Submit your solution on BlackBoard Visual displays of probability distributions
ID: 3050314 • Letter: S
Question
Submit your solution on BlackBoard Visual displays of probability distributions are typically bar graphs Examples 18 2: The probability distributions above can be represented by bar graphs shown below Probahity Distribution for Number of Heads Probability Distribution fer Dice Ouico 0.4 0.2 03 02 0.1 0.15 0.1 0.05 234519 Nunber of Head Sum of wwo dice When probability distributions are symmetric, the expected value will occur at the center of the distribution. In the images above, a dashed line shows the center. The first image confirms the above computation of 1.5, and the second image shows that the expected value when rolling two dice is 7 Expected value is a useful computation when analyzing games of chance. The example below shows how Example 3: A local charity is selling raffe tickets. In total, 1000 tickets were sold for $1 each, and one ticket holder wins $350. Find the expected value of the raffle Event (gain) pfx) Solution Method 1: We can set up a probability distribution for Lose1 the amount gained by ticket holders (x). The expected value of the game is then win | 349 | Solution Method 2 Fundamentally, the expected value is the mean amount gained or lost on the raffle. Since $999 was lost on all of the losing tickets and $349 was gained by the single winner, from the perspective of alt 1000 ticket hoilders that is a total loss of $650, or $0.65 each. The expected value is-$0.65 DO IT NOW (3) 5-3.1: A game is played as follows: A player places a bet. Then, she roils a 20-sided die with the numbers 1-20 on it. If the die lands on an even number, the player wins $5. If the de lands on a 1, 5,7 or 13, the player wins$ 10Otherwse·tho player loses.u it costs $3.50 to betwhat is the expected value of this game? Subrnit your solutions on BlackBoardExplanation / Answer
20 sided die
even number - probability - 1/2 , wins 5 dollar
1,5,7 ,13 - probability - 4/20 = 1/5 - wins 10 dollar
rest odd number - probability - 6/20 = 3/10- losses
hence
E(X) = 5*1/2 + 10 *1/5 + 0 - 3.5
= 1