Chapter 4: Discrete Random Variables
4.2 Measures of General Discrete Random Variables
Learning Objectives
By the end of this section, the student should be able to:
- Calculate and interpret expected values of general random variables
- Calculate and interpret the variance and standard deviation of general random variables
Once we know how to work with Discrete Random Variables we may be interested in some other measures such as the mean, variance, and standard deviation. The ideas here are slightly different than we have seen before within our new context of Random Variables.
The Expected Value (Mean) of a Discrete Random Variable
The Law of Large Numbers states: as the number of trials in a probability experiment increases, our results become closer to what we “expect.” For instance, the student who was guessing on the true-false quiz in the chapter introduction would expect to get about half of the questions correct, since there are two options.
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 random variable and is denoted by the Greek letter [latex]\mu[/latex], or in the context of random variables, [latex]E[X][/latex]. In other words, after conducting many trials of an experiment, you would expect this average value.
To find the expected value, we multiply each value of the random variable by its probability, then add the products.
Mean or Expected Value: [latex]\mu={\sum\limits_{x \in X} }^{\text{}}xP\left(x\right).[/latex]
Example
A university 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, [latex]\mu[/latex], of the number of days per week the team plays soccer.
We first let the random variable [latex]X =[/latex] the number of days the team plays soccer per week. Then [latex]x[/latex] takes on the values 0, 1, 2. Construct a PDF table adding a column [latex]x \cdot P(x)[/latex]. In this column, you multiply each value by its probability.
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x \cdot P(x)[/latex] |
|---|---|---|
| 0 | 0.2 | (0)(0.2) = 0 |
| 1 | 0.5 | (1)(0.5) = 0.5 |
| 2 | 0.3 | (2)(0.3) = 0.6 |
What is the expected value?
Solution
Add the last column [latex]xP(x)[/latex] to find the long term average or expected value: [latex](0)(0.2)+(1)(0.5)+(2)(0.3)=0+0.5+0.6=1.1[/latex].
The expected value is 1.1. The university soccer team would, on the average, expect to play soccer 1.1 days per week. The number 1.1 is the long-term average or expected value if the soccer team plays soccer week after week after week. We say [latex]\mu = 1.1[/latex].
Your Turn!
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?
| [latex]x[/latex] | [latex]P(x)[/latex] |
|---|---|
| 0 | [latex]P(0)=\frac{4}{50}[/latex] |
| 1 | [latex]P(1) = \frac{8}{50}[/latex] |
| 2 | [latex]P(2) = \frac{16}{50}[/latex] |
| 3 | [latex]P(3) = \frac{14}{50}[/latex] |
| 4 | [latex]P(4) = \frac{6}{50}[/latex] |
| 5 | [latex]P(5) = \frac{2}{50}[/latex] |
Solution
Since [latex]0 \cdot \frac{4}{50} + 1 \cdot \frac{8}{50} + 2 \cdot \frac{16}{50} + 3 \cdot \frac{14}{50} + 4 \cdot \frac{6}{50} + 5 \cdot \frac{2}{50}=2.32[/latex], the expected value is 2.32. This means that over many days and weeks in the hospital, post-op patients will ring the nurse an average of 2.32 times per 12-hour shift.
We write the mean as [latex]\mu=2.32[/latex], and when referring to the random variable [latex]X=[/latex] the number of times the average post-op patient will ring the nurse during a 12-hour shift, we write [latex]E[X]=2.32[/latex].
The Variance and Standard Deviation of a Discrete Random Variable
Like data, probability distributions have standard deviations. The standard deviation ([latex]\sigma[/latex]) of a probability distribution is the sum of the squares of the difference between its outcomes and expected value weighted by the probability of the outcome.
Computing the variance, [latex]\sigma^2[/latex] or [latex]V[X][/latex], and standard deviation, [latex]\sigma[/latex] or [latex]SD[X][/latex]. of a random variable starts similar to what we have seen before for data but differs at step 4:
- Find the mean [latex]\mu[/latex].
- Subtract the mean from each value of x to get your deviations.
- Square each deviation.
- Multiply each squared deviation by its probability, [latex]P(x)[/latex].
- Sum each of the products.
At this point you now have the variance then can of course take the square root of the variance to get your standard deviation. The formula looks like this:
[latex]\sigma = \sqrt{\underset{x\in X}{{\sum }^{\text{}}}{\left(x-\mu \right)}^{2}P\left(x\right)}[/latex]
Example
Find the expected value of the number of times a newborn baby’s crying wakes one of their parents after midnight. The expected value is the expected number of times per week a newborn baby’s crying wakes one of their parents after midnight. Calculate the standard deviation of the variable as well.
a. Add the values in the third column of the table to find the expected value of [latex]X[/latex].
