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

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

Answer

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: Fill in the blanks

The first blank corresponds to the first payout value which is $1$. The second blank corresponds to the probability of the second - payout, which is $0.15$. The third blank corresponds to the fourth - payout value which is $7$, and the fourth blank corresponds to the probability of the fifth - payout, which is $0.30$. So the expression is $1\times0.10 + 3\times0.15+5\times0.20 + 7\times0.25+9\times0.30$.

Step3: Calculate each product

$1\times0.10=0.10$, $3\times0.15 = 0.45$, $5\times0.20=1.00$, $7\times0.25 = 1.75$, $9\times0.30=2.70$.

Step4: Sum up the products

$0.10 + 0.45+1.00 + 1.75+2.70=6.00$.

Answer:

$6.00$