4.2 Mean or Expected Value and Standard Deviation
SLO 3
Describe and apply probability concepts and distributions.
A distribution is only useful once you can summarize it, so here you build the expected value and standard deviation from a PDF table and read them as a long-run average gain or loss - the tool that tells you whether a bet, a raffle, or an investment is worth taking.
Learning Objectives
By the end of this section, you will be able to:
- explain what the Law of Large Numbers says about the difference between short-term results and long-term averages;
- compute the expected value \(\mu\) of a discrete random variable by building an expected value table;
- compute the standard deviation \(\sigma\) of a discrete probability distribution from its deviations;
- interpret an expected value as a long-run average gain or loss, and use it to decide whether a game or a bet is worth playing;
- recognize when an expected value table has been filled in incorrectly.
Section 4.1 gave you a probability distribution: a list of the values a discrete random variable can take and how likely each one is. That is a complete description, but it is a lot to hold in your head at once. What people usually want from a distribution is a single number that says "here is roughly what to expect."
This section builds that number, and then builds a second one that says how far from it you should expect to land.
4.2.1 Expected Value: The Long-Term Average
Let \(X\) be a discrete random variable with probability distribution function \(P(x)\). The expected value, or mean, of \(X\) is
$$ \mu = \sum \left( x \cdot P(x) \right) $$where the sum runs over every value \(x\) that \(X\) can take.
A family can have an expected number of 2.3 children, and no family has 2.3 children. The expected value is the balance point of the distribution, not a prediction of any single outcome. Read it as "the average over a very long run," never as "the result I should expect tonight."
Definition 4.2.1 — The expected value is where the x times P(x) contributions land you on the same axis the outcomes live on.
In words: multiply each value of the random variable by its probability, then add all the products. The bookkeeping for that calculation is easy to lose track of in your head, so we set it up in a table with one row per value of \(X\) and one column for each piece of the formula. That table is called an expected value table, and it is the workhorse of this entire section. Every problem you will meet here — soccer schedules, lottery tickets, biased coins, earthquake bets — is the same three columns filled with different numbers. Build the table first and the arithmetic takes care of itself.
Toss a coin and record the result. What is the probability the result is heads? If you flip a coin twice, does probability promise you one head and one tail? It does not. You might toss a fair coin ten times and record nine heads. As you saw in Probability Topics, probability says nothing about the short-term results of an experiment. It tells you what to expect in the long run.
Karl Pearson once tossed a fair coin 24,000 times to make exactly this point. He recorded 12,012 heads — a relative frequency of 0.5005, almost dead on the theoretical 0.5. Ten tosses can land anywhere. Twenty-four thousand cannot.
That is the Law of Large Numbers: as the number of trials in a probability experiment increases, the difference between the theoretical probability of an event and its relative frequency approaches zero. The two get closer and closer together.
Once you accept that, a natural question follows. If you ran an experiment over and over, what would the average outcome settle down to? That long-term average is called the mean or expected value of the experiment, and it is written with the Greek letter \(\mu\).
A customer orders 1, 2, or 3 bottles of water. The probability of ordering 1 bottle is 0.6, the probability of ordering 2 bottles is 0.3, and the probability of ordering 3 bottles is 0.1. Find the long-term average or expected value, \(\mu\), of the number of bottles a customer will order.
Solution
Step 1 — Build the expected value table. Let \(X\) = the number of bottles a customer orders.
| x | P(x) | x · P(x) |
|---|---|---|
| 1 | 0.6 | (1)(0.6) = 0.6 |
| 2 | 0.3 | (2)(0.3) = 0.6 |
| 3 | 0.1 | (3)(0.1) = 0.3 |
Step 2 — Add the last column.
$$ \mu = 0.6 + 0.6 + 0.3 = 1.5 $$Answer: \(\mu = 1.5\) bottles. Over many customers, the store sells an average of 1.5 bottles per order.
Coach Dana Whitfield is planning next season around her team's schedule. Her soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value, \(\mu\), of the number of days per week the soccer team plays soccer.
Solution
Step 1 — Define the random variable. Let \(X\) = the number of days the team Dana coaches plays soccer per week. \(X\) takes on the values 0, 1, and 2.
Step 2 — Build the expected value table. Construct the PDF table and add a column for \(x \cdot P(x)\). In that column you multiply each \(x\) value by its probability.
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.2 | (0)(0.2) = 0 |
| 1 | 0.5 | (1)(0.5) = 0.5 |
| 2 | 0.3 | (2)(0.3) = 0.6 |
Step 3 — Add the \(x \cdot P(x)\) column.
$$ \mu = (0)(0.2) + (1)(0.5) + (2)(0.3) = 0 + 0.5 + 0.6 = 1.1 $$Answer: \(\mu = 1.1\). Dana's team would, on average, expect to play soccer 1.1 days per week. No single week has 1.1 practice days — 1.1 is the long-term average if she runs the season week after week after week.
Try it in rāSHio
Open rāSHio and choose Calculators → Expected Value, then enter Dana's three outcomes (0, 1, 2) against their probabilities (0.2, 0.5, 0.3) to get \(\mu = 1.1\) without building the \(x \cdot P(x)\) column by hand. The same panel reports the standard deviation, so it also does the deviation column of Table 4.2.6 for you — build the table yourself first, then use the calculator to check it.
Figure 4.2.1 — Reading μ and σ straight off a discrete distribution in rāSHio: Calculators → Expected Value. The walkthrough uses its own demo numbers; run the same steps on Dana's table.
4.2.2 The Standard Deviation of a Probability Distribution
Let \(X\) be a discrete random variable with mean \(\mu\). The standard deviation of \(X\) is the square root of its variance:
$$ \sigma = \sqrt{\sum \left[ (x - \mu)^2 \cdot P(x) \right]} $$Values above the mean give positive deviations and values below give negative ones, and if you just added them they would cancel to zero every time. Squaring makes every deviation count as distance, regardless of direction. Taking the square root at the end puts the answer back into the original units.
Definition 4.2.2 — Each deviation becomes a square, probability shrinks it, and the side of the pooled square is the standard deviation.
To calculate it: find each value's deviation from the expected value, square it, multiply by that value's probability, add the products, and take the square root. When all outcomes in the distribution are equally likely, this formula and the one in Definition 4.2.1 reduce to the ordinary mean and standard deviation of the set of possible outcomes.
The expected value tells you where the distribution balances. It says nothing about how spread out it is. Two distributions can share the same \(\mu\) and behave completely differently: one clustered tightly around the mean, one throwing you far to either side. Just as with data, the number that measures that spread is the standard deviation.
The recipe is the same one you used in Chapter 2, with one change. In Chapter 2 every data point counted equally. Here, each value gets weighted by how likely it is — a rare outcome that sits far from the mean should not stretch the spread as much as a common one does.