Solution
[latex]\mu = \text{Expected Value} = \frac{105}{50} = 2.1[/latex]
b. Use [latex]\mu[/latex] to complete the table. The fourth column of this table will provide the values you need to calculate 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].)
c. Add the values in the fourth column of the table:
Solution
[latex]0.1764 + 0.2662 + 0.0046 + 0.1458 + 0.2888 + 0.1682 = 1.05[/latex]
d. The standard deviation of [latex]X[/latex] is the square root of this sum.
Solution
[latex]\sigma = \sqrt{1.05} \approx 1.0247[/latex]
e. The mean, [latex]\mu[/latex], of a discrete probability function is the expected value.
Solution
[latex]\mu = \Sigma (x \cdot P(x))[/latex]
f. The standard deviation, [latex]\Sigma[/latex], of the PDF is the square root of the variance.
Solution
[latex]\sigma = \sqrt{\Sigma \left[ (x - \mu)^2 \cdot P(x) \right]}[/latex]
When all outcomes in the probability distribution are equally likely, these formulas coincide with the mean and standard deviation of the set of possible outcomes.
Your Turn!
Suppose you are booking a trip two days away that costs $1,000 and choose to purchase travel insurance for an additional $30. If you cannot go on your trip due to serious illness, you will receive back your $1,000 and have spent $30. Suppose the probability of getting ill in the next 48 hours is about 1.08%. Let [latex]X =[/latex] the cost of booking the trip. Find the mean and standard deviation of [latex]X[/latex].
If you travel at this cost frequently and always purchase travel insurance, will you save or lose money vs. if you did not purchase travel insurance?
Solution
We have two potential costs: $1030 if we are able to go on the trip, and $30 if we are not. The probability of not being able to travel is 0.0108, so the probability of going is [latex]1−0.0108=0.9892[/latex].
Therefore the expected value is [latex]E(X)= \mu =1030 \cdot 0.9892+30 \cdot 0.0108=1019.2[/latex]. This means that if we frequently took trips at these prices and always bought travel insurance, we would pay $1019.20 per trip, more than the $1000 cost of the trip.
The variance is [latex]V(X)=\sigma^2=(1030−1019.2)^2 \cdot 0.9892+(30−1019.2)^2 \cdot 0.0108 = 10683.36[/latex],
so the standard deviation is [latex]SD(X)= \sigma = \sqrt{10683.36} \approx 103.36[/latex].
Note
Generally for probability distributions, we use a calculator or a computer to calculate [latex]\mu[/latex] and [latex]\sigma[/latex] to reduce roundoff error. For many special cases of probability distributions, there are short-cut formulas for calculating [latex]\mu[/latex], [latex]\sigma[/latex], and associated probabilities. We will see some of these in the future.
Your Turn!
Toss a fair, six-sided die twice. Let X = the number of faces that show an even number. Construct a table like Figure 4.6 and calculate the mean [latex]\mu[/latex] and standard deviation [latex]\sigma[/latex] of [latex]X[/latex].
Solution
Since we are tossing twice, the possible values of the number of even faces are [latex]x=0,1,2[/latex]. Since each die has 6 sides, there are 6⋅6=36 possible outcomes, such as rolling two 1’s, rolling a 1 and then a 2, and so on. The outcomes with no even faces are (1,1), (1,3), (1,5), (3,1), (3,3), (3,5), (5,1), (5,3), (5,5), for a total of 9. Similarly, there 9 results with two even faces. So there must be 36−9−9=18 outcomes with one even face. Since 936=14 and 1836=12, we get the following table.
| [latex]x[/latex] | [latex]P(x)[/latex] | [latex]x \cdot P(x)[/latex] | [latex](x - \mu)^2 \cdot P(x)[/latex] |
| 0 | [latex]\frac{1}{4}[/latex] | 0 | [latex](0-1)^2 \cdot \frac{1}{4} = \frac{1}{4}[/latex] |
| 1 | [latex]\frac{1}{2}[/latex] | [latex]\frac{1}{2}[/latex] | [latex](1-1)^2 \cdot \frac{1}{2} = 0[/latex] |
| 2 | [latex]\frac{1}{4}[/latex] | [latex]\frac{1}{2}[/latex] | [latex](2-1)^2 \cdot \frac{1}{4} = \frac{1}{4}[/latex] |
Therefore [latex]\mu = 0 + \frac{1}{2} + \frac{1}{2} = 1[/latex] and [latex]\sigma = \sqrt{\frac{1}{4}+0+\frac{1}{4}} = \sqrt{2}[/latex].
