) be the number facing up. Find the expected value and variance of \(Y\).

Short Answer

Expert verified
The expected value of \(Y\) is 3.5 and the variance is 2.9167.

Step by step solution

01

Understand the problem

A fair die is a six-sided die where each side has an equal probability of landing face up when tossed. The problem asks us to calculate the expected value and variance of the number facing up, which we denote as \(Y\).
02

List the possible outcomes of the die roll

The possible outcomes of a die roll are the numbers 1 through 6. These represent all the possible values for \(Y\).
03

Determine the probability distribution

Since the die is fair, each number has an equal chance of appearing. This means that the probability \(P(Y = k)\) for \(k = 1, 2, 3, 4, 5, 6\) is \(\frac{1}{6}\).
04

Calculate the expected value (mean) of \(Y\)

The expected value \(E(Y)\) is calculated as:\[E(Y) = \sum_{k=1}^{6} k \, P(Y = k) = 1 \times \frac{1}{6} + 2 \times \frac{1}{6} + 3 \times \frac{1}{6} + 4 \times \frac{1}{6} + 5 \times \frac{1}{6} + 6 \times \frac{1}{6}\]\[E(Y) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5\]
05

Calculate the variance of \(Y\)

Variance is calculated using the formula:\[\text{Var}(Y) = E(Y^2) - (E(Y))^2\]First, find \(E(Y^2)\):\[E(Y^2) = \sum_{k=1}^{6} k^2 \, P(Y = k) = 1^2 \times \frac{1}{6} + 2^2 \times \frac{1}{6} + 3^2 \times \frac{1}{6} + 4^2 \times \frac{1}{6} + 5^2 \times \frac{1}{6} + 6^2 \times \frac{1}{6}\]\[E(Y^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} \approx 15.1667\]Now calculate the variance:\[\text{Var}(Y) = 15.1667 - (3.5)^2 = 15.1667 - 12.25 = 2.9167\]
06

Conclusion

The expected value of the number facing up when a fair die is rolled is 3.5 and the variance is approximately 2.9167.

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Key Concepts

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

Variance
Variance is a vital statistical concept that helps us measure the spread of a set of numbers. In simple terms, it tells us how far the numbers are from their average value, or mean.
In the context of rolling a die, variance gives us an idea of how much the results can differ. To calculate variance, we first need to understand two components:
  • Expected Value (Given that a fair die has numbers 1 through 6, and each is equally likely to appear, the expected value, or mean, is 3.5 as shown in the calculations above.
  • Second Moment of Expected Value (This is basically the average of the squared results. In the dice example, we calculate this by finding the expected value of the squares of the numbers (1, 4, 9, 16, 25, 36) which results in approximately 15.1667.
To find the variance, use the formula: \[ \text{Var}(Y) = E(Y^2) - (E(Y))^2 \] This equals approximately 2.9167 for a single die roll. The variance tells us that, although we expect the value of the roll to average around 3.5, the results can deviate significantly due to the nature of dice rolls.
Probability Distribution
A probability distribution depicts how probabilities are spread across possible outcomes. It provides a framework for understanding random variables, enabling us to predict outcomes.
In the case of a single fair die, the probability distribution is simple because each face (or number) has equal chances:
  • There are six possible outcomes: 1, 2, 3, 4, 5, and 6.
  • Each number has a probability of \( \frac{1}{6} \), representing an equal chance for each face to appear in any toss.
This uniform distribution is an example of a discrete probability distribution. In each trial, the probabilities sum to 1, affirming that some number must appear when you roll the die.
Understanding probability distributions is essential in statistics since it aids in computing expectations and variances, providing a broader picture of the data we work with.
Discrete Random Variables
Discrete random variables are types of random variables that can take a finite or countably infinite number of distinct values. In our dice example, they allow us to model the outcomes of a single die roll as a sequence of numbers. Each outcome on the die is assigned a probability.
Here鈥檚 why they鈥檙e crucial:
  • With a discrete random variable like the number of rolled dice, you can calculate probabilities for each possible value or outcome鈥攈elpful for understanding the likelihood of events.
  • They differ from continuous random variables, which can take on an infinite number of values within a range. Dice rolls, like our example, only have six possible values, making them discrete.
When analyzing discrete random variables, we focus on probability mass functions (PMF), which describe the probability of every possible outcome. It forms the basis for calculating further properties like the mean and variance.
Discrete random variables thus offer a clear path for statistical calculations, making complex probabilities accessible and understandable.

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Most popular questions from this chapter

Asupplier of heavy construction equipment has found that new customers are normally obtained through customer requests for a sales call and that the probability of a sale of a particular piece of equipment is .3. If the supplier has three pieces of the equipment available for sale, what is the probability that it will take fewer than five customer contacts to clear the inventory?

An oil prospector will drill a succession of holes in a given area to find a productive well. The probability that he is successful on a given trial is \(.2 .\) a. What is the probability that the third hole drilled is the first to yield a productive well? b. If the prospector can afford to drill at most ten wells, what is the probability that he will fail to find a productive well?

Many utility companies promote energy conservation by offering discount rates to consumers who keep their energy usage below certain established subsidy standards. A recent EPA report notes that \(70 \%\) of the island residents of Puerto Rico have reduced their electricity usage sufficiently to qualify for discounted rates. If five residential subscribers are randomly selected from San Juan, Puerto Rico, find the probability of each of the following events: a. All five qualify for the favorable rates. b. At least four qualify for the favorable rates.

If \(Y\) is a geometric random variable, define \(Y^{*}=Y-1 .\) If \(Y\) is interpreted as the number of the trial on which the first success occurs, then \(Y^{*}\) can be interpreted as the number of failures before the first success. If \(Y^{*}=Y-1, P\left(Y^{*}=y\right)=P(Y-1=y)=P(Y=y+1)\) for \(y=0,1,2, \ldots .\) Show that $$P\left(Y^{*}=y\right)=q^{y} p, \quad y=0,1,2, \dots$$ The probability distribution of \(Y^{*}\) is sometimes used by actuaries as a model for the distribution of the number of insurance claims made in a specific time period.

Five cards are dealt at random and without replacement from a standard deck of 52 cards. What is the probability that the hand contains all 4 aces if it is known that it contains at least 3 aces?
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