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 | \(P(x = 5) = \frac{2}{50}\) |
Solution
Step 1 — Convert the probabilities to decimals and build the \(x \cdot P(x)\) column.
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.08 | 0 |
| 1 | 0.16 | 0.16 |
| 2 | 0.32 | 0.64 |
| 3 | 0.28 | 0.84 |
| 4 | 0.12 | 0.48 |
| 5 | 0.04 | 0.20 |
Step 2 — Add the last column.
$$ \mu = 0 + 0.16 + 0.64 + 0.84 + 0.48 + 0.20 = 2.32 $$Answer: \(\mu = 2.32\). A post-op patient rings the nurse an average of about 2.3 times per 12-hour shift.
Mateo Reyes and his husband have been keeping a log since their daughter was born. Find the expected value of the number of times a newborn baby's crying wakes its parents after midnight, using the distribution Mateo recorded below. The expected value is the expected number of times per week a newborn baby's crying wakes its parents after midnight. Calculate the standard deviation of the variable as well.
| x | P(x) |
|---|---|
| 0 | \(P(x = 0) = \frac{2}{50}\) |
| 1 | \(P(x = 1) = \frac{11}{50}\) |
| 2 | \(P(x = 2) = \frac{23}{50}\) |
| 3 | \(P(x = 3) = \frac{9}{50}\) |
| 4 | \(P(x = 4) = \frac{4}{50}\) |
| 5 | \(P(x = 5) = \frac{1}{50}\) |
Solution
Step 0 — Add the two working columns. Fill in \(x \cdot P(x)\) for the mean, then the deviation column once the mean is known.
| x | P(x) | x · P(x) | (x − μ)2 · P(x) |
|---|---|---|---|
| 0 | \(\frac{2}{50}\) | \((0)\left(\frac{2}{50}\right) = 0\) | \((0 - 2.1)^2 \cdot 0.04 = 0.1764\) |
| 1 | \(\frac{11}{50}\) | \((1)\left(\frac{11}{50}\right) = \frac{11}{50}\) | \((1 - 2.1)^2 \cdot 0.22 = 0.2662\) |
| 2 | \(\frac{23}{50}\) | \((2)\left(\frac{23}{50}\right) = \frac{46}{50}\) | \((2 - 2.1)^2 \cdot 0.46 = 0.0046\) |
| 3 | \(\frac{9}{50}\) | \((3)\left(\frac{9}{50}\right) = \frac{27}{50}\) | \((3 - 2.1)^2 \cdot 0.18 = 0.1458\) |
| 4 | \(\frac{4}{50}\) | \((4)\left(\frac{4}{50}\right) = \frac{16}{50}\) | \((4 - 2.1)^2 \cdot 0.08 = 0.2888\) |
| 5 | \(\frac{1}{50}\) | \((5)\left(\frac{1}{50}\right) = \frac{5}{50}\) | \((5 - 2.1)^2 \cdot 0.02 = 0.1682\) |
Step 1 — Add the values in the third column to find the expected value.
$$ \mu = \text{Expected Value} = \frac{105}{50} = 2.1 $$You expect a newborn to wake its parents after midnight 2.1 times per week, on average.
Step 2 — Use \(\mu\) to complete the fourth column. For each value \(x\), multiply the square of its deviation by its probability. Each deviation has the form \(x - \mu\), which is why the mean has to be computed first.
Step 3 — Add the values in the fourth column.
$$ 0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05 $$Step 4 — Take the square root.
$$ \sigma = \sqrt{1.05} \approx 1.0247 $$Answer: \(\mu = 2.1\) wakings per week with \(\sigma \approx 1.0247\). A typical week lands roughly one waking above or below 2.1, which matches what Mateo and his husband see in their log.
4.2.3 Expected Value in Games of Chance
Expected value earns its keep fastest when the random variable is money. Set \(X\) to your profit — what you walk away with minus what you paid — and \(\mu\) tells you your average gain or loss per play over the long run. A positive \(\mu\) means the game pays you to play it. A negative \(\mu\) means it does not, no matter how good the jackpot sounds.
A $100,000 prize looks enormous next to a $2 ticket, and it is. But 0.00001 is a very small number, and multiplying an enormous prize by a tiny probability can easily land under the price of the ticket. Expected value makes those two quantities comparable.
The only subtlety in these problems is deciding what \(X\) measures. In a game of chance the values of \(X\) are almost never the numbers printed on the cards or the dice — they are the dollar amounts attached to those outcomes, and mixing the two up is the single most common way these problems go wrong. Ask yourself what you actually walk away with, not what the machine displays. There is a second habit worth building at the same time: write the profit from the player's point of view, so money you hand over is negative and money you receive is positive. A game that charges $2 and pays $100,000 has profit values of \(-2\) and \(100000\), not \(2\) and \(100{,}002\), because the $2 you paid comes back to you along with the prize. Get that translation right and the expected value table looks exactly like the ones you have already built — three columns, one row per outcome, and a sum at the bottom that answers the question.
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?
Solution
Step 1 — Find the probability of winning. There are four suits, so the probability of guessing one card's suit correctly is \(\frac{1}{4}\). Because the cards are replaced, the four guesses are independent:
$$ P(\text{win}) = \left(\frac{1}{4}\right)^4 = \frac{1}{256} \approx 0.0039 $$So \(P(\text{lose}) = 1 - \frac{1}{256} = \frac{255}{256}\).
Step 2 — Define the profit. Let \(X\) = your profit. If you win you get your $1 back plus $256, so \(x = 256\). If you lose you are out the $1, so \(x = -1\).
Step 3 — Build the expected value table and add the last column.
| Outcome | x | P(x) | x · P(x) |
|---|---|---|---|
| Win | 256 | \(\frac{1}{256}\) | 1 |
| Lose | −1 | \(\frac{255}{256}\) | \(-\frac{255}{256} \approx -0.9961\) |
Answer: Your expected profit is about $0.004 per game — less than half a cent. This game is very close to fair; over the long run you neither gain nor lose in any meaningful way.
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?
Solution
Step 1 — Decide what \(X\) measures. Let \(X\) = the amount of money you profit. The values of \(X\) are not 0 through 9 — those are the numbers being drawn. Since you care about profit or loss, the values of \(X\) are $100,000 and \(-\$2\).
Step 2 — Find the probability of winning. 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, and you may choose a number more than once. So
$$ \left(\frac{1}{10}\right)\left(\frac{1}{10}\right)\left(\frac{1}{10}\right)\left(\frac{1}{10}\right)\left(\frac{1}{10}\right) = (1)(10^{-5}) = 0.00001 $$The probability of winning is 0.00001, so the probability of losing is
$$ 1 - 0.00001 = 0.99999 $$Step 3 — Build the expected value table.
| Outcome | x | P(x) | x · P(x) |
|---|---|---|---|
| Loss | −2 | 0.99999 | (−2)(0.99999) = −1.99998 |
| Profit | 100,000 | 0.00001 | (100,000)(0.00001) = 1 |
Step 4 — Add the last column.
$$ \mu = -1.99998 + 1 = -0.99998 $$Answer: Since \(-0.99998\) is about \(-1\), you would on average expect to lose approximately $1 for each game you play. Notice what that number is not: each time you play you either lose $2 or profit $100,000, and neither of those is $1. The $1 is the average loss per game after playing this game over and over.
