# Solved:Imagine you are in a game show. There are 4 prizes hidden on a game board with 10 spaces. One prize is worth

#### ByDr. Raju Chaudhari

Aug 23, 2020

Imagine you are in a game show. There are 4 prizes hidden on a game board with 10 spaces. One prize is worth \$100, another is worth \$50, and two are worth \$10. You have to pay \$20 to the host if your choice is not correct. Let the random variable $X$ be the winning. Show all work.

(a) What is your expected winning in this game?
(b) Determine the standard deviation of $X$. (Round the answer to two decimal places)

#### Solution

(a) $X:$ amount of winning = \$100, \$50, \$10, -\$20

The probabilities are

$P(X=100) = \frac{1}{10}$, $P(X=50) = \frac{1}{10}$, $P(X=10) =\frac{2}{10}$ and $P(X=-20) =\frac{6}{10}$.

Thus the expected value is

\begin{aligned} E(X) &= \sum_x x*P(X=x)\\ &=\bigg[100\times \frac{1}{10} + 50\times \frac{1}{10}+ 10\times \frac{2}{10}+ (-20)\times \frac{6}{10} \bigg]\\ &= \frac{1}{10}\big(100+ 50 + 20- 120\big)\\ &= \5 \end{aligned}

Thus the expected winning of the game is \$5. (b) To determine standard deviation we need to compute$E(X^2)\begin{aligned} E(X^2) &= \sum_x x^2*P(X=x)\\ &=\bigg[(100)^2\times \frac{1}{10} + (50)^2\times \frac{1}{10}+ (10)^2\times \frac{2}{10}+ (-20)^2\times \frac{6}{10} \bigg]\\ &= \frac{1}{10}\big(10000+ 2500 + 200+ 2400\big)\\ &= 1510 \end{aligned} Thus the standard deviation ofXis \begin{aligned} \sigma &= \sqrt{E(X^2) - \big(E(X)\big)^2}\\ &= \sqrt{1510- 5^2}\\ &= \sqrt{1510- 25}\\ &=\sqrt{1485}\\ &= 38.54 \end{aligned} Thus the standard deviation ofX$is \$ 38.54.