4.3 Expected Value and Standard Deviation for a Discrete Probability Distribution
LEARNING OBJECTIVES
- Calculate and interpret the expected value of a probability distribution.
- Calculate the standard deviation for a probability distribution.
Expected Value of a Probability Distribution
The expected value is often referred to as the “long-term” average or mean. That is, over the long term of repeatedly doing an experiment, the expected outcome is this average.
Suppose we toss a coin and record the result. What is the probability that the result is heads? If we flip a coin two times, does the probability tell us that these flips will result in one head and one tail? We might toss a fair coin ten times and record nine heads. Probability does not describe the short-term results of an experiment. Probability gives information about what can be expected in the long term. To demonstrate this, Karl Pearson once tossed a fair coin 24,000 times! He recorded the results of each toss, obtaining heads 12,012 times, meaning that roughly half of the flips resulted in heads. In his experiment, Pearson illustrated the Law of Large Numbers.
The Law of Large Numbers states that, as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and the relative frequency approaches zero—the theoretical probability and the relative frequency get closer and closer together. This is exactly what Karl Pearson observed when he tossed the coin 24,000 times. When evaluating the long-term results of statistical experiments, we often want to know the “average” outcome. This “long-term average” is known as the mean or expected value of the experiment. In other words, after conducting many trials of an experiment, we would expect this average value.
The expected value, denoted by [latex]\mu[/latex] or [latex]E(x)[/latex], is a weighted average where each value of the random variable is weighted by the value’s corresponding probability.
[latex]\begin{eqnarray*}E(x)&=&\sum\left(x\times P(x)\right)\\\\\end{eqnarray*}[/latex]
EXAMPLE
A men’s soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is [latex]0.2[/latex], the probability that they play one day is [latex]0.5[/latex], and the probability that they play two days is [latex]0.3[/latex]. Find the long-term average or expected value of the number of days per week the men’s soccer team plays soccer.
Solution
First, let the random variable [latex]X[/latex] be the number of days the men’s soccer team plays soccer per week. [latex]X[/latex] takes on the values [latex]0,1,2[/latex]. The table below shows the probability distribution for [latex]X[/latex], and includes an additional column [latex]x\times P(x)[/latex] that we will use to calculate the expected value. In this new column, we will multiply each [latex]x[/latex] value by its corresponding probability.
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x\times P(x)[/latex] |
|---|---|---|
| [latex]0[/latex] | [latex]0.2[/latex] | [latex]0\times 0.2=0[/latex] |
| [latex]1[/latex] | [latex]0.5[/latex] | [latex]1\times 0.5=0.5[/latex] |
| [latex]2[/latex] | [latex]0.3[/latex] | [latex]2\times 0.3=0.6[/latex] |
Add the last column to find the long-term average or expected value:
[latex]\begin{eqnarray*}E(x)&=&(0\times 0.2)+(1\times 0.5)+(2\times 0.3)\\&=&0+0.5+0.6\\&=&1.1\\\end{eqnarray*}[/latex]
The expected value is [latex]1.1[/latex]. The men’s soccer team would, on the average, expect to play soccer [latex]1.1[/latex] days per week. The number [latex]1.1[/latex] is the long-term average or expected value if the men’s soccer team plays soccer week after week after week.
NOTE
The expected value does not represent a value that the random variable takes on. The expected value is an average. In this case, the expected value of [latex]1.1[/latex] is the average times the team plays per week. To understand what this means, imagine that each week, we record the number of times the soccer team played that week. We do this repeatedly for many, many, many weeks. Then, we calculate the mean of the numbers we recorded (using the techniques we learned previously)—the mean of these numbers equals [latex]1.1[/latex], the expected value. The number of trials must be very, very large in order for the mean of the values recorded from the trials to equal the expected value calculated using the expected value formula.
EXAMPLE
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 [latex]\$2[/latex] to play and could profit [latex]\$100,000[/latex] if you match all five numbers in order (you get your [latex]\$2[/latex] back plus [latex]\$100,000[/latex]). Over the long term, what is your expected profit from playing the game?