Try it in rāSHio
A wager is the same calculation with money in the \(x\) column: in Calculators → Expected Value, enter \(-2\) and \(100{,}000\) against 0.99999 and 0.00001 to watch the lottery's expected profit come out at about \(-\$1\) — then change the payout or the odds and see how far they have to move before the game stops favouring the house.
4.2.4 Reading a Game's Expected Value Table
Sometimes a problem hands you a partly filled expected value table and asks you to finish it. That is the same work as before, done in a different order — you are given some cells and have to reconstruct the rest before you can add up the last column.
Two habits make this reliable. First, always check that the \(P(x)\) column sums to 1; if it does not, either a probability is missing or one is wrong. Second, keep the sign of each \(x\) honest. Money you pay is negative, money you win is positive, and a sign error in one row flips the conclusion of the whole problem.
Suppose you play a game with a spinner. You play each game by spinning the spinner once. \(P(\text{red}) = \frac{2}{5}\), \(P(\text{blue}) = \frac{2}{5}\), and \(P(\text{green}) = \frac{1}{5}\). If you land on red, you pay $10. If you land on blue, you don't pay or win anything. If you land on green, you win $10. Complete the following expected value table.
| Outcome | x | P(x) | x · P(x) |
|---|---|---|---|
| Red | \(-\frac{20}{5}\) | ||
| Blue | \(\frac{2}{5}\) | ||
| Green | 10 |
Solution
Step 1 — Fill in the \(x\) values from the payouts. Landing on red costs you $10, so \(x = -10\). Blue pays nothing, so \(x = 0\). Green wins $10, so \(x = 10\).
Step 2 — Fill in the probabilities and compute the products.
| Outcome | x | P(x) | x · P(x) |
|---|---|---|---|
| Red | −10 | \(\frac{2}{5}\) | \(-\frac{20}{5} = -4\) |
| Blue | 0 | \(\frac{2}{5}\) | 0 |
| Green | 10 | \(\frac{1}{5}\) | \(\frac{10}{5} = 2\) |
Step 3 — Add the last column.
$$ \mu = -4 + 0 + 2 = -2 $$Answer: \(\mu = -\$2\). You lose an average of $2 every time you spin, so this is not a game to play for money.
Nora Whitaker runs a booth at her campus queer-student-union fundraiser, and she is testing a game with a biased coin before she puts it out. You play each game by tossing the coin once. \(P(\text{heads}) = \frac{2}{3}\) and \(P(\text{tails}) = \frac{1}{3}\). 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 \(X\).
b. Complete the following expected value table.
| Outcome | x | P(x) | x · P(x) |
|---|---|---|---|
| WIN | 10 | \(\frac{1}{3}\) | |
| LOSE | \(-\frac{12}{3}\) |
c. What is the expected value, \(\mu\)? Do you come out ahead?
Solution
a. \(X\) = the amount of profit, in dollars, from one toss.
b. You win $10 on a tail, which has probability \(\frac{1}{3}\). You pay $6 on a head, which has probability \(\frac{2}{3}\), so that row's \(x\) value is \(-6\).
| Outcome | x | P(x) | x · P(x) |
|---|---|---|---|
| WIN | 10 | \(\frac{1}{3}\) | \(\frac{10}{3}\) |
| LOSE | −6 | \(\frac{2}{3}\) | \(-\frac{12}{3}\) |
c. Add the last column of the table.
$$ \mu = \frac{10}{3} - \frac{12}{3} = -\frac{2}{3} $$Answer: \(\mu = -\frac{2}{3}\). You lose, on average, about 67 cents each time you play the game, so you do not come out ahead — which is exactly why Nora is happy to keep it at her booth.
4.2.5 Building the Standard Deviation Column
Like data, probability distributions have standard deviations. To calculate the standard deviation \(\sigma\) 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.
To see the calculation laid out step by step, look again at the table for the number of days per week the soccer team plays. Its expected value is \(\mu = 1.1\), which we found in Example 4.2.1. To find the standard deviation, add the entries in the column labeled \((x - \mu)^2 P(x)\) and take the square root.
| x | P(x) | x · P(x) | (x − μ)2 · P(x) |
|---|---|---|---|
| 0 | 0.2 | (0)(0.2) = 0 | \((0 - 1.1)^2 (0.2) = 0.242\) |
| 1 | 0.5 | (1)(0.5) = 0.5 | \((1 - 1.1)^2 (0.5) = 0.005\) |
| 2 | 0.3 | (2)(0.3) = 0.6 | \((2 - 1.1)^2 (0.3) = 0.243\) |
Add the last column in the table: \(0.242 + 0.005 + 0.243 = 0.490\). The standard deviation is the square root of 0.49, so \(\sigma = \sqrt{0.49} = 0.7\).
Generally, for probability distributions we use a calculator or a computer to calculate \(\mu\) and \(\sigma\), which reduces round-off error. For some probability distributions there are short-cut formulas for calculating \(\mu\) and \(\sigma\), and you will meet several of them in the sections that follow.
Two stacks of cards are placed on a table. Each stack has cards numbered 1 to 8. One card from each stack is picked up. Let \(X\) = the number of cards showing an odd number. Construct a table and calculate the mean \(\mu\) and standard deviation \(\sigma\) of \(X\).
Solution
Step 1 — Find the probabilities. Each stack has four odd cards (1, 3, 5, 7) out of eight, so the probability that one card is odd is \(\frac{4}{8} = \frac{1}{2}\). The two picks are independent, so
$$ P(0) = \left(\tfrac{1}{2}\right)\left(\tfrac{1}{2}\right) = \tfrac{1}{4}, \qquad P(1) = 2\left(\tfrac{1}{2}\right)\left(\tfrac{1}{2}\right) = \tfrac{1}{2}, \qquad P(2) = \left(\tfrac{1}{2}\right)\left(\tfrac{1}{2}\right) = \tfrac{1}{4} $$(There are two ways to get exactly one odd card: odd-then-even, or even-then-odd.)
