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What are the variance and standard deviation for the sum of the tops of \(n\) dice that you roll?

Short Answer

Expert verified
The variance of the sum is \(n \, \times \, 2.9167\) and the standard deviation is \(\sqrt{n \, \times \, 2.9167}\).

Step by step solution

01

Understand the Problem

We are asked to find the variance and standard deviation of the sum of the tops of \(n\) dice. This means we need to analyze the statistical properties of a sum of random variables, each representing one die roll.
02

Define Random Variables

Let \(X_i\) be the random variable representing the result of the \(i^{th}\) die roll, where \(i = 1, 2, \, \dots, n\). The total sum \(S\) is given by \(S = X_1 + X_2 + \cdots + X_n\).
03

Calculate the Expected Value of Each Die Roll

Since each die is fair and has outcomes \(1\) through \(6\), the expected value \(E(X_i)\) of one die roll is \(E(X_i) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5\).
04

Calculate the Variance of Each Die Roll

The variance \(Var(X_i)\) of a single die roll is calculated by \(Var(X_i) = E(X_i^2) - [E(X_i)]^2\). First, \(E(X_i^2) = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6} = 15.1667\). Then, the variance is \(Var(X_i) = 15.1667 - 3.5^2 = 2.9167\).
05

Calculate the Variance of the Sum

Using the property of variance for independent random variables, \(Var(S) = Var(X_1 + X_2 + \cdots + X_n) = \sum_{i=1}^n Var(X_i) = n \, Var(X_i) = n \, \times \, 2.9167\).
06

Calculate the Standard Deviation of the Sum

The standard deviation is the square root of the variance. Therefore, \(SD(S) = \sqrt{Var(S)} = \sqrt{n \, \times \, 2.9167}\).

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Key Concepts

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

Random Variables
When we talk about random variables in statistics, we're referring to variables that can take on a set of different values, each with an associated probability. For example, when you roll a die, the outcome can be any number from 1 to 6, each with an equal chance of occurring. In this context, each die roll is considered a random variable because we don't know the outcome in advance.
A random variable is denoted by symbols like \(X\) or \(Y\), and in the exercise above, each individual die roll is represented by \(X_i\). This means the result of the \(i^{th}\) roll of the die.
Understanding random variables is crucial because they form the basis of probability and statistics, helping us quantify and model uncertainty.
Expected Value
The expected value of a random variable gives us a measure of the 'central tendency' or average value we expect from the variable. For instance, when rolling a single six-sided die, the numbers 1 to 6 are equally likely, so the expected value represents the mean of these outcomes.
The expected value \(E(X)\) for one die roll is calculated as the sum of each possible outcome multiplied by its probability, which gives \(E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5\).
This tells us that, over a large number of rolls, the average result will tend toward 3.5. The concept of expected value is key to making predictions based on random variables and plays a crucial role in fields such as economics, finance, and risk management.
Independent Random Variables
In the context of probability and statistics, two random variables are considered independent if the outcome of one does not affect the outcome of the other. This is an important property, especially when performing operations like addition and multiplication with these variables.
In our exercise, each die roll is considered an independent random variable because what number comes up on one die doesn't influence what number comes up on another die. This means the probability distribution of each roll remains the same, no matter what has happened in other rolls.
The independence of random variables simplifies calculations, particularly when dealing with variance and expected values, as it ensures that combined outcomes follow predictable statistical rules.
Sum of Independent Variables
When you're dealing with several independent random variables, like multiple die rolls, the sum of these variables follows certain statistical rules.
If all the variables are independent, the expected value of the sum is the sum of their expected values. For example, if \(S = X_1 + X_2 + \cdots + X_n\), then \(E(S) = E(X_1) + E(X_2) + \cdots + E(X_n)\). This makes it straightforward because, for \(n\) dice, you'll simply do \(n \times 3.5\) to get the total expected value.
The variance of the sum of independent variables is also easy to calculate: it's the sum of their variances, \(Var(S) = Var(X_1) + Var(X_2) + \cdots + Var(X_n) = n \times Var(X_i)\), which simplifies to \(n \times 2.9167\) for our dice rolls.
Understanding how to sum independent variables lets us predict the behavior of complex systems by analyzing simpler, individual components.

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Most popular questions from this chapter

You are a contestant on the TV game show Let’s Make a Deal. In this game show, there are three curtains. Behind one of the curtains is a new car, and behind the other two are cans of Spam. You get to pick one of the curtains. After you pick one of the curtains, the emcee, Monty Hall, who we assume knows where the car is, reveals what is behind one of the curtains that you did not pick, showing you some cans of Spam. He then asks you if you would like to switch your choice of curtain. Should you switch? Why or why not? Please answer this question carefully. You have all the tools needed to answer it, but several math Ph.D.s are on record (in Parade magazine) giving the wrong answer.

In three flips of a coin, is the event of getting at most one tail independent of the event that not all flips are identical?

Evaluate the sum $$ \sum_{i=0}^{10} i\left(\begin{array}{c} 10 \\ i \end{array}\right)(.9)^{i}(.1)^{10-i} $$ which arose in computing the expected number of right answers a person would have on a 10 -question test with probability \(.9\) of answering each question correctly. First, use the binomial theorem and calculus to show that $$ 10(.1+x)^{9}=\sum_{i=0}^{10} i\left(\begin{array}{c} 10 \\ i \end{array}\right)(.1)^{10-i} x^{i-1} $$ Substituting \(x=.9\) almost gives the sum you want on the right side of the equation, except that in every term of the sum, the power on 9 is one too small. Use some simple algebra to fix this and then explain why the expected number of right answers is 9 .

Suppose you have a recurrence of the form $$ T(n) \leq T\left(a_{n} n\right)+T\left(\left(1-a_{n}\right) n\right)+b n, \text { if } n>1, $$ where \(a_{n}\) is between \(1 / 5\) and \(4 / 5\). Show that all solutions to this recurrence are of the form \(T(n)=O(n \log n)\).

Give an example of a random variable on the sample space \(\left\\{S, F S, F F S, \ldots, F^{i} S, \ldots\right\\}\) with an infinite expected value, using a geometric distribution for probabilities of \(F^{i} S\).

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