Solution
To solve this problem, set up an expected value table for the amount of money you can profit. Let [latex]X[/latex] be the amount of money you profit. The values of [latex]x[/latex] are not 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Because you are interested in your profit (or loss), the values of [latex]x[/latex] are [latex]\$100,000[/latex] and [latex]−\$2[/latex] dollars.
To win, you must get all five numbers correct, in order. Because there are ten numbers, the probability of choosing one correct number is [latex]\displaystyle{\frac{1}{10}=0.1}[/latex]. You may choose a number more than once. The probability of choosing all five numbers correctly and in order is
[latex]\displaystyle{\frac{1}{10}\times\frac{1}{10}\times\frac{1}{10}\times\frac{1}{10}\times\frac{1}{10}=0.00001}[/latex]
Therefore, the probability of winning is [latex]0.00001[/latex] and the probability of losing is [latex]\displaystyle{1-0.00001=0.99999}[/latex].
The expected value is as follows:
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x\times P(x)[/latex] |
|---|---|---|
| [latex]-2[/latex] | [latex]0.99999[/latex] | [latex]-1.99998[/latex] |
| [latex]100,000[/latex] | [latex]0.00001[/latex] | [latex]1[/latex] |
[latex]\begin{eqnarray*}\\E(x)&=&-1.99998+1\\&=&-0.99998\\\\\end{eqnarray*}[/latex]
Because [latex]–0.99998[/latex] is about [latex]–1[/latex], you would, on average, expect to lose approximately [latex]\$1[/latex] for each game you play. However, each time you play, you either lose [latex]\$2[/latex] or profit [latex]\$100,000[/latex]. The [latex]\$1[/latex] is the average (or expected) LOSS per game after playing this game over and over.
EXAMPLE
Suppose you play a game with a biased coin where the probability of heads is [latex]\displaystyle{\frac{2}{3}}[/latex]. You play each game by tossing the coin once. If you toss a head, you pay [latex]\$6[/latex]. If you toss a tail, you win [latex]\$10[/latex]. If you play this game many times, will you come out ahead?
- Define a random variable [latex]X[/latex].
- Construct the probability distribution for [latex]X[/latex].
- What is the expected value? Do you come out ahead?
Solution
- Let [latex]X[/latex] be the amount of profit per game. The values of [latex]x[/latex] are [latex]–\$6[/latex] (for a loss) and [latex]\$10[/latex] (for a win).
-
[latex]x[/latex] [latex]P(x)[/latex] [latex]10[/latex] [latex]\displaystyle{\frac{1}{3}}[/latex] [latex]-6[/latex] [latex]\displaystyle{\frac{2}{3}}[/latex] -
[latex]x[/latex] [latex]P(x)[/latex] [latex]x\times P(x)[/latex] [latex]10[/latex] [latex]\displaystyle{\frac{1}{3}}[/latex] [latex]\displaystyle{\frac{10}{3}}[/latex] [latex]-6[/latex] [latex]\displaystyle{\frac{2}{3}}[/latex] [latex]-4[/latex] [latex]\begin{eqnarray*}\\E(x)&=&\frac{10}{3}+(-4)\\&=&-0.67\\\\\end{eqnarray*}[/latex]
On average, you lose [latex]\$0.67[/latex] each time you play the game, so you do not come out ahead.
TRY IT
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 [latex]\$1[/latex] to play. If you guess the right suit every time, you get your money back and [latex]\$256[/latex]. What is your expected profit from playing the game over the long term?
Click to see Solution
Let [latex]X[/latex] be the amount of money you profit. The values of [latex]x[/latex] are [latex]–\$1[/latex] (for a loss) and [latex]\$256[/latex] (for a win).