Step 2 — Build the table.
| x | P(x) | x · P(x) | (x − μ)2 · P(x) |
|---|---|---|---|
| 0 | \(\frac{1}{4}\) | 0 | \((0-1)^2 \cdot \frac{1}{4} = \frac{1}{4}\) |
| 1 | \(\frac{1}{2}\) | \(\frac{1}{2}\) | \((1-1)^2 \cdot \frac{1}{2} = 0\) |
| 2 | \(\frac{1}{4}\) | \(\frac{1}{2}\) | \((2-1)^2 \cdot \frac{1}{4} = \frac{1}{4}\) |
Step 3 — Add the columns.
$$ \mu = 0 + \tfrac{1}{2} + \tfrac{1}{2} = 1, \qquad \sigma = \sqrt{\tfrac{1}{4} + 0 + \tfrac{1}{4}} = \sqrt{0.5} \approx 0.7071 $$Answer: \(\mu = 1\) odd card and \(\sigma \approx 0.7071\).
Toss a fair, six-sided die twice. Let \(X\) = the number of faces that show an even number. Construct a table and calculate the mean \(\mu\) and standard deviation \(\sigma\) of \(X\).
Solution
Step 1 — Write out the sample space. Tossing one fair six-sided die twice has the same sample space as tossing two fair six-sided dice. The sample space has 36 outcomes:
| (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
| (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
| (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
| (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
| (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
Three of the six faces are even, so exactly 9 of the 36 outcomes have no even face, 18 have one, and 9 have two.
Step 2 — Use the sample space to complete the table.
| x | P(x) | x · P(x) | (x − μ)2 · P(x) |
|---|---|---|---|
| 0 | \(\frac{9}{36}\) | 0 | \((0-1)^2 \cdot \frac{9}{36} = \frac{9}{36}\) |
| 1 | \(\frac{18}{36}\) | \(\frac{18}{36}\) | \((1-1)^2 \cdot \frac{18}{36} = 0\) |
| 2 | \(\frac{9}{36}\) | \(\frac{18}{36}\) | \((2-1)^2 \cdot \frac{9}{36} = \frac{9}{36}\) |
Step 3 — Add the third column to find the expected value.
$$ \mu = \frac{36}{36} = 1 $$Use this value to complete the fourth column.
Step 4 — Add the values in the fourth column and take the square root.
$$ \sigma = \sqrt{\frac{18}{36}} \approx 0.7071 $$Answer: \(\mu = 1\) even face and \(\sigma \approx 0.7071\).
4.2.6 When the Odds Come From Data
Every game so far handed you its probabilities cleanly: a fair coin, a spinner with known sectors, a lottery with known digits. Real decisions are rarely that tidy. More often the probability is an estimate that came from data — a forecast, a survey, a historical rate — and the expected value you compute is only as good as that estimate.
The arithmetic does not change. What changes is how you report the answer: state the assumption the probability rests on, and remember that a large standard deviation means the long-run average is a poor description of any single outcome. That combination — a modest expected loss and a huge spread — is exactly the shape of most bets people find tempting.
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 1.08%. Using the same style of bet as the Iran example in this subsection, you bet that a moderate earthquake will occur in Japan during this period. If you win the bet, you win $100. If you lose the bet, you pay $10. Let \(X\) = the amount of profit from a bet. Find the mean and standard deviation of \(X\).
Solution
Step 1 — Build the table. \(P(\text{win}) = 0.0108\) and \(P(\text{lose}) = 1 - 0.0108 = 0.9892\).
| Outcome | x | P(x) | x · P(x) | (x − μ)2 · P(x) |
|---|---|---|---|---|
| win | 100 | 0.0108 | 1.08 | \((100 - (-8.812))^2 (0.0108) = 127.87\) |
| loss | −10 | 0.9892 | −9.892 | \((-10 - (-8.812))^2 (0.9892) = 1.40\) |
Step 2 — Add the \(x \cdot P(x)\) column.
$$ \mu = 1.08 + (-9.892) = -8.812 $$Step 3 — Add the deviation column and take the square root.
$$ \sigma = \sqrt{127.87 + 1.40} = \sqrt{129.27} \approx 11.37 $$Answer: \(\mu \approx -\$8.81\) and \(\sigma \approx \$11.37\). If you make this bet many times under the same conditions, you lose about $8.81 per bet on average. Japan's 1.08% is far lower than Iran's 21.42%, so the same style of bet is a much worse deal here.
Kai Nakamura studies seismic forecasts, and they want to know whether a bet on one is worth taking. 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 Iran was about 21.42%. Suppose Kai makes a bet that a moderate earthquake will occur in Iran during this period. If they win the bet, they win $50. If they lose the bet, they pay $20. Let \(X\) = the amount of profit from a bet.
$$ P(\text{win}) = P(\text{one moderate earthquake will occur}) = 21.42\% $$ $$ P(\text{loss}) = P(\text{one moderate earthquake will not occur}) = 100\% - 21.42\% = 78.58\% $$If Kai bets many times, will they come out ahead? Explain your answer in a complete sentence using numbers. What is the standard deviation of \(X\)?
Solution
Step 1 — Build the table.
| Outcome | x | P(x) | x · P(x) | (x − μ)2 · P(x) |
|---|---|---|---|---|
| win | 50 | 0.2142 | 10.71 | \((50 - (-5.006))^2 (0.2142) = 648.0964\) |
| loss | −20 | 0.7858 | −15.716 | \((-20 - (-5.006))^2 (0.7858) = 176.6636\) |
Step 2 — Add the \(x \cdot P(x)\) column.
$$ \mu = \text{Expected Value} = 10.71 + (-15.716) = -5.006 $$Step 3 — Add the deviation column and take the square root.
$$ \sigma = \sqrt{648.0964 + 176.6636} = \sqrt{824.76} \approx 28.7186 $$Answer: \(\mu \approx -\$5.01\), so if Kai makes this bet many times under the same conditions their long-term outcome is an average loss of about $5.01 per bet. They do not come out ahead. The standard deviation is about $28.72 — far larger than the average loss, which is what makes a bet like this feel winnable even though it is not.
Some of the more common discrete probability functions are binomial, geometric, hypergeometric, and Poisson. Most elementary courses do not cover the geometric, hypergeometric, and Poisson; your instructor will let you know if they wish to cover these distributions.
A probability distribution function is a pattern. You try to fit a probability problem into a pattern or distribution so you can perform the necessary calculations. These distributions are tools that make solving probability problems easier. Each distribution has its own special characteristics, and learning those characteristics is what lets you tell the distributions apart.
Problem Set 4.2
Problem 1. Complete the expected value table.
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.2 | |
| 1 | 0.2 | |
| 2 | 0.4 | |
| 3 | 0.2 |
Solution
Multiply each \(x\) by its probability to fill the third column, then add.
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.2 | (0)(0.2) = 0 |
| 1 | 0.2 | (1)(0.2) = 0.2 |
| 2 | 0.4 | (2)(0.4) = 0.8 |
| 3 | 0.2 | (3)(0.2) = 0.6 |
Answer: the completed column is 0, 0.2, 0.8, 0.6, and \(\mu = 0 + 0.2 + 0.8 + 0.6 = 1.6\).
