5 Discrete Probability Distributions
5.01: Introduction
A variable or what will be called the random variable from now on, is represented by the letter X and it represents a quantitative (numerical) variable that is measured or observed in an experiment.
A random variable (usually X) is a numeric description of an event. Recall that a sample space is all the possible outcomes and an event is a subset of the sample space.
Usually we use capital letters from the beginning of the alphabet to represent events, A, B, C, etc. We use capital letters from the end of the alphabet to represent random variables, X, Y, and Z. The possible outcomes of X are labeled with a corresponding lower-case letter x and subscripts like x1, x2, etc.
For instance, if we roll two 6-sided dice the sample space is S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2),…, (6,6)} and the event E the sum of the two rolls is five, then E = {(1,4), (2,3), (3,2), (4,1)}.
Now, we could define the random variable X to denote the sum of the two rolls, then X = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} and event E corresponds to x = 5.
There are different types of quantitative variables, called discrete or continuous. Discrete random variables can only take on particular values in a range. Continuous random variables can take on any value in a range. Discrete random variables usually arise from counting while continuous random variables usually arise from measuring
A discrete random variable is a variable that is finite or infinitely countable.
A continuous random variable is a variable that has an infinite number of possible values in an interval of numbers.
5.02: Discrete Probability Distributions
In many cases, the random variable is what you are measuring, but when it comes to discrete random variables, it is usually what you are counting. For the example of height, the random variable is the height of the child. For the example of how many fleas are on prairie dogs in a colony, the random variable is the number of fleas on a prairie dog in a colony.
Now suppose you put all the possible values of the random variable together with the probability that the random variable would occur. You could then have a distribution like before, but now it is called a probability distribution since it involves probabilities. A probability distribution is an assignment of probabilities to all the possible values of the random variable. The abbreviation of pdf is used for a probability density (distribution) function in your calculators.
The probability distribution of X lists all the possible values of x and their corresponding probabilities.
A valid discrete probability distribution has to satisfy two criteria:
1. The probability of x is between 0 and 1, 0 ≤ P(xi) ≤ 1.
2. The probability of all x values adds up to 1, ∑ P(xi) = 1.
Two books are assigned for a statistics class: a textbook and its corresponding study guide. No students buy just the study guide. The university bookstore determined 20% of enrolled students do not buy either book, 55% buy the textbook only, 25% buy both books, and these percentages are relatively constant from one term to another. Is this a valid discrete probability distribution?
Solution
Each probability is a number between 0 and 1; 0 ≤ 0.2 ≤ 1, 0 ≤ 0.55 ≤ 1, and 0 ≤ 0.25 ≤ 1. The sum of the probabilities adds up to 1; ∑ P(xi) = 0.2 + 0.55 + 0.25 = 1. Yes, this is a valid discrete probability distribution.
A random experiment consists of flipping a fair coin three times. Let X = the number of heads that show up. Create the probability distribution of X.
Solution
When you flip a coin three times, you can get 0 heads, 1 head, 2 heads or 3 heads, the corresponding values of x are 0, 1, 2, 3. We can find the probabilities for each of these values by finding the sample space and using the classical method of computing probability.
The sample space is S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
The event x = 0 can happen one way, namely {TTT}. Thus P(X = x) = P(X = 0) = \(\frac{1}{8}\).
The event x = 1 can happen three ways, namely {HTT, THT, TTH}. Thus P(X = 1) = \(\frac{3}{8}\).
The event x = 2 can happen three ways, namely {HHT, HTH, THH}. Thus P(X = 2) = \(\frac{3}{8}\).
The event x = 3 can happen one way, namely {HHH}. Thus P(X = 3) = \(\frac{1}{8}\).
Therefore, the probability distribution of X is:
The following table shows the probability of winning an online video game, where X = the net dollar winnings for the video game. Is the following table a valid discrete probability distribution?
Words Are Important!
When finding probabilities pay close attention to the following phrases.
Figure 5-2
The following is a valid discrete probability distribution of X = the net dollar winnings for an online video game. Find the probability of winning at most $2.50.
