Question

Find the expected value of the winningsfrom a game that has the followingpayout probability distribution:Payout ($) 02 48 10Probability 0.45 0.3 0.1 0.1 0.05Expected Value = [?]Round to the nearest hundredth.

Answer 1

Solution:

Given:

Using the formula:

Therefore:

$\begin{gathered} E(x)=(0\cdot0.45)+(2\cdot0.3)+(4\cdot0.1)+(8\cdot0.1)+(10\cdot0.05) \\ \\ E(x)=2.30 \end{gathered}$

ANSWER: 2.30