Problem 2. Find the expected value from the expected value table.
| x | P(x) | x · P(x) |
|---|---|---|
| 2 | 0.1 | 2(0.1) = 0.2 |
| 4 | 0.3 | 4(0.3) = 1.2 |
| 6 | 0.4 | 6(0.4) = 2.4 |
| 8 | 0.2 | 8(0.2) = 1.6 |
Solution
The third column is already filled in, so just add it.
$$ \mu = 0.2 + 1.2 + 2.4 + 1.6 = 5.4 $$Answer: \(\mu = 5.4\).
Problem 3. Find the standard deviation.
| x | P(x) | x · P(x) | (x − μ)2 · P(x) |
|---|---|---|---|
| 2 | 0.1 | 2(0.1) = 0.2 | \((2 - 5.4)^2 (0.1) = 1.156\) |
| 4 | 0.3 | 4(0.3) = 1.2 | \((4 - 5.4)^2 (0.3) = 0.588\) |
| 6 | 0.4 | 6(0.4) = 2.4 | \((6 - 5.4)^2 (0.4) = 0.144\) |
| 8 | 0.2 | 8(0.2) = 1.6 | \((8 - 5.4)^2 (0.2) = 1.352\) |
Solution
The fourth column is already computed from \(\mu = 5.4\). Add it, then take the square root.
$$ 1.156 + 0.588 + 0.144 + 1.352 = 3.24 $$ $$ \sigma = \sqrt{3.24} = 1.8 $$Answer: \(\sigma = 1.8\).
Problem 4. Identify the mistake in the probability distribution table.
| x | P(x) | x · P(x) |
|---|---|---|
| 1 | 0.15 | 0.15 |
| 2 | 0.25 | 0.50 |
| 3 | 0.30 | 0.90 |
| 4 | 0.20 | 0.80 |
| 5 | 0.15 | 0.75 |
Solution
Check the two things a probability distribution table has to satisfy. Every product \(x \cdot P(x)\) is correct: \((1)(0.15) = 0.15\), \((2)(0.25) = 0.50\), \((3)(0.30) = 0.90\), \((4)(0.20) = 0.80\), \((5)(0.15) = 0.75\). So the arithmetic is not the problem.
Now add the \(P(x)\) column:
$$ 0.15 + 0.25 + 0.30 + 0.20 + 0.15 = 1.05 $$Answer: the probabilities sum to 1.05, not 1. For a valid probability distribution the \(P(x)\) column must total exactly 1, so at least one of the listed probabilities is wrong.
Problem 5. Identify the mistake in the probability distribution table.
| x | P(x) | x · P(x) |
|---|---|---|
| 1 | 0.15 | 0.15 |
| 2 | 0.25 | 0.40 |
| 3 | 0.25 | 0.65 |
| 4 | 0.20 | 0.85 |
| 5 | 0.15 | 1 |
Problems 4.2.6–4.2.10: Professor Amara Okonkwo wants to know what percent of her physics majors will spend the next several years doing post-graduate research. She has the following probability distribution.
| x | P(x) | x · P(x) |
|---|---|---|
| 1 | 0.35 | |
| 2 | 0.20 | |
| 3 | 0.15 | |
| 4 | ||
| 5 | 0.10 | |
| 6 | 0.05 |
Solution
This time the \(P(x)\) column is fine: \(0.15 + 0.25 + 0.25 + 0.20 + 0.15 = 1\).
The third column is the problem. Its entries are 0.15, 0.40, 0.65, 0.85, 1 — those are the running totals of \(P(x)\) (\(0.15\), then \(0.15 + 0.25 = 0.40\), then \(0.40 + 0.25 = 0.65\), and so on), not the products \(x \cdot P(x)\).
The correct products are \((1)(0.15) = 0.15\), \((2)(0.25) = 0.50\), \((3)(0.25) = 0.75\), \((4)(0.20) = 0.80\), \((5)(0.15) = 0.75\).
Answer: the third column holds the cumulative sum of \(P(x)\) instead of \(x \cdot P(x)\). With the correct products, \(\mu = 0.15 + 0.50 + 0.75 + 0.80 + 0.75 = 2.95\).
Problem 6. Define the random variable \(X\).
Solution
Answer: \(X\) = the number of years a physics major spends doing post-graduate research.
Problem 7. Define \(P(x)\), or the probability of \(x\).
Solution
Answer: \(P(x)\) = the probability that a physics major spends exactly \(x\) years doing post-graduate research.
Problem 8. Find the probability that a physics major will do post-graduate research for four years. \(P(x = 4) = \) ______
Solution
The probabilities in the table have to total 1, so the missing entry is whatever is left over.
$$ 0.35 + 0.20 + 0.15 + 0.10 + 0.05 = 0.85 $$ $$ P(x = 4) = 1 - 0.85 = 0.15 $$Answer: \(P(x = 4) = 0.15\).
Problem 9. Find the probability that a physics major will do post-graduate research for at most three years. \(P(x \le 3) = \) ______
Solution
"At most three years" means 1, 2, or 3 years, so add those three probabilities.
$$ P(x \le 3) = 0.35 + 0.20 + 0.15 = 0.70 $$Answer: \(P(x \le 3) = 0.70\).
Problem 10. On average, how many years would you expect a physics major to spend doing post-graduate research?
Problems 4.2.11–4.2.17: Rosa Delgado, 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 \(X\) = the number of years a student will study ballet with her, and let \(P(x)\) = the probability that a student will study ballet \(x\) years.
| x | P(x) | x · P(x) |
|---|---|---|
| 1 | 0.10 | |
| 2 | 0.05 | |
| 3 | 0.10 | |
| 4 | ||
| 5 | 0.30 | |
| 6 | 0.20 | |
| 7 | 0.10 |
Solution
Use \(P(x = 4) = 0.15\) from the earlier problem, then build the \(x \cdot P(x)\) column and add it.
$$ \mu = 1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) $$ $$ \mu = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.60 $$Answer: \(\mu = 2.6\) years of post-graduate research, on average.
Problem 11. Complete the table above using the data provided.
Solution
Step 1 — Find the missing probability. The given entries total \(0.10 + 0.05 + 0.10 + 0.30 + 0.20 + 0.10 = 0.85\), so \(P(x = 4) = 1 - 0.85 = 0.15\).
Step 2 — Fill in the \(x \cdot P(x)\) column.
| x | P(x) | x · P(x) |
|---|---|---|
| 1 | 0.10 | 0.10 |
| 2 | 0.05 | 0.10 |
| 3 | 0.10 | 0.30 |
| 4 | 0.15 | 0.60 |
| 5 | 0.30 | 1.50 |
| 6 | 0.20 | 1.20 |
| 7 | 0.10 | 0.70 |
Answer: the missing probability is 0.15, and the \(x \cdot P(x)\) column totals 4.50.