The 2010 United States Census found the chance of a household being a certain size.
Rare Events
The reason probability is studied in statistics is to help in making decisions in inferential statistics.
To understand how making decisions is done, the concept of a rare event is needed.
Rare Event Rule for Inferential Statistics: If, under a given assumption, the probability of a particular observed event is extremely small, then you can conclude that the assumption is probably not correct.
An example of this is suppose you roll an assumed fair die 1,000 times and get six 600 times, when you should have only rolled a six around 160 times, then you should believe that your assumption about it being a fair die is untrue.
Determining if an event is unusual: If you are looking at a value of X for a discrete variable, and the P(the variable has a value of x or more) ≤ 0.05, then you can consider the x an unusually high value. Another way to think of this is if the probability of getting such a high value is less than or equal to 0.05, then the event of getting the value x is unusual.
Similarly, if the P(the variable has a value of x or less) ≤ 0.05, then you can consider this an unusually low value. Another way to think of this is if the probability of getting a value as small as x is less than or equal to 0.05, then the event x is considered unusual.
Why is it "x or more" or "x or less" instead of just "x" when you are determining if an event is unusual?
Consider this example: you and your friend go out to lunch every day. Instead of each paying for their own lunch, you decide to flip a coin, and the loser pays for both lunches. Your friend seems to be winning more often than you would expect, so you want to determine if this is unusual before you decide to change how you pay for lunch (or accuse your friend of cheating). The process for how to calculate these probabilities will be presented in a later section on the binomial distribution.
If your friend won 6 out of 10 lunches, the probability of that happening turns out to be about 20.5%, not unusual. The probability of winning 6 or more is about 37.7%, still not unusual.
However, what happens if your friend won 501 out of 1,000 lunches? That does not seem so unlikely! The probability of winning 501 or more lunches is about 47.8%, and that is consistent with your hunch that this probability is not so unusual. Nevertheless, the probability of winning exactly 501 lunches is much less, only about 2.5%.
That is why the probability of getting exactly that value is not the right question to ask: you should ask the probability of getting that value or more (or that value or less on the other side).
The value 0.05 will be explained later, and it is not the only value you can use for deciding what is unusual.
The 2010 United States Census found the chance of a household being a certain size.
What was that voice?" shouted Arthur. "I don't know," yelled Ford,
"I don't know. It sounded like a measurement of probability."
"Probability? What do you mean?"
"Probability. You know, like two to one, three to one, five to four against. It said two to the power of one hundred thousand to one against. That's pretty improbable you know."
A million-gallon vat of custard upended itself over them without warning.
"But what does it mean?" cried Arthur.
"What, the custard?" "No, the measurement of probability!"
"I don't know. I don't know at all. I think we're on some kind of spaceship."
(Adams, 2002)
5.2.1 Mean of a Discrete Probability Distribution
The mean of a discrete random variable X is an average of the possible values of x, which considers the fact that not all outcomes are equally likely. The mean of a random variable is the value of x that one would expect to see after averaging a large number of trials. The mean of a random variable does not need to be a possible value of x.
Two books are assigned for a statistics class: a textbook and its corresponding study guide. No students buy just the study guide. The university bookstore determined 20% of enrolled students do not buy either book, 55% buy the textbook only, 25% buy both books, and these percentages are relatively constant from one term to another. If there are 100 students enrolled, how many books should the bookstore expect to sell to this class?
Solution
It is expected that 0.20 × 100 = 20 students will not buy either book (0 books total), that 0.55 × 100 = 55 will buy just the textbook (55 books total), and 0.25 × 100 = 25 will buy both books (totaling 50 books for these 25 students). The bookstore should expect to sell about 105 total books for this class.
The textbook costs $137 and the study guide $33. How much revenue should the bookstore expect from this class of 100 students? Use the results from the previous example.
Solution
It is expected that 55 students will just buy a textbook, providing revenue of $137 × 55 = $7,535. The roughly 25 students who buy both the textbook and the study guide would pay a total of ($137 + $33)25 = $170 × 25 = $4,250. Thus, the bookstore should expect to generate about $7,535 + $4,250 = $11,785 from these 100 students for this one class. However, there might be some sampling variability so the actual amount may differ by a little bit.