The probability of winning (guessing the correct suit on each draw) is
[latex]\displaystyle{\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}=0.0039}[/latex]
The probability of losing is
[latex]\displaystyle{1-0.0039=0.9961}[/latex]
The expected value is as follows:
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x\times P(x)[/latex] |
|---|---|---|
| [latex]-1[/latex] | [latex]0.99961[/latex] | [latex]-0.9961[/latex] |
| [latex]256[/latex] | [latex]0.0039[/latex] | [latex]0.9984[/latex] |
[latex]\begin{eqnarray*}\\E(x)&=&-0.9961+0.9984\\&=&0.0023\\\\\end{eqnarray*}[/latex]
Playing the game over and over again means you would average [latex]\$0.0023[/latex] in profit per game.
TRY IT
Suppose you play a game with a spinner that has three colours on it: red, green, and blue. The probability of landing on red is [latex]40\%[/latex], and the probability of landing on green is [latex]20\%[/latex]. You play a game by spinning the spinner once. If you land on red, you pay [latex]\$10[/latex]. If you land on blue, you do not pay or win anything. If you land on green, you win [latex]\$10[/latex]. What is the expected value of this game? Do you come out ahead?
Click to see Solution
Let [latex]X[/latex] be the amount won in a game. The values of [latex]x[/latex] are [latex]–\$10[/latex] (for red), [latex]\$0[/latex] (for blue) and [latex]\$10[/latex] (for a green).
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x\times P(x)[/latex] |
|---|---|---|
| [latex]-10[/latex] | [latex]0.4[/latex] | [latex]-4[/latex] |
| [latex]0[/latex] | [latex]0.4[/latex] | [latex]0[/latex] |
| [latex]10[/latex] | [latex]0.2[/latex] | [latex]2[/latex] |
[latex]\begin{eqnarray*}\\E(x)&=&-4+0+2\\&=&-2\\\\\end{eqnarray*}[/latex]
On average, you lose [latex]\$2[/latex] per game. So you do not come out ahead.
Standard Deviation of a Probability Distribution
Like data, probability distributions have standard deviations. The standard deviation, denoted [latex]\sigma[/latex], of a probability distribution for a random variable [latex]X[/latex] describes the spread or variability of the probability distribution. The standard deviation is the standard deviation we expect when doing an experiment over and over.
[latex]\begin{eqnarray*}\sigma&=&\sqrt{\sum\left((x-\mu)^2\times P(x)\right)}\end{eqnarray*}[/latex]
To calculate the standard deviation of a probability distribution, find each deviation from its expected value, square it, multiply it by its probability, add the products, and take the square root.
EXAMPLE
Let [latex]X[/latex] be the number of times per week a newborn baby’s crying wakes its mother after midnight. The probability distribution for [latex]X[/latex] is:
| [latex]x[/latex] | [latex]P(x)[/latex] |
|---|---|
| [latex]0[/latex] | [latex]0.04[/latex] |
| [latex]1[/latex] | [latex]0.22[/latex] |
| [latex]2[/latex] | [latex]0.46[/latex] |
| [latex]3[/latex] | [latex]0.18[/latex] |
| [latex]4[/latex] | [latex]0.08[/latex] |
| [latex]5[/latex] | [latex]0.02[/latex] |
Find the expected value and standard deviation of the number of times a newborn baby’s crying wakes its mother after midnight.
Solution
For the expected value:
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x\times P(x)[/latex] |
|---|---|---|
| [latex]0[/latex] | [latex]0.04[/latex] | [latex]0[/latex] |
| [latex]1[/latex] | [latex]0.22[/latex] | [latex]0.22[/latex] |
| [latex]2[/latex] | [latex]0.46[/latex] | [latex]0.92[/latex] |
| [latex]3[/latex] | [latex]0.18[/latex] | [latex]0.54[/latex] |
| [latex]4[/latex] | [latex]0.08[/latex] | [latex]0.32[/latex] |
| [latex]5[/latex] | [latex]0.02[/latex] | [latex]0.1[/latex] |
[latex]\begin{eqnarray*}\\\mu&=&0+0.22+0.92+0.54+0.32+0.1\\&=&2.1\\\\\end{eqnarray*}[/latex]
On average, a newborn wakes its mother after midnight [latex]2.1[/latex] times per week.
