Consider the following card game with a well-shuffled deck of cards. If you draw a red card, you win nothing. If you get a spade, you win $5. For any club, you win $10 plus an extra $20 for the ace of clubs.
Create a probability model for the amount you win at this game. Also, find the expected winnings for a single game and the standard deviation of the winnings.
X = amount you can win at this game (# of dollars you can win)
x = (0, 5, 10, 20)
P(X=0) = P(probability of picking a red card) = 26/52 = .50
P(X=5) = P(probability of picking a spade) = 13/52 = .25
P(X=10) = P(proability of picking a club, but not the ace) = 12/52 \(\approx\) .231
P(X=20) = P(probability of picking the ace of clubs) = 1/52 \(\approx\) .019
Probability Distribution of X
| x | p(X = x) | Decimal |
|---|---|---|
| 0 | 26/52 | 0.50 |
| 5 | 13/52 | 0.25 |
| 10 | 12/52 | 0.231 |
| 20 | 1/52 | 0.019 |
| Total | 52/52 | 1.00 |
Expected winnings = mean = \(\mu\) = E[X] = \(\sum_{X}\) * P(X=x)
\(\mu\) = (0 * 26/52) + (5 * 13/52) + (10 * 12/52) + (20 * 1/52)
\(\mu\) = 0 + 65/52 + 120/52 + 20/52 = 205/52 \(\approx\) 3.94
Answer to 2.34 (a) part 1: Expected winnings = 3.94
Standard Deviation = Squareroot of the variance = \(\sigma\)
\(\sigma^2\) = V[X] = \(\sum_{X} (x - \mu)^2\) * P(X = x)
\(\sigma^2 = (0 - 205/52)^2 * (26/52) + (5 - 205/52)^2 * (13/52) + (10 - 205/52)^2 * (12/52) + (20 - 205/52)^2 * (1/52)\)
\(\sigma^2\) = 42,025/5,408 + 3,025/10,816 + 297,675/35,152 + 697,225/140,608
\(\sigma^2\) = 87,075/10,816 + 145,225/10,816 = 232,300/10,816 = 58,075/2,704 \(\approx\) 21.477
Standard Deviation = \(\sigma = \sqrt{\sigma^2} = \sqrt{21.477} \approx 4.6\)
Answer to 2.34 (a) part 2: Standard Deviation = 4.6
What is the maximum amount you would be willing to pay to play this game? Explain.
Answer to 2.34 (b) = I would pay (bet) upto $5, because this distribution means you win half the time. Out of that half you make back $5 half the time and more the other half. Much better odds than Vegas. In other words, the expected win is 3.94 or 5 (out of available values) \(\pm\) 4.6 or 5.
An airline charges the following baggage fees: $25 for the first bag and $35 for the second. Suppose 54% of passengers have no checked luggage, 34% have one piece of checked luggage and 12% have two pieces. We suppose a negligible portion of people check more than two bags.
Build a probability model, compute the average revenue per passenger, and compute the corresponding standard deviation.
| Event | X | P(X) | X * P(X) |
|---|---|---|---|
| No checked bag | 0 | .54 | 0 |
| 1 checked bag | $25 | .34 | 8.5 |
| 2 checked bags | $60 | .12 | 7.2 |
| E(X) = 15.7 |
Answer to 2.40 (a) part 1: average revenue per passenger = E(x) = $15.70
\(\sigma^2\) = V[X] = \(\sum_{X} (x - \mu)^2\) * P(X = x)
\(\sigma^2 = (0 - 15.7)^2 * .54 + (25 - 15.7)^2 * .34 + (60 - 15.7)^2 * .12\)
\(\sigma^2\) = 133.1 + 29.4 + 235.5 = 398
Standard Deviation = \(\sigma = \sqrt{\sigma^2} = \sqrt{398} \approx 19.95\)
Answer to 2.40 (a) part 2: corresponding standard deviation = \(\sigma \approx 19.95\)
About how much revenue should the airline expect for a flight of 120 passengers? With what standard deviation? Note any assumptions you make and if you think they are justified.
Quick Way: revenue = 120 * $15.70 = $1,884 \(\pm\) $20 (or 19.95 rounded up)
Longer Way: (120 * .54 * 0) + (120 * .34 * 25) + (120 * .12 * 60) = $1,884, use same \(\sigma\)
Answer to 2.40 (b): how much revenue = # of passengers * average revenue = $1,884 \(\pm\) $20
Marcie has been tracking the following two items on Ebay:
Marcie wants to sell the video game and buy the textbook. How much net money (profits - losses) would she expect to make or spend? Also compute the standard deviation of how much she would make or spend.
\(X_{game} - Y_{book}\) = $38 - $110 = -$72, or she will spend $72
SD = \(\sigma = \sqrt{\sigma^2}\) = \(\sqrt{4^2 + 5^2} = \sqrt{16 + 25} \approx 6.40\)
Answer to 2.42 (a): She should expect to spend $72 with a SD of $6.40
Lucy is selling the textbook on Ebay for a friend, and her friend is giving her a 10% commission (Lucy keeps 10% of the revenue). How much money should she expect to make? With what standard deviation?
E(X) = $110 * .10 Commisson = $11; SD = $4 * .10 = $.4
Answer to 2.42 (b): Lucy should expect $11 with an SD of $.40 (40 cents).
The relative frequency table below displays the distribution of annual total personal income (in 2009 in ation-adjusted dollars) for a representative sample of 96,420,486 Americans. These data come from the American Community Survey for 2005-2009. This sample is comprised of 59% males and 41% females.
Describe the distribution of total personal income.
Answer to 2.46 (a) This is a smooth distribution rising to the middle value ($35,000 to $49,999), dropping quickly after this for the next two sections, before rising slowly for the last two sections. I would call it roller coaster like.
What is the probability that a randomly chosen US resident makes less than $50,000 per year?
P(<$50k/yr) = 21.2% + 18.3% + 15.8% + 4.7% + 2.2% = 62.2%
Answer to 2.46 (b): P(<$50k/yr) = 62.2%
What is the probability that a randomly chosen US resident makes less than $50,000 per year and is female? Note any assumptions you make.
P(<$50k/yr and female) = P(<$50k/yr) * P(female) = we would expect the percentage of females to be evenly distributed throughout this distribution if these things were independent = 41% (females) of the 62.2% (<$50k/yr) = .622 * .41 = ,255 = 25.5%
Answer to 2.46 (c): P(<$50k/yr and female) = 25.5%, if these two events are independent
The same data source indicates that 71.8% of females make less than $50,000 per year. Use this value to determine whether or not the assumption you made in part (c) is valid.
Answer to 2.46 (d) No, my assumption is not valid. Being a woman amd making less money is NOT independent.