Problem 12. In words, define the random variable \(X\).
Solution
Answer: \(X\) = the number of years a student studies ballet with this instructor.
Problem 13. \(P(x = 4) = \) ______
Solution
The table's probabilities must total 1, and the listed ones total 0.85.
Answer: \(P(x = 4) = 1 - 0.85 = 0.15\).
Problem 14. \(P(x < 4) = \) ______
Solution
"Fewer than 4" means 1, 2, or 3 years. Note that \(x = 4\) is not included.
$$ P(x < 4) = 0.10 + 0.05 + 0.10 = 0.25 $$Answer: \(P(x < 4) = 0.25\).
Problem 15. On average, how many years would you expect a child to study ballet with this teacher?
Solution
Add the \(x \cdot P(x)\) column built earlier.
$$ \mu = 0.10 + 0.10 + 0.30 + 0.60 + 1.50 + 1.20 + 0.70 = 4.50 $$Answer: \(\mu = 4.5\) years. A student stays with this instructor an average of four and a half years.
Problem 16. What does the column \(P(x)\) sum to, and why?
Solution
Answer: the \(P(x)\) column sums to 1. The values 1 through 7 are all the outcomes that can happen, exactly one of them must happen, and the probability of something in the whole list happening is 1.
Problem 17. What does the column \(x \cdot P(x)\) sum to, and why?
Solution
Answer: the \(x \cdot P(x)\) column sums to the expected value, \(\mu = 4.5\). That sum is the definition of expected value — each outcome weighted by how likely it is.
Problem 18. 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 $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. What is the expected value of playing the game?
Solution
Step 1 — Find the probabilities. There are 12 face cards in 52, so \(P(\text{face card}) = \frac{12}{52}\) and \(P(\text{not a face card}) = \frac{40}{52}\).
Step 2 — Set the profit values. A face card wins $30, so \(x = 30\). Anything else costs $2, so \(x = -2\).
Step 3 — Build and add the table.
$$ \mu = 30\left(\frac{12}{52}\right) + (-2)\left(\frac{40}{52}\right) = \frac{360 - 80}{52} = \frac{280}{52} \approx 5.38 $$Answer: \(\mu \approx \$5.38\) per play.
Problem 19. 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 $30. If it is not a face card, you pay $2. There are 12 face cards in a deck of 52 cards. Should you play the game?
Solution
Answer: yes. The expected value is about $5.38 per play, which is positive, so over the long run this game pays you an average of $5.38 every time you draw.
Problem 20. A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 apiece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of $150.
a) What are you interested in here?
b) In words, define the random variable \(X\).
c) List the values that \(X\) may take on.
d) Construct a PDF.
e) If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle?
Solution
a. You are interested in your long-term average winnings per raffle — the expected value of your net gain.
b. \(X\) = your net gain, in dollars, from buying four tickets in one raffle.
c. You spend \(4 \times \$5 = \$20\). If one of your tickets wins you receive the $150 prize, so your net gain is \(150 - 20 = 130\). Otherwise you are out the $20. So \(X\) takes on the values 130 and \(-20\).
d. You hold 4 of the 100 tickets, so \(P(\text{win}) = \frac{4}{100} = 0.04\) and \(P(\text{lose}) = 0.96\).
| x | P(x) | x · P(x) |
|---|---|---|
| 130 | 0.04 | 5.20 |
| −20 | 0.96 | −19.20 |
e.
$$ \mu = 5.20 + (-19.20) = -14 $$Answer: your expected average winnings are \(-\$14\) — that is, an average loss of $14 per raffle.
Problem 21. 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 $6.
- If the card is a face card, and the coin lands on Tails, you win $2.
- If the card is not a face card, you lose $2, no matter what the coin shows.
a) Find the expected value for this game (expected net gain or loss).
b) Explain what your calculations indicate about your long-term average profits and losses on this game.
c) Should you play this game to win money?
Solution
a. The variable of interest is \(X\), the gain or loss in dollars. The face cards are the jack, queen, and king, so there are \((3)(4) = 12\) face cards and \(52 - 12 = 40\) cards that are not face cards. Build the distribution from the card-and-coin events, but use the dollar values for \(x\).
| Card event | x (net gain/loss) | P(x) |
|---|---|---|
| Face card and Heads | 6 | \(\left(\frac{12}{52}\right)\left(\frac{1}{2}\right) = \frac{6}{52}\) |
| Face card and Tails | 2 | \(\left(\frac{12}{52}\right)\left(\frac{1}{2}\right) = \frac{6}{52}\) |
| Not a face card, either coin face | −2 | \(\left(\frac{40}{52}\right)(1) = \frac{40}{52}\) |
The expected value is \(-\$0.62\), rounded to the nearest cent.
b. If you play this game repeatedly, over a long string of games you would expect to lose about 62 cents per game, on average.
c. No. You should not play this game to win money, because the expected value is an expected average loss.
Problem 22. You buy a lottery ticket to a lottery that costs $10 per ticket. There are only 100 tickets available to be sold in this lottery. In this lottery there are one $500 prize, two $100 prizes, and four $25 prizes. Find your expected gain or loss.
Solution
Step 1 — Work out each net gain. You pay $10 for the ticket, so subtract 10 from every prize.
| x (net gain) | P(x) | x · P(x) |
|---|---|---|
| 490 | \(\frac{1}{100} = 0.01\) | 4.90 |
| 90 | \(\frac{2}{100} = 0.02\) | 1.80 |
| 15 | \(\frac{4}{100} = 0.04\) | 0.60 |
| −10 | \(\frac{93}{100} = 0.93\) | −9.30 |
Step 2 — Add the last column.
$$ \mu = 4.90 + 1.80 + 0.60 - 9.30 = -2.00 $$Answer: you expect to lose $2 per ticket. Sanity check: the lottery pays out \(500 + 2(100) + 4(25) = \$800\) and takes in \(100 \times \$10 = \$1{,}000\), so the $200 it keeps, spread over 100 tickets, is exactly $2 a ticket.
Problem 23. Complete the PDF and answer the questions.
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.3 | |
| 1 | 0.2 | |
| 2 | ||
| 3 | 0.4 |
a) Find the probability that \(X = 2\).
b) Find the expected value.
Solution
a. The probabilities must total 1, and the given entries total \(0.3 + 0.2 + 0.4 = 0.9\), so \(P(X = 2) = 1 - 0.9 = 0.1\).
b. Fill in the \(x \cdot P(x)\) column and add it.
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.3 | 0 |
| 1 | 0.2 | 0.2 |
| 2 | 0.1 | 0.2 |
| 3 | 0.4 | 1.2 |
Answer: \(P(X = 2) = 0.1\) and \(\mu = 1.6\).
