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Random Variables, Expectation & Variance · Question 53 of 68

What is the variance of the player's winnings X in the dice game, without the charge?

Question 53

HardOriginal question

At a fair a player rolls a fair die once. A 6 wins ₹30, a 5 wins ₹12, and any other number loses ₹6. Let XX be the player's winnings in rupees.

What is the variance of the player's winnings XX in the dice game, without the charge?

Options for question 53

Show the answer and explanation
CorrectNot quiteAnswer C

Correct answer: C

Explanation

  1. E(X2)=900+144+4(36)6=1,1886=198E(X^{2}) = \frac{900 + 144 + 4(36)}{6} = \frac{1{,}188}{6} = 198.

  2. Var(X)=198−32=189\text{Var}(X) = 198 - 3^{2} = 189.

Worth noting

207 adds the square of the mean instead of subtracting it.

Concept
Variance of a game's winnings
Typical time
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