Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). Two dice are thrown; \(x=\) sum of the numbers on the dice.

Short Answer

Expert verified
Define the sample space with 36 outcomes, calculate sums and their probabilities, find the mean, variance, standard deviation, and plot the cumulative distribution function.

Step by step solution

01

Define the Sample Space

When two dice are thrown, each die has 6 faces with values from 1 to 6. The total number of possible outcomes is 6 * 6 = 36. Each outcome can be represented as an ordered pair \(a, b\), where \(a\) and \(b\) are the values on the first and second die, respectively.
02

Calculate the Sum for Each Outcome

List all possible outcomes and calculate the sum\( s = a + b \). For each pair \( (a, b) \) in the sample space, compute \( s \) and write it next to the outcome. For example, if the outcome is \( (1, 1) \), then \( s = 1 + 1 = 2 \). Continue this process for all 36 outcomes.
03

Determine the Probability for Each Sum

Calculate the frequency of each possible sum of the dice, ranging from 2 to 12. Since there are 36 possible outcomes, the probability of each sum is given by \[ P(s) = \frac{\text{Number of occurrences of } s}{36} \].
04

Create a Table for the Random Variable \(X\)

Tabulate the different possible sums \( s_i \) as \( x_i \) and their corresponding probabilities \( P(x_i) \.\) Using the previously calculated values, create a table with columns for \( x_i \) and \( P(x_i) \).
05

Compute the Mean of \(X\)

The mean of \( X \) is given by \[ \text{Mean } ( \mu \ ) = \sum x_i \cdot P(x_i) \. \] Calculate this using the table values.
06

Compute the Variance of \(X\)

The variance of \( X \) is given by \[ \text{Variance } ( \sigma^2 \ ) = \sum (x_i - \mu)^2 \cdot P(x_i) \. \] Use the mean computed in step 5 and the table values to find the variance.
07

Compute the Standard Deviation of \(X\)

The standard deviation of \( X \) is the square root of the variance: \[ \text{Standard Deviation } ( \sigma \ ) = \sqrt{\text{Variance}} \. \]
08

Calculate and Plot the Cumulative Distribution Function \(F(x)\)

The cumulative distribution function \( F(x) \) gives the probability that \( X \) takes a value less than or equal to \( x \). Compute \( F(x) \) as \[ F(x) = \sum_{x_i \leq x} P(x_i) \] and plot it. This involves summing the probabilities from the table up to and including \( x \).

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Key Concepts

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

Random Variables
A random variable is a variable that takes on different numerical values, each associated with a probability, depending on the outcome of a random event. In the exercise, the random variable is the sum of the numbers on two dice. Each possible sum (ranging from 2 to 12) is a potential value of the random variable. For example:

If the dice show (1,1), the random variable (sum) is 2. If the dice show (6,6), the random variable (sum) is 12.

Each of these sums corresponds to a specific probability, which can be calculated by analyzing the frequency of occurrence of each sum.
Mean
The mean of a random variable provides a measure of the central tendency of the distribution. It's often referred to as the expected value. For a given random variable \(X\), the mean is calculated by summing the products of each possible value \(x_i\) and its associated probability \(P(x_i)\).

Mathematically, it’s given by:
\[ \text{Mean } ( \mu \ ) = \sum_{i} x_i \cdot P(x_i) \]

In our dice example, we would calculate the mean by considering all possible sums (from 2 to 12) and their corresponding probabilities. For instance:
  • Mean contribution from sum of 2: \(2 \cdot P(2)\)
  • Mean contribution from sum of 3: \(3 \cdot P(3)\)
And so on, until sum of 12.
Variance
The variance of a random variable measures the spread or dispersion of the possible values around the mean. For a random variable \(X\), it’s calculated by summing the squared differences between each value \(x_i\) and the mean \(\mu\), each weighted by its probability:

\[ \text{Variance } ( \sigma^2 \ ) = \sum (x_i - \mu)^2 \cdot P(x_i) \]

Variance helps us understand how much the values of the random variable deviate from the mean. In the dice example:
  • Compute \( (2 - \mu)^2 \cdot P(2) \) for sum 2
  • Compute \( (3 - \mu)^2 \cdot P(3) \) for sum 3
And continue this for all possible sums up to 12.
Standard Deviation
The standard deviation is the square root of the variance. While variance gives a measure of spread in squared units, standard deviation provides it in the same units as the mean, making it easier to interpret.