Probability distribution for the bookstore’s revenue from a single student. The distribution balances on a triangle representing the average revenue per student represented in Figure 5-4.
Figure 5-4
What is the average revenue per student for this course?
Solution
The expected total revenue is $11,785, and there are 100 students. Therefore, the expected revenue per student is $11,785/100 = $117.85.
Mean of a Discrete Random Variable: Suppose that X is a discrete random variable with values x1, x2, …, xk. Then the mean of X is, μ = Σ(xi ∙ P(xi)) = x1 ∙ P(x1) + x2 ∙ P(x2) + ‧‧∙ + xk ∙ P(xk). The mean is also referred to as the expected value of x denoted as E(X).
Two books are assigned for a statistics class: a textbook and its corresponding study guide. The university bookstore determined 20% of enrolled students do not buy either book, 55% buy the textbook only, 25% buy both books, and these percentages are relatively constant from one term to another. Use the formula for the mean to find the average textbook cost.
Solution
Let X = the revenue from statistics students for the bookstore, then x1 = $0, x2 = $137, and x3 = $170, which occur with probabilities 0.20, 0.55, and 0.25. The distribution of X is summarized in the table below.
Note using this method we are not dividing the answer by the total number of students since the probabilities are found by dividing the frequency by the total so we would not divide again.
It may have been tempting to set up the table with the x values 0, 1 and 2 books. This would be fine if the question was asking for the average number of books sold. Since the books are different prices, we would not be able to get an average cost. Make sure X represents the variable that you are using to calculate the mean.
The following is a valid discrete probability distribution of X = the net dollar winnings for an online video game. Find the mean net earnings for playing the game.
The 2010 United States Census found the chance of a household being a certain number of people (household size). Compute the mean household size.
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | 0.267 | 0.336 | 0.158 | 0.137 | 0.063 | 0.024 | 0.015 |
| x∙P(x) | 0.267 | 0.672 | 0.474 | 0.548 | 0.315 | 0.144 | 0.105 |
Add up the new row and you get the answer 2.525. This is the mean or the expected value, μ = 2.525 people.
This means that you expect a household in the United States to have 2.525 people in it. Keep your answer as a decimal. Now of course you cannot have part of a person, but what this tells you is that you expect a household to have either 2 or 3 people, with a little more 3-person households than 2-person households.
Just as with any data set, you can calculate the mean and standard deviation. In problems involving a table of probabilities, the probability distribution represents a population. The probability distribution in most cases comes from repeating an experiment many times. This is because you are using the data from repeated experiments to estimate the true probability.
Since a probability distribution represents a population, the mean and standard deviation that are calculated are actually the population parameters and not the sample statistics. The notation used is for the population mean μ “mu” and population standard deviation σ “sigma.”
Note, the mean can be thought of as the expected value. It is the value you expect to get on average, if the trials were repeated an infinite number of times. The mean or expected value does not need to be a whole number, even if the possible values of x are whole numbers.
In the Oregon lottery called Pick-4, a player pays $1 and then picks a four-digit number. If those four numbers are picked in that specific order, the person wins $2,000. What is the expected value in this game?
Solution
To find the expected value, you need to first create the probability distribution. In this case, the random variable x = $ net winnings. If you pick the right numbers in the right order, then you win $2,000, but you paid $1 to play, so you actually win a net amount of $1,999. If you did not pick the right numbers, you lose the $1, the x net value is –$1.
You also need the probability of winning and losing. Since you are picking a four-digit number, and for each digit, there are 10 possible numbers to pick from, with each independent of the others, you can use the fundamental counting rule. To win, you have to pick the right numbers in the right order. The first digit, you pick 1 number out of 10, the second digit you pick 1 number out of 10, and the third digit you pick 1 number out of 10. The probability of picking the right number in the right order is \(\frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10} \cdot \frac{1}{10}=\frac{1}{10000}=0.0001\)
The probability of losing (not winning) would be the complement rule 1 – 0.0001 = 0.9999.