For the standard deviation: For each value [latex]x[/latex], multiply the square of its deviation by its probability (each deviation has the format [latex]x-\mu[/latex]).
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex](x-\mu)^2\times P(x)[/latex] |
|---|---|---|
| [latex]0[/latex] | [latex]0.04[/latex] | [latex](0-2.1)^2\times 0.04=0.1764[/latex] |
| [latex]1[/latex] | [latex]0.22[/latex] | [latex](1-2.1)^2\times 0.22=0.2662[/latex] |
| [latex]2[/latex] | [latex]0.46[/latex] | [latex](2-2.1)^2\times 0.46=0.0046[/latex] |
| [latex]3[/latex] | [latex]0.18[/latex] | [latex](3-2.1)^2\times 0.18=0.1458[/latex] |
| [latex]4[/latex] | [latex]0.08[/latex] | [latex](4-2.1)^2\times 0.08=0.2888[/latex] |
| [latex]5[/latex] | [latex]0.02[/latex] | [latex](5-2.1)^2\times 0.02=0.1682[/latex] |
| Sum | [latex]1.05[/latex] |
Add the values in the third column of the table and then take the square root of this sum:
[latex]\begin{eqnarray*}\sigma&=&\sqrt{1.05}\\&=&1.024...\end{eqnarray*}[/latex]
TRY IT
A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. Let [latex]X[/latex] be the number of times a post-op patient rings for the nurse. For a random sample of [latex]50[/latex] patients, the following information was obtained. What is the expected value? What is the standard deviation?
| [latex]x[/latex] | [latex]P(x)[/latex] |
|---|---|
| [latex]0[/latex] | [latex]0.08[/latex] |
| [latex]1[/latex] | [latex]0.16[/latex] |
| [latex]2[/latex] | [latex]0.32[/latex] |
| [latex]3[/latex] | [latex]0.28[/latex] |
| [latex]4[/latex] | [latex]0.12[/latex] |
| [latex]5[/latex] | [latex]0.04[/latex] |
Click to see Solution
For the expected value:
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x\times P(x)[/latex] |
|---|---|---|
| [latex]0[/latex] | [latex]0.08[/latex] | [latex]0[/latex] |
| [latex]1[/latex] | [latex]0.16[/latex] | [latex]0.16[/latex] |
| [latex]2[/latex] | [latex]0.32[/latex] | [latex]0.64[/latex] |
| [latex]3[/latex] | [latex]0.28[/latex] | [latex]0.84[/latex] |
| [latex]4[/latex] | [latex]0.12[/latex] | [latex]0.48[/latex] |
| [latex]5[/latex] | [latex]0.04[/latex] | [latex]0.2[/latex] |
[latex]\begin{eqnarray*}\\\mu&=&0+0.16+0.64+0.84+0.48+0.2\\&=&2.32\end{eqnarray*}[/latex]
For the standard deviation:
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex](x-\mu)^2\times P(x)[/latex] |
|---|---|---|
| [latex]0[/latex] | [latex]0.08[/latex] | [latex]0.430592[/latex] |
| [latex]1[/latex] | [latex]0.16[/latex] | [latex]0.278784[/latex] |
| [latex]2[/latex] | [latex]0.32[/latex] | [latex]0.032768[/latex] |
| [latex]3[/latex] | [latex]0.28[/latex] | [latex]0.129472[/latex] |
| [latex]4[/latex] | [latex]0.12[/latex] | [latex]0.338688[/latex] |
| [latex]5[/latex] | [latex]0.04[/latex] | [latex]0.287296[/latex] |
[latex]\begin{eqnarray*}\\\sigma&=&\sqrt{0.430592+0.27878784+0.032768+0.129472+0.338688+0.287296}\\&=&1.22....\end{eqnarray*}[/latex]
TRY IT
On May 11, 2013, at 9:30 PM, the probability that moderate seismic activity (one moderate earthquake) would occur in the next 48 hours in Japan was about [latex]1.08\%[/latex]. You bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win [latex]\$100[/latex]. If you lose the bet, you pay [latex]\$10[/latex]. Let [latex]X[/latex] be the amount of profit from a bet. Find the mean and standard deviation of [latex]X[/latex].
