Suppose you make the following bet with a friend:
A six-sided die is rolled. You get 6 dollars if it comes up 1, 9 dollars if it comes up 2 or 3, but you lose 4 dollars if it comes up 4, 5 or 6.
So, you have a 1/6 chance of getting 6 dollars, a 1/3 chance of getting 9 dollars, and a 1/2 chance of getting -4 dollars. We could represent this bet as a random variable X using the following table:
x 6 9 -4
p(x) 1/6 1/3 1/2
(a) Using the formula in BSNS112 Slide Set 2, calculate the expected value of this random variable. Show your working.
(b) Using the formula in BSNS112 Slide Set 2, calculate the variance of this random variable. Show your working.

Let's tackle the problem step by step.

### (a) Calculating the Expected Value (E(X))

The expected value \(E(X)\) of a random variable \(X\) is given by:

\[ E(X) = \sum (x_i \cdot p(x_i)) \]

Using the values provided:

- For \(x = 6\), \(p(x) = \frac{1}{6}\)
- For \(x = 9\), \(p(x) = \frac{1}{3}\)
- For \(x = -4\), \(p(x) = \frac{1}{2}\)

Substitute these values into the formula:

\[ E(X) = 6 \cdot \frac{1}{6} + 9 \cdot \frac{1}{3} + (-4) \cdot \frac{1}{2} \]

Calculate each term separately:

\[ 6 \cdot \frac{1}{6} = 1 \]
\[ 9 \cdot \frac{1}{3} = 3 \]
\[ -4 \cdot \frac{1}{2} = -2 \]

Now, sum these values:

\[ E(X) = 1 + 3 - 2 = 2 \]

So, the expected value \(E(X)\) is 2 dollars.

### (b) Calculating the Variance (Var(X))

The variance \(\text{Var}(X)\) of a random variable \(X\) is given by:

\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]

First, we need to calculate \(E(X^2)\), which is the expected value of \(X^2\):

\[ E(X^2) = \sum (x_i^2 \cdot p(x_i)) \]

Using the values provided:

- For \(x = 6\), \(p(x) = \frac{1}{6}\)
- For \(x = 9\), \(p(x) = \frac{1}{3}\)
- For \(x = -4\), \(p(x) = \frac{1}{2}\)

Calculate \(x_i^2 \cdot p(x_i)\) for each \(x_i\):

\[ 6^2 \cdot \frac{1}{6} = 36 \cdot \frac{1}{6} = 6 \]
\[ 9^2 \cdot \frac{1}{3} = 81 \cdot \frac{1}{3} = 27 \]
\[ (-4)^2 \cdot \frac{1}{2} = 16 \cdot \frac{1}{2} = 8 \]

Sum these values to get \(E(X^2)\):

\[ E(X^2) = 6 + 27 + 8 = 41 \]

Now, use the expected value \(E(X) = 2\) to find the variance:

\[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]
\[ \text{Var}(X) = 41 - (2)^2 \]
\[ \text{Var}(X) = 41 - 4 \]
\[ \text{Var}(X) = 37 \]

So, the variance \(\text{Var}(X)\) is 37 dollars\(^2\).

### Summary

(a) The expected value \(E(X)\) is 2 dollars.

(b) The variance \(\text{Var}(X)\) is 37 dollars\(^2\).

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Would you like any additional details or have any questions about this calculation?

Here are some related questions for further exploration:

1. How does the expected value change if the payout for rolling a 1 increases to 8 dollars?
2. What is the standard deviation of this random variable?
3. If the bet were modified such that rolling a 4, 5, or 6 resulted in a different loss amount, how would you calculate the new expected value and variance?
4. How would the expected value and variance change if the probability distribution of the outcomes changed?
5. Can the expected value be used to predict the outcome of a single roll of the die? Why or why not?
6. How does the variance help in understanding the risk associated with the bet?
7. What is the significance of having a negative expected value in a different betting scenario?
8. How would you simulate this bet using a computer program to estimate the expected value and variance?

**Tip:** Understanding the expected value and variance of a random variable helps in assessing both the average outcome and the risk (variability) associated with different scenarios, which is crucial in decision-making processes.

Learn how to calculate the expected value and variance of a random variable from a given probability distribution. This problem involves a scenario with a six-sided die and different payouts for each outcome, covering key concepts in probability theory.

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