Putting this information into a table will help to calculate the expected value.
| Game Outcome | Win | Lose | |
| x | $1,999 | –$1 | |
| P(x) | 0.0001 | 0.9999 | Total |
| x∙P(x) | $0.1999 | –$0.9999 | –$0.80 |
Now sum the last row and you have the expected value of $0.1999 + (–$0.9999) = –$0.80. If you kept playing this game, in the long run, you will expect to lose $0.80 per game. Since the expected value is not 0, this game is not fair.
Most lottery and casino games such as craps, 21, roulette, etc. and insurance policies are built with negative expectations for the consumer. This is how casinos and insurance companies stay in business.
5.2.2 Variance & Standard Deviation of Discrete Probability Distributions
Suppose you ran the university bookstore. Besides how much revenue you expect to generate, you might also want to know the volatility (variability) in your revenue.
The variance and standard deviation can be used to describe the variability of a random variable. When we first introduced a method for finding the variance and standard deviation for a data set we had sample data and found sample statistics. We first computed deviations from the mean (xi − μ), squared those deviations (xi − μ)2 , and took an average to get the variance. In the case of a random variable, we again compute squared deviations. However, we take their sum weighted by their corresponding probabilities, just as we did for the expectation. This weighted sum of squared deviations equals the variance, and we calculate the standard deviation by taking the square root of the variance, just as we did for a sample variance. We also are using notation for the population parameters σ and σ2, instead of the sample statistics s and s2.
Variance of a Discrete Random Variable X:
Suppose that X is a discrete random variable with values x1, x2, …, xk.
Then the variance of X is, σ2 = ∑(xi – μ)2 ∙P(xi) or an easier to compute, algebraically equivalent formula σ2 = ∑(xi2 ∙P(xi)) – μ2 = (x12 ∙P(x1) + x22 ∙P(x2) + ∙∙∙ + xk2 ∙P(xk)) – μ2 .
Standard Deviation of a Discrete Random Variable X:
The standard deviation of X is the positive square root of the variance, \(\sigma=\sqrt{\sigma^{2}}\)
There are many situations that we can model a discrete distribution with a formula. This will make finding probabilities with large sample sizes much easier than making a table. There are many types of discrete distributions. We will just be covering a few of them.
The following is a valid discrete probability distribution of X = the net dollar winnings for an online video game. Find the standard deviation of the net earnings for playing the game.
The 2010 United States Census found the chance of a household being a certain size. Compute the variance and standard deviation.
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | 0.267 | 0.336 | 0.158 | 0.137 | 0.063 | 0.024 | 0.015 |
| x2*P(x) | 0.267 | 1.344 | 1.422 | 2.192 | 1.575 | 0.864 | 0.735 |
Add this new row up to get the beginning of the variance formula ∑(xi2 ∙P(xi)) = 8.399.
In a previous example we found the mean household size to be μ = 2.525 people.
To finish finding the variance we need to square the mean and subtract from the sum to get σ2 = ∑(xi2 ∙P(xi)) – μ2 = 8.399 – 2.5252 = 2.023375 people2.
Having a measurement in squared units is not very helpful when trying to interpret so we take the square root to find the standard deviation, which is back in the original units.
The standard deviation of the number of people in a household is σ = \(\sqrt{2.023375}\) = 1.4225 people. This means that you can expect an average United States household to have 2.525 people in it, with an average spread or standard deviation of 1.42 people.
TI-84: Press [STAT], choose 1:Edit. For x and P(x) data pairs, enter all x-values in one list. Enter all corresponding P(x) values in a second list. Press [STAT]. Use cursor keys to highlight CALC. Select 1:1-Var Stats.
Enter list 1 for List and list 2 for frequency list. Press Enter to calculate the statistics. For TI-83, you will just see 1-Var Stats on the screen. Enter each list separated with a comma by pressing [2nd], then press the number 1 key corresponding to your x list, then a comma, then [2nd] and the number 2 key corresponding to your P(x) values. The home screen should look like this 1-Var Stats L1,L2.