Click to see Solution
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x\times P(x)[/latex] | [latex](x-\mu)^2\times P(x)[/latex] |
|---|---|---|---|
| [latex]100[/latex] | [latex]0.0108[/latex] | [latex]1.08[/latex] | [latex]127.8726[/latex] |
| [latex]-10[/latex] | [latex]0.9892[/latex] | [latex]-9.892[/latex] | [latex]1.3961[/latex] |
[latex]\begin{eqnarray*}\\\mu&=&1.08+(-9.892)\\&=&-8.812\\\\\sigma&=&\sqrt{127.7826+1.3961}\\&=&11.3696....\end{eqnarray*}[/latex]
Video: “Mean (expected value) of a discrete random variable | AP Statistics | Khan Academy” by Khan Academy [4:32] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.
Video: “Variance and standard deviation of a discrete random variable | AP Statistics | Khan Academy” by Khan Academy [6:26] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.
Exercises
- A company wants to evaluate its attrition rate, in other words, how long new hires stay with the company. Over the years, they have established the following probability distribution. Let [latex]X[/latex] be the number of years a new hire will stay with the company.
[latex]x[/latex] [latex]P(x)[/latex] [latex]0[/latex] [latex]0.12[/latex] [latex]1[/latex] [latex]0.18[/latex] [latex]2[/latex] [latex]0.30[/latex] [latex]3[/latex] [latex]0.15[/latex] [latex]4[/latex] [latex]0.10[/latex] [latex]5[/latex] [latex]0.10[/latex] [latex]6[/latex] [latex]0.05[/latex] - On average, how long would you expect a new hire to stay with the company?
- Calculate the standard deviation for the probability distribution.
Click to see Answer
- [latex]2.43[/latex] years
- [latex]1.65[/latex] years
- A baker is deciding how many batches of muffins to make to sell in his bakery. He wants to make enough to sell every one and no fewer. Through observation, the baker has established a probability distribution. The random variable [latex]X[/latex] is the number of batches of muffins the baker sells.
[latex]x[/latex] [latex]P(x)[/latex] [latex]1[/latex] [latex]0.15[/latex] [latex]2[/latex] [latex]0.35[/latex] [latex]3[/latex] [latex]0.40[/latex] [latex]4[/latex] [latex]0.10[/latex] - On average, how many batches should the baker make?
- Calculate the standard deviation for the probability distribution.
Click to see Answer
- [latex]2.45[/latex] batches
- [latex]0.86[/latex] batches
- A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. Let the random variable be the number of years a physics major will spend doing post-graduate research. The professor has the following probability distribution.
[latex]x[/latex] [latex]P(x)[/latex] [latex]1[/latex] [latex]0.35[/latex] [latex]2[/latex] [latex]0.20[/latex] [latex]3[/latex] [latex]0.15[/latex] [latex]4[/latex] [latex]0.15[/latex] [latex]5[/latex] [latex]0.10[/latex] [latex]6[/latex] [latex]0.05[/latex] - On average, how many years would you expect a physics major to spend doing post-graduate research?
- Calculate the standard deviation for the probability distribution.
Click to see Answer
- [latex]2.6[/latex] years
- [latex]1.56[/latex] years
- A ballet instructor is interested in knowing what percent of each year’s class will continue on to the next, so that she can plan what classes to offer. Over the years, she has established the following probability distribution. Let the random variable be the number of years a student will study ballet with the teacher.