Mathematically, it’s given by:
\[ \text{Standard Deviation } ( \sigma \ ) = \sqrt{\text{Variance}} \]

For the dice example, once we have the variance, the standard deviation is simply:
\[ \sigma = \sqrt{ \sigma^2 } \]

This value tells us, in general, how much the sums (2 to 12) deviate from the calculated mean.
Cumulative Distribution Function
The cumulative distribution function (CDF) of a random variable \(X\) gives the probability that \(X\) will take a value less than or equal to \(x\). It provides a complete description of the distribution.

Mathematically, for a value \(x\), it’s defined as:
\[ F(x) = \sum_{x_i \leq x} P(x_i) \]

In the dice example, the CDF can be built by summing the probabilities of all sums (from 2 to the current sum):
  • \( F(2) = P(2) \)
  • \( F(3) = P(2) + P(3) \)
  • \( F(4) = P(2) + P(3) + P(4) \)
And so on, until \(x = 12\). Plotting the CDF gives a graphical representation of the probability that the sum of the dice is less than or equal to any given value.

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Most popular questions from this chapter

Five cards are dealt from a shuffled deck. What is the probability that they are all of the same suit? That they are all diamond? That they are all face cards? That the five cards are a sequence in the same suit (for example, 3,4,5,6,7 of hearts)?

(a) Three typed letters and their envelopes are piled on a desk. If someone puts the letters into the envelopes at random (one letter in each), what is the probability that each letter gets into its own envelope? Call the envelopes \(A, B, C,\) and the corresponding letters \(a, b, c,\) and set up the sample space. Note that " \(a\) in \(C\) \(b\) in \(B, c\) in \(A "\) is one point in the sample space. (b) What is the probability that at least one letter gets into its own envelope? Hint: What is the probability that no letter gets into its own envelope? (c) Let \(A\) mean that \(a\) got into envelope \(A\), and so on. Find the probability \(P(A)\) that \(a\) got into \(A\). Find \(P(B)\) and \(P(C)\). Find the probability \(P(A+B)\) that either \(a\) or \(b\) or both got into their correct envelopes, and the probability \(P(A B)\) that both got into their correct envelopes. Verify equation (3.6)

Set up sample spaces for Problems 1 to 7 and list next to each sample point the value of the indicated random variable \(x,\) and the probability associated with the sample point. Make a table of the different values \(x_{i}\) of \(x\) and the corresponding probabilities \(p_{i}=f\left(x_{i}\right)\) Compute the mean, the variance, and the standard deviation for \(x\). Find and plot the cumulative distribution function \(F(x)\). A weighted coin with probability \(p\) of coming down heads is tossed three times; \(x=\) number of heads minus number of tails.

What is the probability that you and a friend have different birthdays? (For simplicity, let a year have 365 days.) What is the probability that three people have three different birthdays? Show that the probability that \(n\) people have \(n\) different birthdays is $$p=\left(1-\frac{1}{365}\right)\left(1-\frac{2}{365}\right)\left(1-\frac{3}{365}\right) \cdots\left(1-\frac{n-1}{365}\right).$$ Estimate this for \(n \ll 365\) by calculating \(\ln p\) [recall that \(\ln (1+x)\) is approximately \(x\) for \(x \ll 1]\). Find the smallest (integral) \(n\) for which \(p<\frac{1}{2}\). Hence, show that for a group of 23 people or more, the probability is greater than \(\frac{1}{2}\) that two of them have the same birthday. (Try it with a group of friends or a list of people such as the presidents of the United States.)

(a) Suppose that Martian dice are regular tetrahedra with vertices labeled 1 to 4 Two such dice are tossed and the sum of the numbers showing is even. Let \(x\) be this sum. Set up the sample space for \(x\) and the associated probabilities. (b) Find \(E(x)\) and \(\sigma_{x}\) (c) Find the probability of exactly fifteen 2 's in 48 tosses of a Martian die using the binomial distribution. (d) Approximate (c) using the normal distribution. (e) Approximate (c) using the Poisson distribution.

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