Where the calculator says \(\overline{ x }\) is µ the population mean and σx is the population standard deviation (square this number to get the population variance).
TI-89: Go to the [Apps] Stat/List Editor, and type the x values into List 1 and P(x) values into List2. Select F4 for the Calc menu. Use cursor keys to highlight 1:1-Var Stats.
Under List, press [2nd] Var-Link, then select list1. Under Freq, press [2nd] Var-Link, then select list2. Press enter twice and the statistics will appear in a new window.
Use the cursor keys to arrow up and down to see all of the values. Note: \(\overline{ x }\) this is µ the population mean and σx is the population standard deviation; square this value to get the variance.
Bernoulli trial
The focus of the previous section was on discrete probability distribution tables. To find the table for those situations, we usually need to actually conduct the experiment and collect data. Then one can calculate the experimental probabilities.
If certain conditions are met, we can instead use theoretical probabilities. One of these theoretical probabilities can be used when we have a Bernoulli trial.
Properties of a Bernoulli trial or binomial trial:
- Trials are independent, which means that what happens on one trial does not influence the outcomes of other trials.
- There are only two outcomes, which are called a success and a failure.
- The probability of a success does not change from trial to trial, where p = probability of success and q = probability of failure, the complement of p, q = 1 – p.
If you know you have a Bernoulli trial, then you can calculate probabilities using some theoretical probability formulas. This is important because Bernoulli trials come up often in real life.
Examples of Bernoulli experiments are:
- Toss a fair coin twice, and find the probability of getting two heads.
- Question twenty people in class, and look for the probability of more than half are business majors.
- A patient has a virus.
- What is the probability of passing a multiple-choice test if you have not studied?
Bernoulli trials are used in both the geometric and binomial distributions.
5.03: Geometric Distributions
The geometric distribution is a discrete probability distribution used to find the probability of success when there are two outcomes to each trial, and the trials are independent with the same probability of occurrence. A geometric probability distribution results from a random experiment that meets all of the following requirements.
- Repeat the trials until you get a success.
- The trials must be independent.
- Each trial must have exactly two categories that can be labeled “success” and “failure.”
- The probability of a “success,” denoted by p, remains the same in all trials. The probability of “failure” is often denoted by q, thus q = 1– p.
- Random Variable, X, counts the number of trials until your first success
If a random experiment satisfies all of the above, the distribution of the random variable X, where X counts the number of trials until the first success, is called a geometric distribution. If a discrete random variable X has a geometric distribution with the probability of success is p, we write X ~ G(p).
The geometric distribution is P(X = x) = p ∙ q(x – 1) , x = 1, 2, 3, … where x is the number of trials up to the first success that you are trying to find the probability for, p is the probability of a success for one trial and q = 1 – p is the probability of a failure for one trial.
Be careful, a “success” is not always a “good” thing. Sometimes a success is something that is “bad,” like finding a defect or getting in a car crash. The success will be the event from the probability question.
A game of chance has only two possible outcomes to win or lose for each time you play. The probability of losing the game is 65%. You play this game until your first loss. What is the probability of playing the game exactly 6 times in a row?
Solution
We have independent trials with two outcomes, no set sample size and the same probability of success each time we play so we can use the geometric distribution. Let p = 0.65, which means q = 1 – 0.65 = 0.35 and x = 6. P(X = 6) = 0.65 ∙ 0.355 = 0.0034. This is the probability of winning the first 5 games and losing on the 6th game in which you would stop playing.
TI-84: Press [2nd] [DISTR]. This will get you a menu of probability distributions. Press 0 or arrow down to geometpdf( and press [ENTER]. This puts geometpdf( on the home screen. Enter the values for p and x with a comma between each. Press [ENTER]. This is the probability density function and will return you the probability of exactly x successes. For the previous example, we would use geometpdf(0.65,6).
TI-89: Go to the [Apps] Stat/List Editor, then select F5 [DISTR]. This will get you a menu of probability distributions. Arrow down to Geometric Pdf and press [ENTER]. Enter the values for p and x into each cell. Press [ENTER]. This gives the probability density function and will return you the probability of exactly x successes.