Section 4.2 Review
The expected value, or mean, of a discrete random variable predicts the long-term results of a statistical experiment that has been repeated many times. The standard deviation of a probability distribution is used to measure the variability of possible outcomes.
Formula Review
Mean or Expected Value: [latex]\mu =\sum\limits_{x\in X} xP\left(x\right)[/latex]
Standard Deviation: [latex]\sigma =\sqrt{\sum\limits_{x\in X}{\left(x-\mu \right)}^{2}P\left(x\right)}[/latex]
Section 4.2 Practice
Complete the expected value table.
| x | P(x) | x*P(x) |
|---|---|---|
| 0 | 0.2 | |
| 1 | 0.2 | |
| 2 | 0.4 | |
| 3 | 0.2 |
Solution
| 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 |
2. Find the expected value from the expected value table.
Solution
[latex]0.2 + 1.2 + 2.4 + 1.6 = 5.4[/latex]
3. Find the standard deviation.
| x | P(x) | x*P(x) | (x – μ)2P(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
[latex]\sigma = 1.156+0.588+0.144+1.352 = 3.24[/latex]
[latex]\sqrt{3.24}=1.8[/latex]
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
The values of [latex]P(x)[/latex] do not sum to one.
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 |
Solution
The values of [latex]x*P(x)[/latex] are not correct.
Use the following information to answer the next five exercises: A physics professor wants to know what percent of physics majors will spend the next several years doing post-graduate research. He 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 |
1. Define the random variable [latex]X[/latex].
Solution
Let [latex]X =[/latex] the number of years a physics major will spend doing post-graduate research.
2. Define [latex]P(x)[/latex], or the probability of [latex]x[/latex].
Solution
Let [latex]P(x) =[/latex] the probability that a physics major will do post-graduate research for [latex]x[/latex] years.
3. Find the probability that a physics major will do post-graduate research for four years. [latex]P(x = 4) = \underline{\hspace{2cm}}[/latex]
Solution
[latex]1 – 0.35 – 0.20 – 0.15 – 0.10 – 0.05 = 0.15[/latex]
4. Find the probability that a physics major will do post-graduate research for at most three years. [latex]P(x \le 3) =\underline{\hspace{2cm}}[/latex]
Solution
[latex]0.35 + 0.20 + 0.15 = 0.70[/latex]
5. On average, how many years would you expect a physics major to spend doing post-graduate research?
Solution
[latex]1(0.35) + 2(0.20) + 3(0.15) + 4(0.15) + 5(0.10) + 6(0.05) = 0.35 + 0.40 + 0.45 + 0.60 + 0.50 + 0.30 = 2.6[/latex] years
Use the following information to answer the next seven exercises: 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 [latex]X =[/latex] the number of years a student will study ballet with the teacher.
- Let [latex]P(x) =[/latex] the probability that a student will study ballet x years.
Complete the table using the data provided.
| 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 |
1. In words, define the random variable [latex]X[/latex].
Solution
[latex]X[/latex] is the number of years a student studies ballet with the teacher.
2. [latex]P(x = 4) = \underline{\hspace{2cm}}[/latex]
Solution
[latex]1 – 0.10 – 0.05 – 0.10 – 0.30 – 0.20 – 0.10 = 0.15[/latex]
3. [latex]P(x \lt 4) = \underline{\hspace{2cm}}[/latex]
Solution
[latex]0.10 + 0.05 + 0.10 = 0.25[/latex]
4. On average, how many years would you expect a child to study ballet with this teacher?
Solution
[latex]1(0.10) + 2(0.05) + 3(0.10) + 4(0.15) + 5(0.30) + 6(0.20) + 7(0.10) = 4.5[/latex] years
5. What does the column “[latex]P(x)[/latex]” sum to and why?
Solution
The sum of the probabilities sum to one because it is a probability distribution.
6. What does the column “[latex]x*P(x)[/latex]” sum to and why?
Solution
The sum of [latex]x*P(x) = 4.5[/latex]; it is the mean of the distribution.
