91Ó°ÊÓ

Consider a game in which a red die and a blue die are rolled. Let \(x_{R}\) denote the value showing on the uppermost face of the red die, and define \(x_{B}\) similarly for the blue die. a. The probability distribution of \(x_{R}\) is $$ \begin{array}{lrrrrrr} x_{R} & 1 & 2 & 3 & 4 & 5 & 6 \\ p\left(x_{R}\right) & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 & 1 / 6 \end{array} $$ Find the mean, variance, and standard deviation of \(x_{R}\). b. What are the values of the mean, variance, and standard deviation of \(x_{B} ?\) (You should be able to answer this question without doing any additional calculations.) c. Suppose that you are offered a choice of the following two games: Game 1: Costs $$\$ 7$$ to play, and you win \(y_{1}\) dollars, where \(y_{1}=x_{R}+x_{B}\) Game 2: Doesn't cost anything to play initially, but you "win" \(3 y_{2}\) dollars, where \(y_{2}=x_{R}-x_{B}\). If \(y_{2}\) is negative, you must pay that amount; if it is positive, you receive that amount. For Game 1, the net amount won in a game is \(w_{1}=\) \(y_{1}-7=x_{R}+x_{B}-7 .\) What are the mean and standard deviation of \(w_{1}\) ? d. For Game 2, the net amount won in a game is \(w_{2}=\) \(3 y_{2}=3\left(x_{R}-x_{B}\right) .\) What are the mean and standard deviation of \(w_{2}\) ? e. Based on your answers to Parts (c) and (d), if you had to play, which game would you choose and why?

Step by step solution

01

Calculating the mean, variance, and standard deviation of \(x_{R}\)

We know that for a probability distribution, the mean or expected value \(E(x)\) is given by \(\sum{x*p(x)}\), the variance \(Var(x)\) is \(\sum{(x-E(x))^2*p(x)}\) and the standard deviation \(SD(x)\) is the square root of the variance. Calculating these for \(x_{R}\) we get: \[E(x_{R})=\sum_{i=1}^{6}{i*\frac{1}{6}}=3.5\] \[Var(x_{R})=\sum_{i=1}^{6}{(i-3.5)^2*\frac{1}{6}}=2.92\] \(SD(x_{R})=\sqrt{2.92}\approx 1.71\].

02

Mean, Variance, and Standard Deviation of \(x_{B}\)

Since we have symmetric and identical distribution for both dies, the mean, variance and standard deviation of \(x_{B}\) occupy the same values as \(x_{R}\) thus \(E(x_{B})=3.5\), \(Var(x_{B})=2.92\), and \(SD(x_{B})=1.71\).

03

Mean and Standard Deviation of \(w_{1}\)

For Game 1, we need to calculate the mean and standard deviation for \(w_{1}\) where \(w_{1}=x_{R}+x_{B}-7\). Using the properties of means and variances we know that: \[E(w_{1})=E(x_{R}+x_{B}-7)=E(x_{R})+E(x_{B})-7=3.5+3.5-7=0\] For the standard deviation, because die rolls are independent, variance of \(w_{1}\) is: \(Var(w_{1})=Var(x_{R}+x_{B})=Var(x_{R})+Var(x_{B})=2.92+2.92=5.83\). Thus, \(SD(w_{1})=\sqrt{5.83}\approx2.41\).

04

Mean and Standard Deviation of \(w_{2}\)

For Game 2, we compute the mean and standard deviation for \(w_{2}\) given by \(w_{2}=3(x_{R}-x_{B})\). Now, \(E(w_{2})=E(3x_{R}-3x_{B})=3E(x_{R}-x_{B})=3(E(x_{R})-E(x_{B}))=3(3.5-3.5)=0\). Again, since the die rolls are independent, \(Var(w_{2})=9Var(x_{R}+x_{B})=9(2.92+2.92)=52.56\). Hence, \(SD(w_{2})=\sqrt{52.56}\approx7.25\).

05

Choosing the Game to Play

Based on the calculations, both games have the same mean of 0 but Game 1 has a smaller standard deviation of 2.41 compared to Game 2 which has a standard deviation of 7.25. A lower standard deviation means less risk or variability, so if one had to play, Game 1 would be a more prudent choice if one aims to minimize the risk.

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Key Concepts

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

Expected Value

The expected value, often referred to as the mean, is a central cornerstone in probability distributions. It essentially represents the average outcome you would expect if you repeated a random experiment an infinite number of times. For any given variable, the expected value is calculated by multiplying each possible outcome by its probability and then summing up all these products. In mathematical terms, for a discrete random variable \(X\), the expected value \(E(X)\) can be expressed as: \[ E(X) = \sum{x_i \cdot p(x_i)} \]where \(x_i\) represents each outcome and \(p(x_i)\) is the probability of \(x_i\).
In the context of our dice problem, for the red die, the expected value calculates to 3.5, meaning that, over many rolls, the average outcome on the die will hover around 3.5.

• Expected value helps determine "central" tendencies
• It offers a statistical long-term average

Variance

Variance is a measure that quantifies the amount by which values in a probability distribution can deviate or spread out from the mean. In practical terms, it gives insight into the variability or the spread of outcomes. Variance is particularly useful because it provides a square sum of these deviations, which can be positive or negative.
The formula for variance \(Var(X)\) of a discrete random variable \(X\) is: \[ Var(X) = \sum{(x_i - E(X))^2 \cdot p(x_i)} \]Here, \(x_i\) is each outcome, \(E(X)\) is the expected value found previously, and \(p(x_i)\) is the probability.
In our exercise, the variance for both dice, red and blue, is computed to be 2.92. This value tells us about the spread of the dice results around the mean of 3.5.

• High variance indicates a larger spread of values
• Low variance suggests data values are close to the mean

Standard Deviation

The standard deviation is simply the square root of the variance and gives a sense of how much, on average, each outcome deviates from the mean in the units of the variable itself. It is one of the most commonly used metrics in statistics because it provides an easily interpretable measure of variability.
For a random variable \(X\), if the variance is \(Var(X)\), then the standard deviation \(SD(X)\) is given by:\[ SD(X) = \sqrt{Var(X)} \]In our dice game, the standard deviation for each die ends up being approximately 1.71, telling us that most outcomes will deviate from the expected value by about 1.71 on average.

• Standard deviation offers insights into data dispersion
• A smaller standard deviation indicates less variability

Independent Events

Independent events in probability refer to scenarios where the occurrence of one event does not affect the probability of another event. In terms of our exercise, the rolls of the red die and the blue die are independent. This independence allows us to use properties for calculating combined variances and other statistical measures for composite games.
When events are independent, we can combine their probability measures as follows:

• The total probability of both events occurring is the product of their individual probabilities.
• The variance of a sum or difference of independent random variables can simply be the sum of their respective variances.

This characteristic of independence is crucial in determining the variance and standard deviation of outcomes like \(w_1 = x_R + x_B - 7\) and \(w_2 = 3(x_R - x_B)\) in our games. In both cases, since the red and blue die rolls are independent, we are free to use straightforward arithmetic with their variances, which simplifies our calculations immensely.