Flip a fair coin until you get a tail. What is the probability of flipping the coin exactly 8 times until you get a tail?
Solution
In this case the probability of a success is getting a tail on any one toss which is p = ½ = 0.5. We are interested in a success on the eighth flip so x = 8.
P(X = 8) = 0.5 ∙ 0.57 = 0.00391.
Figure 5-5 shows a graph of the discrete probability distribution.
Figure 5-5
When looking at a person’s eye color, it turns out that only 2% of people in the world have green eyes (not to be confused with hazel colored eyes). Randomly select people and look at their eye color. What is the probability that you get someone with green eyes on the 5th person?
Solution
The probability of a success is 0.02. We are looking for P(X = 5) = 0.02 ∙ 0.984 = 0.0184.
Mean, Variance & Standard Deviation of a Geometric Distribution
For a geometric distribution, μ, the expected number of successes, σ2 , the variance, and σ, the standard deviation for the number of success are given by the formulas, where p is the probability of success and q = 1 – p.
\(\mu=\frac{1}{p} \quad \sigma^{2}=\frac{1-p}{p^{2}} \quad \sigma=\sqrt{\frac{1-p}{p^{2}}}\)
5.04: Binomial Distribution
The binomial distribution is a discrete probability distribution used to find the probability of success when there are two outcomes to each trial, and there are a set number of independent trials with the same probability of occurrence. To develop the process for calculating the probabilities in a binomial experiment, consider the following example.
Suppose you are given a three-question multiple-choice test. Each question has four responses and only one is correct. Suppose you want to find the probability that you can just guess at the answers and get two questions correct. To help with the idea that you are going to guess, suppose the test is on string theory.
- Is this a binomial experiment? Since all of the properties are met, this is a binomial experiment.
- What is the probability of getting two questions right?
- What is the probability of getting zero right, one right, two right and all three right?
Solution
a) The random variable is x = number of correct answers. 1. There are three questions, and each question is a trial, so there are a fixed number of trials. In this case, n = 3. 2. Getting the first question right has no effect on getting the second or third question correct, thus the trials are independent. 3. Either you get the question right or you get it wrong, so there are only two outcomes. In this case, the success is getting the question right. 4. The probability of getting a question right is one out of four. This is the same for every trial since each question has 4 responses. In this case, p = ¼ and q = 1 – ¼ = ¾.
b) To answer this question, start with the sample space. SS = {RRR, RRW, RWR, WRR, WWR, WRW, RWW, WWW}, where RRW means you get the first question right, the second question right, and the third question wrong.
Now the event space for getting 2 right is {RRW, RWR, WRR}. What you did in chapter four was just to find three divided by eight to compute the probability outcome. However, this would not be correct in this case, because the probability of getting a question right is different from getting a question wrong. What else can you do?
Look at just P(RRW) for the moment. Again, that means P(RRW) = P(R on 1st, R on 2nd, and W on 3rd). Since the trials are independent, then P(RRW) = P(R on 1st, R on 2nd, and W on 3rd) = P(R on 1st) * P(R on 2nd) * P(W on 3rd). Just multiply p · p · q = P(RRW) = ¼ ∙ ¼ ∙ ¾ = (¼)2 ∙ (¾)1.
Similarly, you can compute the P(RWR) and P(WRR). To find the probability of 2 correct answers, just add these three probabilities together.
P(2 correct answers) = P(RRW) + P(RWR) + P(WRR) = (¼)2 ∙ (¾)1 + (¼)2 ∙ (¾)1 + (¼)2 ∙ (¾)1 = 3∙(¼)2 ∙ (¾)1 .
c) You could go through the same argument that you did above and come up with the following:
Do you see the resulting pattern? You can now write the general formula for the probabilities for a binomial experiment.
First, the random variable in a binomial experiment is x = number of successes.
A binomial probability distribution results from a random experiment that meets all of the following requirements.
- The procedure has a fixed number of trials (or steps), which is denoted by n.