Problem 24. Suppose that you are offered the following "deal." You roll a die. If you roll a six, you win $10. If you roll a four or five, you win $5. If you roll a one, two, or three, you pay $6.
a) What are you ultimately interested in here (the value of the roll or the money you win)?
b) In words, define the random variable \(X\).
c) List the values that \(X\) may take on.
d) Construct a PDF.
e) Over the long run of playing this game, what are your expected average winnings per game?
f) Based on numerical values, should you take the deal? Explain your decision in complete sentences.
Solution
a. The money you win. The number on the die is only the mechanism that decides the payout.
b. \(X\) = your net gain, in dollars, from one roll.
c. \(X\) takes on the values 10, 5, and \(-6\).
d. A six has probability \(\frac{1}{6}\); a four or five has probability \(\frac{2}{6} = \frac{1}{3}\); a one, two, or three has probability \(\frac{3}{6} = \frac{1}{2}\).
| x | P(x) | x · P(x) |
|---|---|---|
| 10 | \(\frac{1}{6}\) | \(\frac{10}{6}\) |
| 5 | \(\frac{1}{3}\) | \(\frac{10}{6}\) |
| −6 | \(\frac{1}{2}\) | \(-\frac{18}{6}\) |
e.
$$ \mu = \frac{10}{6} + \frac{10}{6} - \frac{18}{6} = \frac{2}{6} = \frac{1}{3} \approx 0.33 $$f. Yes, take the deal. The expected value is about $0.33 per roll, which is positive, so over many rolls you gain roughly 33 cents each time you play. Any single roll can still cost you $6 — the 33 cents is only the long-run average.
Problem 25. Priya Raman, a venture capitalist willing to invest $1,000,000, has three investments she can choose from. The first investment, a software company, has a 10% chance of returning $5,000,000 profit, a 30% chance of returning $1,000,000 profit, and a 60% chance of losing the million dollars. The second company, a hardware company, has a 20% chance of returning $3,000,000 profit, a 40% chance of returning $1,000,000 profit, and a 40% chance of losing the million dollars. The third company, a biotech firm, has a 10% chance of returning $6,000,000 profit, a 70% chance of no profit or loss, and a 20% chance of losing the million dollars.
a) Construct a PDF for each investment.
b) Find the expected value for each investment.
c) Which is the safest investment? Why do you think so?
d) Which is the riskiest investment? Why do you think so?
e) Which investment has the highest expected return, on average?
Solution
a. Each company gets its own PDF, with \(x\) measured in dollars of profit.
| Software company: x | P(x) |
|---|---|
| 5,000,000 | 0.10 |
| 1,000,000 | 0.30 |
| −1,000,000 | 0.60 |
| Hardware company: x | P(x) |
|---|---|
| 3,000,000 | 0.20 |
| 1,000,000 | 0.40 |
| −1,000,000 | 0.40 |
| Biotech firm: x | P(x) |
|---|---|
| 6,000,000 | 0.10 |
| 0 | 0.70 |
| −1,000,000 | 0.20 |
b.
$$ \mu_{\text{software}} = 500{,}000 + 300{,}000 - 600{,}000 = 200{,}000 $$ $$ \mu_{\text{hardware}} = 600{,}000 + 400{,}000 - 400{,}000 = 600{,}000 $$ $$ \mu_{\text{biotech}} = 600{,}000 + 0 - 200{,}000 = 400{,}000 $$So the expected values are $200,000, $600,000, and $400,000.
c. The third investment, the biotech firm — it has the lowest probability of losing the million dollars (0.20).
d. The first investment, the software company — it has the highest probability of loss (0.60).
e. The second investment, the hardware company, at $600,000 expected profit.
Problem 26. Suppose that 20,000 married adults in the United States were randomly surveyed as to the number of children they have. The results are compiled and are used as theoretical probabilities. Let \(X\) = the number of children married people have.
| x | P(x) | x · P(x) |
|---|---|---|
| 0 | 0.10 | |
| 1 | 0.20 | |
| 2 | 0.30 | |
| 3 | ||
| 4 | 0.10 | |
| 5 | 0.05 | |
| 6 (or more) | 0.05 |
a) Find the probability that a married adult has three children.
b) In words, what does the expected value in this example represent?
c) Find the expected value.
d) Is it more likely that a married adult will have two to three children or four to six children? How do you know?
Solution
a. The listed probabilities total \(0.10 + 0.20 + 0.30 + 0.10 + 0.05 + 0.05 = 0.80\), so \(P(x = 3) = 1 - 0.80 = 0.20\).
b. It represents the average number of children a married adult in the United States has, taken over the whole surveyed population.
c. Treating "6 or more" as 6:
$$ \mu = 0(0.10) + 1(0.20) + 2(0.30) + 3(0.20) + 4(0.10) + 5(0.05) + 6(0.05) $$ $$ \mu = 0 + 0.20 + 0.60 + 0.60 + 0.40 + 0.25 + 0.30 = 2.35 $$d. \(P(2 \text{ or } 3) = 0.30 + 0.20 = 0.50\), while \(P(4 \text{ to } 6) = 0.10 + 0.05 + 0.05 = 0.20\). Two to three children is more likely, because 0.50 is larger than 0.20.
Problem 27. Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given as follows. On average, how many years do you expect it to take for an individual to earn a B.S.?
| x | P(x) |
|---|---|
| 3 | 0.05 |
| 4 | 0.40 |
| 5 | 0.30 |
| 6 | 0.15 |
| 7 | 0.10 |
Solution
Build the \(x \cdot P(x)\) column and add it.
$$ \mu = 3(0.05) + 4(0.40) + 5(0.30) + 6(0.15) + 7(0.10) $$ $$ \mu = 0.15 + 1.60 + 1.50 + 0.90 + 0.70 = 4.85 $$Answer: \(\mu = 4.85\) years.
Problem 28. People visiting video game rental stores often rent more than one game at a time. The probability distribution for game rentals per customer at Game Stop is given in the following table. There is a five-game limit per customer at this store, so nobody ever rents more than five video games.
| x | P(x) |
|---|---|
| 0 | 0.03 |
| 1 | 0.50 |
| 2 | 0.24 |
| 3 | |
| 4 | 0.07 |
| 5 | 0.04 |
a) Describe the random variable \(X\) in words.
b) Find the probability that a customer rents three video games.
c) Find the probability that a customer rents at least four video games.
d) Find the probability that a customer rents at most two video games.
Another shop, Games Galore, also rents video games. The probability distribution for video game rentals per customer at this shop is given as follows. Games Galore also has a five-game limit per customer.
| x | P(x) |
|---|---|
| 0 | 0.35 |
| 1 | 0.25 |
| 2 | 0.20 |
| 3 | 0.10 |
| 4 | 0.05 |
| 5 | 0.05 |
e) At which store is the expected number of video games rented per customer higher?
f) If Game Stop estimates that they will have 300 customers next week, how many video games do they expect to rent next week? Answer in sentence form.
g) If Game Stop expects 300 customers next week, and Games Galore projects that they will have 420 customers, for which store is the expected number of video game rentals for next week higher? Explain.
h) Which of the two stores experiences more variation in the number of video games per customer? How do you know that?