[latex]x[/latex] [latex]P(x)[/latex] [latex]1[/latex] [latex]0.10[/latex] [latex]2[/latex] [latex]0.05[/latex] [latex]3[/latex] [latex]0.10[/latex] [latex]4[/latex] [latex]0.15[/latex] [latex]5[/latex] [latex]0.30[/latex] [latex]6[/latex] [latex]0.20[/latex] [latex]7[/latex] [latex]0.10[/latex] - On average, how many years would you expect a child to study ballet with this teacher?
- Calculate the standard deviation for the probability distribution.
Click to see Answer
- [latex]4.5[/latex] years
- [latex]1.72[/latex] years
- You are playing a game by drawing a card from a standard deck and replacing it. If the card is a face card, you win [latex]\$30[/latex]. If it is not a face card, you pay [latex]\$2[/latex]. There are 12 face cards in a deck of 52 cards. Let the random variable be the amount of money you win/lose for each draw from the deck.
- Construct the probability distribution for this random variable.
- What is the expected value of playing the game?
- Should you play the game? Explain.
Click to see Answer
-
[latex]x[/latex] [latex]P(x)[/latex] [latex]30[/latex] [latex]\frac{12}{52}[/latex] [latex]-2[/latex] [latex]\frac{40}{52}[/latex] - [latex]\$5.38[/latex]
- Yes, because, on average, you will win [latex]\$5.38[/latex] for each draw.
- A theatre group holds a fund-raiser. It sells 100 raffle tickets for [latex]\$5[/latex] apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of [latex]\$150[/latex]. Let the random variable be the amount of money you spent/earned in the raffle.
- Construct the probability distribution for this random variable.
- If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle?
- Calculate the standard deviation for the probability distribution.
Click to see Answer
-
[latex]x[/latex] [latex]P(x)[/latex] [latex]-20[/latex] [latex]0.96[/latex] [latex]130[/latex] [latex]0.04[/latex] - [latex]-\$14[/latex]
- [latex]\$29.39[/latex]
- A game involves selecting a card from a regular 52-card deck and tossing a coin. The coin is a fair coin and is equally likely to land on heads or tails.
- If the card is a face card and the coin lands on Heads, you win [latex]\$6[/latex].
- If the card is a face card and the coin lands on Tails, you win [latex]\$2[/latex].
- If the card is not a face card, you lose [latex]\$2[/latex], no matter what the coin shows.
Let the random variable be the amount of money you win/lose for each play.
- Construct the probability distribution for this random variable.
- Find the expected value for this game (expected net gain or loss).
- Explain what your calculations indicate about your long-term average profits and losses on this game.
- Should you play this game to win money?
Click to see Answer
-
[latex]x[/latex] [latex]P(x)[/latex] [latex]6[/latex] [latex]0.1154[/latex] [latex]2[/latex] [latex]0.1154[/latex] [latex]-2[/latex] [latex]0.7692[/latex] - [latex]-\$0.62[/latex]
- If you play the game repeatedly a large number of times, you will lose an average of [latex]\$0.62[/latex] per play.
- You should not play the game because, on average, you will lose [latex]\$0.62[/latex] per play.
- You buy a lottery ticket to a lottery that costs [latex]\$10[/latex] per ticket. There are only [latex]100[/latex] tickets available to be sold in this lottery. In this lottery there are one [latex]\$500[/latex] prize, two [latex]\$100[/latex] prizes, and four [latex]\$25[/latex] prizes. Find your expected gain or loss.
Click to see Answer
[latex]\$0[/latex]
- Suppose that you are offered the following “deal.” You roll a die. If you roll a six, you win [latex]\$10[/latex]. If you roll a four or five, you win [latex]\$5[/latex]. If you roll a one, two, or three, you pay [latex]\$6[/latex]. Let the random variable be the amount of money you win/lose on each play.
- Construct a probability distribution for this random variable.
- Over the long run of playing this game, what are your expected average winnings per game?
- Based on numerical values, should you take the deal? Explain your decision in complete sentences.