- The trials must be independent.
- Each trial must have exactly two categories that can be labeled “success” and “failure.”
- The probability of a “success,” denoted by p, remains the same in all trials. The probability of “failure” is often denoted by q, thus q = 1 – p.
- Random Variable, X, counts the number of “successes.”
If a random experiment satisfies all of the above, the distribution of the random variable X, where X counts the number of successes, is called a binomial distribution. A binomial distribution is described by the population proportion p and the sample size n. If a discrete random variable X has a binomial distribution with population proportion p and sample size n, we write X ~ B(n, p).
Be careful, a “success” is not always a “good” thing. Sometimes a success is something that is “bad,” like finding a defect or getting in a car crash. The success will be the event from the probability question.
The geometric and binomial distributions are easy to mix up. Keep in mind that the binomial distribution has a given sample size, whereas the geometric is sampling until you get a success.
Excel Formula for Binomial Distribution: For exactly P(X = x) use =binom.dist(x,n,p,false).
For P(X ≤ x) use =binom.dist(x,n,p,true).
TI-84: Press [2nd] [DISTR]. This will get you a menu of probability distributions. Press 0 or arrow down to 0:binompdf( and press [ENTER]. This puts binompdf( on the home screen. Enter the values for n, p and x with a comma between each. Press [ENTER]. This is the probability density function and will return you the probability of exactly x successes. If you leave off the x value and just enter n and p, you will get all the probabilities for each x from 0 to n. Press [ALPHA] A or arrow down to A:binomcdf( and press [ENTER]. This puts binomcdf( on the home screen. Enter the values for n, p and x with a comma between each. If you have the newer operating system on the TI84, the screen will prompt you for each value. Press [ENTER]. This is the cumulative distribution function and will return you the probability of at most (≤) x successes. If you have at least x success (≥), use the complement rule. If you have < or > adjust x to get ≤ or ≥.
TI-89: Go to the [Apps] Stat/List Editor, then select F5 [DISTR]. This will get you a menu of probability distributions. Arrow down to binomial Pdf and press [ENTER]. Enter the values for n, p and x into each cell. Press [ENTER]. This is the probability density function and will return you the probability of exactly x successes. If you leave off the x value and just enter n and p, you will get all the probabilities for each x from 0 to n. Arrow down to binomial Cdf and press [ENTER]. Enter the values for n, p and lower and upper value of x into each cell. Press [ENTER]. This is the cumulative distribution function and will return you the probability between the lower and upper x-values, inclusive.
When looking at a person’s eye color, it turns out that only 2% of people in the world have green eyes (not to be confused with hazel colored eyes). Consider a randomly selected group of 20 people.
- Is this a binomial experiment?
- Compute the probability that none have green eyes.
- Compute the probability that nine have green eyes.
- There are 20 people, and each person is a trial, so there are a fixed number of trials.
- If you assume that each person in the group is chosen at random the eye color of one person does not affect the eye color of the next person, thus the trials are independent.
- Either a person has green eyes or they do not have green eyes, so there are only two outcomes. In this case, the success is a person has green eyes.
- The probability of a person having green eyes is 0.02. This is the same for every trial since each person has the same chance of having green eyes.
Solution
a) Yes, since all the requirements are met:
b) You are looking for P(X = 0), since this problem is asking for none x = 0. There are 20 people so n = 20. The success is selecting someone with green eyes, so the probability of a success p = 0.02. Then the probability of not selecting someone with green eyes is q = 1 – p = 1 – 0.02 = 0.98. Using the formula: P(X = 0) = 20C0· 0.020 · 0.98(20-0) = 0.6676. Thus, there is a 66.76% chance that in a group of 20 people none of them will have green eyes.
c) P(X = 9) = binom.dist(9,20,0.02,false) = 6.8859E-11 = 0.00000000006859 which is zero rounded to four decimal places. The probability that out of 20 people, nine of them have green eyes is a very small chance and would be considered a rare event.
As you read through a problem look for some of the following key phrases in Figure 5-6. Once you find the phrase then match up to what sign you would use and then use the table to walk you through the computer or calculator formula. The same idea about signs applies to all the discrete probabilities that follow.
