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A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value?
x | P ( x ) |
---|---|
0 | P ( x = 0) = $\frac{4}{50}$ |
1 | P ( x = 1) = $\frac{8}{50}$ |
2 | P ( x = 2) = $\frac{16}{50}$ |
3 | P ( x = 3) = $\frac{14}{50}$ |
4 | P ( x = 4) = $\frac{6}{50}$ |
5 | P ( x = 5) = $\frac{2}{50}$ |
The expected value is 2.24
(0) $\frac{4}{50}$ + (1) $\frac{4}{50}$ + (2) $\frac{16}{50}$ + (3) $\frac{14}{50}$ + (4) $\frac{6}{50}$ + (5) $\frac{2}{50}$ = 0 + $\frac{8}{50}$ + $\frac{32}{50}$ + $\frac{42}{50}$ + $\frac{24}{50}$ + $\frac{10}{50}$ = $\frac{116}{50}$ = 2.24
Suppose you play a game of chance in which five numbers are chosen from 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. A computer randomly selects five numbers from zero to nine with replacement. You pay $2 to play and could profit $100,000 if you match all five numbers in order (you get your $2 back plus $100,000). Over the long term, what is your expected profit of playing the game?
To do this problem, set up an expected value table for the amount of money you can profit.
Let X = the amount of money you profit. The values of x are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Since you are interested in your profit (or loss), the values of x are 100,000 dollars and −2 dollars.
To win, you must get all five numbers correct, in order. The probability of choosing one correct number is $\frac{1}{10}$ because there are ten numbers. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is
Therefore, the probability of winning is 0.00001 and the probability of losing is
The expected value table is as follows:
x | P ( x ) | x * P ( x ) | |
---|---|---|---|
Loss | –2 | 0.99999 | (–2)(0.99999) = –1.99998 |
Profit | 100,000 | 0.00001 | (100000)(0.00001) = 1 |
Since –0.99998 is about –1, you would, on average, expect to lose approximately $1 for each game you play. However, each time you play, you either lose $2 or profit $100,000. The $1 is the average or expected LOSS per game after playing this game over and over.
You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay $1 to play. If you guess the right suit every time, you get your money back and $256. What is your expected profit of playing the game over the long term?
Let X = the amount of money you profit. The x -values are –$1 and $256.
The probability of guessing the right suit each time is $\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)=\frac{1}{256}$ = 0.0039
The probability of losing is 1 – $\frac{1}{256}$ = $\frac{255}{256}$ = 0.9961
(0.0039)256 + (0.9961)(–1) = 0.9984 + (–0.9961) = 0.0023 or 0.23 cents.
Suppose you play a game with a biased coin. You play each game by tossing the coin once.
P (heads) =
$\frac{2}{3}$ and
P (tails) =
$\frac{1}{3}$ . If you toss a head, you pay $6. If you toss a tail, you win $10. If you play this game many times, will you come out ahead?
a. Define a random variable X .
a.
X = amount of profit
b. Complete the following expected value table.
x | ____ | ____ | |
---|---|---|---|
WIN | 10 | $\frac{1}{3}$ | ____ |
LOSE | ____ | ____ | $\frac{\mathrm{\u201312}}{3}$ |
b.
x | P ( x ) | xP ( x ) | |
---|---|---|---|
WIN | 10 | $\frac{1}{3}$ | $\frac{10}{3}$ |
LOSE | –6 | $\frac{2}{3}$ | $\frac{\mathrm{\u201312}}{3}$ |
c. What is the expected value, μ ? Do you come out ahead?
c. Add the last column of the table. The expected value μ = $\frac{\text{\u2013}2}{3}$ . You lose, on average, about 67 cents each time you play the game so you do not come out ahead.
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