find the expected value of the winnings from a game that has the following payout probability distribution…

Explanation:

Step1: Recall expected - value formula

The formula for the expected value $E(X)$ of a discrete - random variable is $E(X)=\sum_{i}x_ip_i$, where $x_i$ are the possible values and $p_i$ are their corresponding probabilities.

Step2: Calculate the product for each pair

For the first pair: $x_1 = 1$ and $p_1=0.52$, so $x_1p_1=1\times0.52 = 0.52$. For the second pair: $x_2 = 2$ and $p_2 = 0.26$, so $x_2p_2=2\times0.26 = 0.52$. For the third pair: $x_3 = 4$ and $p_3 = 0.13$, so $x_3p_3=4\times0.13 = 0.52$. For the fourth pair: $x_4 = 8$ and $p_4 = 0.06$, so $x_4p_4=8\times0.06 = 0.48$. For the fifth pair: $x_5 = 16$ and $p_5 = 0.03$, so $x_5p_5=16\times0.03 = 0.48$.

Step3: Sum up the products

$E(X)=0.52 + 0.52+0.52 + 0.48+0.48$ $E(X)=(0.52\times3)+(0.48\times2)$ $E(X)=1.56 + 0.96$ $E(X)=2.52$

Answer:

$2.52$