Solution
a. \(X\) = the number of video games a customer rents on one visit to Game Stop, which can be 0, 1, 2, 3, 4, or 5.
b. The listed probabilities total \(0.03 + 0.50 + 0.24 + 0.07 + 0.04 = 0.88\), so \(P(x = 3) = 1 - 0.88 = 0.12\).
c. \(P(x \ge 4) = 0.07 + 0.04 = 0.11\).
d. \(P(x \le 2) = 0.03 + 0.50 + 0.24 = 0.77\).
e. Compute both expected values.
$$ \mu_{\text{Game Stop}} = 0(0.03) + 1(0.50) + 2(0.24) + 3(0.12) + 4(0.07) + 5(0.04) = 1.82 $$ $$ \mu_{\text{Games Galore}} = 0(0.35) + 1(0.25) + 2(0.20) + 3(0.10) + 4(0.05) + 5(0.05) = 1.40 $$Game Stop is higher, at 1.82 games per customer against 1.40.
f. \(300 \times 1.82 = 546\). Game Stop expects to rent about 546 video games next week.
g. Games Galore expects \(420 \times 1.40 = 588\) rentals against Game Stop's 546, so Games Galore's projected total is higher. Its per-customer average is lower, but it expects 120 more customers, and that more than makes up the difference.
h. Compare the standard deviations.
$$ \sigma_{\text{Game Stop}} = \sqrt{1.3476} \approx 1.16 $$ $$ \sigma_{\text{Games Galore}} = \sqrt{2.04} \approx 1.43 $$Games Galore experiences more variation. Its probabilities are spread more evenly across 0 through 5, while half of Game Stop's customers rent exactly one game, which packs that distribution tightly around its mean.
Problem 29. A "friend," Tomás Herrera, offers you the following "deal." For his $10 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 $6.
- Eighty of the coupons are for a free gift worth $8.
- Six of the coupons are for a free gift worth $12.
- Four of the coupons are for a free gift worth $40.
Based upon the financial gain or loss over the long run, should you play the game?
a) Yes, I expect to come out ahead in money.
b) No, I expect to come out behind in money.
c) It doesn't matter. I expect to break even.
Solution
Step 1 — Turn each coupon into a net gain. You pay $10 up front, so subtract 10 from every gift value.
| x (net gain) | P(x) | x · P(x) |
|---|---|---|
| −4 | 0.10 | −0.40 |
| −2 | 0.80 | −1.60 |
| 2 | 0.06 | 0.12 |
| 30 | 0.04 | 1.20 |
Step 2 — Add the last column.
$$ \mu = -0.40 - 1.60 + 0.12 + 1.20 = -0.68 $$Answer: b. No, you expect to come out behind in money — about 68 cents behind per envelope.
Problem 30. Florida State University has 14 statistics classes scheduled for its summer term. One class has space available for 30 students, eight classes have space for 60 students, one class has space for 70 students, and four classes have space for 100 students.
a) What is the average class size assuming each class is filled to capacity?
b) Space is available for 980 students. Suppose that each class is filled to capacity and select a statistics student at random. Let the random variable \(X\) equal the size of the student's class. Define the PDF for \(X\).
c) Find the mean of \(X\).
d) Find the standard deviation of \(X\).
Solution
a. Total capacity is \(30 + 8(60) + 70 + 4(100) = 30 + 480 + 70 + 400 = 980\) seats across 14 classes.
$$ \text{average class size} = \frac{980}{14} = 70 $$b. Now pick a student at random, not a class. A student is more likely to land in a big class simply because big classes hold more students, so weight each size by how many seats it accounts for.
| x (class size) | P(x) |
|---|---|
| 30 | \(\frac{30}{980} \approx 0.0306\) |
| 60 | \(\frac{480}{980} \approx 0.4898\) |
| 70 | \(\frac{70}{980} \approx 0.0714\) |
| 100 | \(\frac{400}{980} \approx 0.4082\) |
c.
$$ \mu = \frac{30(30) + 60(480) + 70(70) + 100(400)}{980} = \frac{74{,}600}{980} \approx 76.12 $$d. Using \(\mu \approx 76.12\):
$$ \sigma = \sqrt{(30 - 76.12)^2(0.0306) + (60 - 76.12)^2(0.4898) + (70 - 76.12)^2(0.0714) + (100 - 76.12)^2(0.4082)} $$ $$ \sigma = \sqrt{65.12 + 127.31 + 2.68 + 232.71} = \sqrt{427.82} \approx 20.68 $$Answer: the average class size is 70, but the average class size experienced by a student is about 76.12 with a standard deviation of about 20.68. The gap between 70 and 76.12 is the whole point of the problem: averaging over classes and averaging over students are different questions.
Problem 31. In a lottery, there are 250 prizes of $5, 50 prizes of $25, and ten prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to charge to break even?
Solution
Step 1 — Find the probability of each prize. There are \(250 + 50 + 10 = 310\) winning tickets among 10,000, so \(10{,}000 - 310 = 9{,}690\) tickets win nothing.
| x (prize won) | P(x) | x · P(x) |
|---|---|---|
| 0 | \(\frac{9{,}690}{10{,}000} = 0.969\) | 0 |
| 5 | \(\frac{250}{10{,}000} = 0.025\) | 0.125 |
| 25 | \(\frac{50}{10{,}000} = 0.005\) | 0.125 |
| 100 | \(\frac{10}{10{,}000} = 0.001\) | 0.100 |
Step 2 — Add the last column.
$$ \mu = 0 + 0.125 + 0.125 + 0.100 = 0.35 $$Answer: the expected winnings per ticket are $0.35, so a fair price — one that makes the lottery break even — is 35 cents per ticket. Sanity check: total prize money is \(250(5) + 50(25) + 10(100) = \$3{,}500\), spread over 10,000 tickets, which is $0.35 each.
Key Terms
expected value — the long-term average of a random variable, computed as \(\mu = \sum (x \cdot P(x))\); also called the mean of the distribution.
mean of a discrete random variable (\(\mu\)) — another name for the expected value; the balance point of the probability distribution.
Law of Large Numbers — as the number of trials increases, the relative frequency of an event approaches its theoretical probability.
expected value table — a table with one row per value of \(X\) and columns for \(x\), \(P(x)\), and \(x \cdot P(x)\), used to organize the expected-value calculation.
standard deviation of a probability distribution (\(\sigma\)) — the square root of \(\sum \left[(x - \mu)^2 P(x)\right]\); it measures how far outcomes typically fall from the expected value.
fair game — a game whose expected profit is zero, so neither player gains or loses money over the long run.