1. 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
[latex]-2\left(\frac{40}{52}\right)+30\left(\frac{12}{52}\right)=-1.54+6.92=5.38[/latex]
2. 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
Yes, because there is a positive expected value, and the more you play, the more likely you are to get closer to the expected value.
A theater group holds a fund-raiser. It sells 100 raffle tickets for $5 a piece. Suppose you purchase four tickets. The prize is two passes to a Broadway show, worth a total of $150.
- What are you interested in here?
- In words, define the random variable [latex]X[/latex].
- List the values that [latex]X[/latex] may take on.
- Construct a PDF.
- If this fund-raiser is repeated often and you always purchase four tickets, what would be your expected average winnings per raffle?
Solution
- I am interested in the average profit or loss.
- Let [latex]X =[/latex] the return from the raffle
- Win($150) or Lose ($0)
-
Net Gain Probability $150 [latex]\frac{4}{100}[/latex] $0 [latex]\frac{0}{100}[/latex] - [latex]150(\frac{4}{100})+0(\frac{99}{100})−20=− \$14[/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 $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.
- 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?
Solution
The variable of interest is [latex]X[/latex], or the gain or loss, in dollars.
The face cards jack, queen, and king. There are [latex](3)(4) = 12[/latex] face cards and [latex]52 – 12 = 40[/latex] cards that are not face cards.
We first need to construct the probability distribution for [latex]X[/latex]. We use the card and coin events to determine the probability for each outcome, but we use the monetary value of [latex]X[/latex] to determine the expected value.
| Card Event | X net gain/loss | P(X) |
|---|---|---|
| Face Card and Heads | 6 | [latex]\left(\frac{12}{52}\right)\left(\frac{1}{2}\right)=\left(\frac{6}{52}\right)[/latex] |
| Face Card and Tails | 2 | [latex]\left(\frac{12}{52}\right)\left(\frac{1}{2}\right)=\left(\frac{6}{52}\right)[/latex] |
| (Not Face Card) and (H or T) | –2 | [latex]\left(\frac{40}{52}\right)\left(1\right)=\left(\frac{40}{52}\right)[/latex] |
- [latex]\text{Expected value}=\left(6\right)\left(\frac{6}{52}\right)+\left(2\right)\left(\frac{6}{52}\right)+\left(-2\right)\left(\frac{40}{52}\right)=–\frac{32}{52}= – \$0.62[/latex]
- If you play this game repeatedly, over a long string of games, you would expect to lose 62 cents per game, on average.
- You should not play this game to win money because the expected value indicates an expected average loss.
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
Start by writing the probability distribution. [latex]X[/latex] is net gain or loss = prize (if any) less $10 cost of ticket
[latex]\text{Expected Value} =(490)(\frac{1}{100})+(90)(\frac{2}{100})+(15)(\frac{4}{100})+(−10)(\frac{93}{100})=−2[/latex]
There is an expected loss of $2 per ticket, on average.
Complete the PDF and answer the questions.
| x | P(x) | xP(x) |
|---|---|---|
| 0 | 0.3 | |
| 1 | 0.2 | |
| 2 | ||
| 3 | 0.4 |
- Find the probability that [latex]x = 2[/latex].
- Find the expected value.
Solution
- 0.1
- 1.6
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.
- What are you ultimately interested in here (the value of the roll or the money you win)?
- In words, define the Random Variable [latex]X[/latex].
- List the values that [latex]X[/latex] may take on.
- Construct a PDF.
- 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.
Solution
- the money won
- [latex]X =[/latex] the amount of money won or lost
- $5, –$6, $10
-
x P(x) 5 [latex]\frac{2}{6}[/latex] -6 [latex]\frac{3}{6}[/latex] 10 [latex]\frac{1}{6}[/latex] - [latex]\text{Expected Value} = (10)(\frac{1}{6})+ (5)(\frac{2}{6})– (6)(\frac{3}{6})= 0.33[/latex]
- Yes, the expected value is 33 cents
A venture capitalist, willing to invest $1,000,000, has three investments to 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% of no profit or loss, and a 20% chance of losing the million dollars.
- Construct a PDF for each investment.
- Find the expected value for each investment.
- Which is the safest investment? Why do you think so?
- Which is the riskiest investment? Why do you think so?
- Which investment has the highest expected return, on average?