Click to see Answer
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[latex]x[/latex] [latex]P(x)[/latex] [latex]-6[/latex] [latex]0.5[/latex] [latex]5[/latex] [latex]0.3333[/latex] [latex]10[/latex] [latex]0.1667[/latex] - [latex]\$0.33[/latex] per play.
- You should play the game because, on average, you will win [latex]\$0.33[/latex] per play.
- A venture capitalist willing to invest [latex]\$1,000,000[/latex] has three investments to choose from.
- The first investment, a software company, has a [latex]10\%[/latex] chance of returning [latex]\$5,000,000[/latex] profit, a [latex]30\%[/latex] chance of returning [latex]\$1,000,000[/latex] profit, and a [latex]60\%[/latex] chance of losing the million dollars
- The second investment, a hardware company, has a [latex]20\%[/latex] chance of returning [latex]\$3,000,000[/latex] profit, a [latex]40\%[/latex] chance of returning [latex]\$1,000,000[/latex] profit, and a [latex]40\%[/latex] chance of losing the million dollars.
- The third investment, a biotech firm, has a [latex]10\%[/latex] chance of returning [latex]\$6,000,000[/latex] profit, a [latex]70\%[/latex] chance of returning [latex]\$1,000,000[/latex], and a [latex]20\%[/latex] chance of losing the million dollars.
Let [latex]X[/latex] be the profit/loss for the first investment, let [latex]Y[/latex] be the profit/loss for the second investment, and let [latex]Z[/latex] be the profit/loss for the third investment.
- Construct a probability distribution for the first investment.
- Construct a probability distribution for the second investment.
- Construct a probability distribution for the third investment.
- Find the expected value for each investment.
- Which investment has the highest expected return, on average?
Click to see Answer
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[latex]x[/latex] [latex]P(x)[/latex] [latex]4,000,000[/latex] [latex]0.1[/latex] [latex]0[/latex] [latex]0.3[/latex] [latex]-1,000,000[/latex] [latex]0.6[/latex] -
[latex]x[/latex] [latex]P(x)[/latex] [latex]2,000,000[/latex] [latex]0.2[/latex] [latex]0[/latex] [latex]0.4[/latex] [latex]-1,000,000[/latex] [latex]0.4[/latex] -
[latex]x[/latex] [latex]P(x)[/latex] [latex]5,000,000[/latex] [latex]0.1[/latex] [latex]0[/latex] [latex]0.7[/latex] [latex]-1,000,000[/latex] [latex]0.2[/latex] - Investment 1: [latex]\$200,000[/latex]; Investment 2: [latex]\$0[/latex]; Investment 3: [latex]\$300,000[/latex].
- Investment 3.
- Suppose that the probability distribution for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given in the following table. On average, how many years do you expect it to take for an individual to earn a B.S.?
[latex]x[/latex] [latex]P(x)[/latex] [latex]3[/latex] [latex]0.05[/latex] [latex]4[/latex] [latex]0.40[/latex] [latex]5[/latex] [latex]0.30[/latex] [latex]6[/latex] [latex]0.15[/latex] [latex]7[/latex] [latex]0.10[/latex] Click to see Answer
[latex]4.85[/latex] years
- A “friend” offers you the following “deal.” For a [latex]\$10[/latex] fee, you may pick an envelope from a box containing 100 seemingly identical envelopes. However, each envelope contains a coupon for a free gift.
- Ten of the coupons are for a free gift worth [latex]\$6[/latex].
- Eighty of the coupons are for a free gift worth [latex]\$8[/latex].
- Six of the coupons are for a free gift worth [latex]\$12[/latex].
- Four of the coupons are for a free gift worth [latex]\$40[/latex].
Based on the financial gain or loss over the long run, should you play the game?
Click to see Answer
No, because, on average, you will lose [latex]-\$0.68[/latex] per play.
“4.3 Expected Value and Standard Deviation for a Discrete Probability Distribution” and “4.6 Exercises” from Introduction to Statistics by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
