Question

Find the expected value of the winnings from a game that has the following payout probability distribution: Skip Payout ($) 1 2 5 8 10 Probability 0.35 0.2 0.1 0.2 0.15 Expected Value = [?] Round to the nearest hundredth.​

Answer 1

Answer:

$4.35

Step-by-step explanation:

The expected value of a random variable X, often denoted as E(X), indicates the probability-weighted average of all possible values/events. The general formula of expected value is

$\mathrm{E(\textit X \mathrm)} = \displaystyle\sum_{\mathclap{i=1}}^{k} \ X \times P(X) \\ \\ \\ = X_{1} \times P(X_{1}) \ + \ X_{2} \times P(X_{2}) \ + \ X_{3} \times P(X_{3}) + \ \cdots \ + X_{k} \times P(X_{k})$.

Therefore, the expected value of the winnings from a game is

$\mathrm{E(\textit X \mathrm)} \ = \ 1 \times 0.35 \ + \ 2 \times 0.2 \ + \ 5 \times 0.1 + \ 8 \times 0.2 \ + 10 \times 0.15 \\ \\ = \ 4.35 \ \ (\mathrm{nearest \ hundredth})$.