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Consider the following probability model where \( X \) depicts the different prizes for a lottery in dollars and the corresponding probabilities.

\[
X = x \ | \ 0 \quad 100 \quad 500 \quad 1000
\]
\[
P(X = x) \ | \ 0.5 \quad 0.2 \quad 0.2 \quad 0.1
\]

a. What is the expected (mean) value of \( X \)? [3 points]

b. What is the standard deviation of \( X \)? [4 points]

Now consider the following probability model where \( Y \) depicts the different prizes for a lottery in dollars and the corresponding probabilities.

\[
Y = y \ | \ 0 \quad 10 \quad 50 \quad 100
\]
\[
P(Y = y) \ | \ 0.5 \quad 0.2 \quad 0.2 \quad 0.1
\]

c. What is the expected (mean) value of \( Y \)? [3 points]

d. What is the standard deviation of \( Y \)? [4 points]

e. Which lottery is riskier: \( X \) or \( Y \)? Why? [2 points]

[Optional bonus question] Would a risk-neutral person buy a lottery ticket for \( X \) and a lottery ticket for \( Y \) if they jointly cost $250? Why? [2 bonus points]

Answer :

Final answer:

The expected value of X is 220, the standard deviation of X is 1031.57, the mean of Y is 22, the standard deviation of Y is 91.34, lottery X is riskier than Y, and a risk neutral person would not buy the lottery tickets.

Explanation:

a. To find the expected value of X, we multiply each value of X by its corresponding probability and add the products: (0)(0.5) + (100)(0.2) + (500)(0.2) + (1000)(0.1) = 0 + 20 + 100 + 100 = 220

b. To find the standard deviation of X, we use the formula: σ = √[∑(x - μ)^2P(x)], where μ represents the mean. Using the mean obtained in part a, we calculate: √[(0 - 220)^2(0.5) + (100 - 220)^2(0.2) + (500 - 220)^2(0.2) + (1000 - 220)^2(0.1)] = √[54400 + 11840 + 121600 + 774400] = √1065240 = 1031.57

c. Using the same process as in part a, we find the mean of Y to be: (0)(0.5) + (10)(0.2) + (50)(0.2) + (100)(0.1) = 0 + 2 + 10 + 10 = 22

d. Again, using the formula for standard deviation, we calculate the standard deviation of Y as: √[(0 - 22)^2(0.5) + (10 - 22)^2(0.2) + (50 - 22)^2(0.2) + (100 - 22)^2(0.1)] = √[484 + 144 + 784 + 5929] = √8339 = 91.34

e. The lottery that is riskier can be determined by comparing the standard deviations of X and Y. The higher the standard deviation, the riskier the lottery. In this case, since the standard deviation of X (1031.57) is higher than the standard deviation of Y (91.34), lottery X is riskier.

For the optional bonus question, a risk neutral person would not buy the lottery tickets because the expected value of both X and Y combined is 220 + 22 = 242, which is less than the cost of the tickets ($250).

Learn more about Probability and Statistics here:

https://brainly.com/question/30448794

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