Solution
-
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 –1,000,00 0.40 Biotech Firm x P(x) 6,00,000 0.10 0 0.70 –1,000,000 0.20 - $200,000; $600,000; $400,000
- third investment because it has the lowest probability of loss
- first investment because it has the highest probability of loss
- second investment
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 [latex]X =[/latex] the number of children married people have.
| x | P(x) | xP(x) |
|---|---|---|
| 0 | 0.10 | |
| 1 | 0.20 | |
| 2 | 0.30 | |
| 3 | ||
| 4 | 0.10 | |
| 5 | 0.05 | |
| 6 (or more) | 0.05 |
- Find the probability that a married adult has three children.
- In words, what does the expected value in this example represent?
- Find the expected value.
- Is it more likely that a married adult will have two to three children or four to six children? How do you know?
Solution
- 0.2
- The average number of children married adults have.
- 2.35
- two of three children
Suppose that the PDF for the number of years it takes to earn a Bachelor of Science (B.S.) degree is given as in the table below.
| x | P(x) |
|---|---|
| 3 | 0.05 |
| 4 | 0.40 |
| 5 | 0.30 |
| 6 | 0.15 |
| 7 | 0.10 |
On average, how many years do you expect it to take for an individual to earn a B.S.?
Solution
4.85 years
People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.
| x | P(x) |
|---|---|
| 0 | 0.03 |
| 1 | 0.50 |
| 2 | 0.24 |
| 3 | |
| 4 | 0.07 |
| 5 | 0.04 |
- Describe the random variable [latex]X[/latex] in words.
- Find the probability that a customer rents three DVDs.
- Find the probability that a customer rents at least four DVDs.
- Find the probability that a customer rents at most two DVDs.
Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer.
x P(x) 0 0.35 1 0.25 2 0.20 3 0.10 4 0.05 5 0.05 - At which store is the expected number of DVDs rented per customer higher?
- If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form.
- If Video to Go expects 300 customers next week, and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain.
- Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that?
Solution
- [latex]X =[/latex]the number of video rentals per customer
- 0.12
- 0.11
- 0.77
- Video To Go (1.82 expected value vs. 1.4 for Entertainment Headquarters)
- The expected number of videos rented to 300 Video To Go customers is 546.
- The expected number of videos rented to 420 Entertainment Headquarters customers is 588. Entertainment Headquarters will rent more videos.
- The standard deviation for the number of videos rented at Video To Go is 1.1609. The standard deviation for the number of videos rented at Entertainment Headquarters is 1.4293. Entertainment Headquarters has more variation.
A “friend” offers you the following “deal.” For a $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.
Choose the best answer below:
a. Based upon the financial gain or loss over the long run, should you play the game?
b. Yes, I expect to come out ahead in money.
c. No, I expect to come out behind in money.
d. It doesn’t matter. I expect to break even.
Solution
c
Florida State University has 14 statistics classes scheduled for its Summer 2013 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.
- What is the average class size assuming each class is filled to capacity?
- 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 [latex]X[/latex] equal the size of the student’s class. Define the PDF for [latex]X[/latex].
- Find the mean of [latex]X[/latex].
- Find the standard deviation of [latex]X[/latex].
Solution
- The average class size is [latex]\frac{30+8(60)+70+4(100)}{14}=70[/latex].
- [latex]P(x=30)=\frac{1}{14}[/latex]; [latex]P(x=60)=\frac{8}{14}[/latex]; [latex]P(x=70)=\frac{1}{14}[/latex]; [latex]P(x=100)=\frac{4}{14}[/latex]
- [latex]\text{Mean of }X=\frac{30}{14}+\frac{480}{14}+\frac{70}{14}+\frac{400}{14}=\frac{980}{14}=70[/latex]
- [latex]\text{Standard Deviation of }X=114.2857+57.1429+0+257.1429=20.702[/latex]
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
Let [latex]X =[/latex] the amount of money to be won on a ticket. The following table shows the PDF for [latex]X[/latex].
| x | P(x) |
|---|---|
| 0 | 0.969 |
| 5 | [latex]\frac{\text{250}}{\text{10,000}}= 0.025[/latex] |
| 25 | [latex]\frac{\text{50}}{\text{10,000}}= 0.005[/latex] |
| 100 | [latex]\frac{\text{10}}{\text{10,000}}= 0.001[/latex] |
Calculate the expected value of [latex]X[/latex].
[latex]0(0.969) + 5(0.025) + 25(0.005) + 100(0.001) = 0.35[/latex]
A fair price for a ticket is $0.35. Any price over $0.35 will enable the lottery to raise money.
A number that measures the central tendency of the data
Mean of a random variable
a number that measures how far the outcomes of a statistical experiment are from the mean of the distribution
