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

find the expected value of the winnings from a game that has the following payout probability distribution: payout ($) 1 3 5 7 9 probability 0.10 0.15 0.20 0.25 0.30 first, enter values into the expression ?·0.10 + 3· + 5·0.20 + ·0.25 + 9·

find the expected value of the winnings from a game that has the following payout probability distribution: payout ($) 1 3 5 7 9 probability 0.10 0.15 0.20 0.25 0.30 first, enter values into the expression ?·0.10 + 3· + 5·0.20 + ·0.25 + 9·

Answer

Answer:

$1\times0.10 + 3\times0.15+5\times0.20 + 7\times0.25+9\times0.30=6$

Explanation:

Step1: Recall expected - value formula

$E(X)=\sum_{i}x_ip_i$

Step2: Identify values for first term

The first payout $x_1 = 1$ and its probability $p_1=0.10$, so the first term is $1\times0.10$.

Step3: Identify values for second term

The second payout $x_2 = 3$ and its probability $p_2 = 0.15$, so the second term is $3\times0.15$.

Step4: Identify values for third term

The third payout $x_3 = 5$ and its probability $p_3=0.20$, so the third term is $5\times0.20$.

Step5: Identify values for fourth term

The fourth payout $x_4 = 7$ and its probability $p_4=0.25$, so the fourth term is $7\times0.25$.

Step6: Identify values for fifth term

The fifth payout $x_5 = 9$ and its probability $p_5=0.30$, so the fifth term is $9\times0.30$.

Step7: Calculate the sum

$1\times0.10+3\times0.15 + 5\times0.20+7\times0.25+9\times0.30=0.10 + 0.45+1+1.75 + 2.7=6$.