Figure 5-6
As of 2018, the Centers for Disease Control and Prevention (CDC) reported that about 1 in 88 children in the United States have been diagnosed with autism spectrum disorder (ASD). A researcher randomly selects 10 children. Compute the probability that 2 children have been diagnosed with ASD.
Solution
The random variable x = number of children with ASD. There are 10 children, and each child is a trial, so there are a fixed number of trials. In this case, n = 10. If you assume that each child in the group is chosen at random, then whether a child has ASD does not affect the chance that the next child has ASD. Thus, the trials are independent. Either a child has been diagnosed with ASD or they have not been diagnosed ASD, so there are two outcomes. In this case, the success is a child has ASD. The probability of a child having ASD is 1/88. This is the same for every trial since each child has the same chance of having ASD, \(p=\frac{1}{88} \text { and } q=1-\frac{1}{88}=\frac{87}{88}\)
Using the formula: \(\mathrm{P}(X=2)={ }_{10} C_{2} \cdot\left(\frac{1}{88}\right)^{2} \cdot\left(\frac{87}{88}\right)^{(10-2)}=0.0053\)
Using the TI-83/84 Calculator: P(X = 2) = binompdf(10,1/88,2) = 0.0053.
Using Excel: =BINOM.DIST(2,10,1/88,FALSE) = 0.0053.
Flip a fair coin exactly 10 times.
a) What is the probability of getting exactly 8 tails?
b) What is the probability of getting 8 or more tails?
Solution
a) There are only two outcomes to each trial, heads or tails. The coin flips are independent and the probability of a success, flipping a tail, p = ½ = 0.5 is the same for each trial. This is a binomial experiment with a sample size n = 10. Using the formula: P(X = 8) = 10C8·0.58 ·0.5(10-8) = 0.0439
b) We still have p = 0.5 and n = 10, however we need to look at the probability of 8 or more. The P(X ≥ 8) = P(X = 8) + P(X =9) + P(X = 10). We can stop at x = 10 since the coin was only flipped 10 times. P(X ≥ 8) = 10C8· 0.58 · 0.5(10-8) + 10C9· 0.59 · 0.5(10-9) + 10C10· 0.510 · 0.5(10-10) = 0.0439 + 0.0098 + 0.0010 = 0.0547.
So far, most of the examples for the binomial distribution were for exactly x successes. If we want to find the probability of accumulation of x values then we would use the cumulative distribution function (cdf) instead of the pdf. Phrases such as “at least,” “more than,” or “below” can drastically change the probability answers.
Approximately 10.3% of American high school students drop out of school before graduation. Choose 10 students entering high school at random. Find the following probabilities.
a) No more than two drop out.
b) At least 6 students graduate.
c) Exactly 10 stay in school and graduate.
Solution
a) There is a set sample size of independent trials. The person either has or has not dropped out of school before graduation. A “success” is what we are finding the probability for, so in this case a success is to drop out so p = 0.103 and q = 1 – 0.103 = 0.897.
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 10C0 · 0.1030 · 0.89710 + 10C1 · 0.1031 ·0.8979 + 10C2 · 0.1032 · 0.8978 = 0.3372 + 0.3872 + 0.2001 = 0.9245
Calculator shortcut use binompdf(10,0.103,0) + binompdf(10,0.103,1) + binompdf(10,0.103,2) = 0.9245 or the binomcdf(10,0.103,2) = 0.9245. On the TI-89 cdf enter the lower value of x as 0 and upper value of x as 2.
In Excel use the function =BINOM.DIST(2,10,0.103,TRUE). Note that if you choose False under cumulative this would return just the P(X = 2) not P(X ≤ 2).
b) A success is to graduate so p = 0.897 and q = 0.103.
P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 10C6 · 0.8976 · 0.1034 + 10C7 · 0.8977 · 0.1033 + 10C8 · 0.8978 · 0.1032 + 10C9 · 0.8979 · 0.1031 + 10C10 · 0.89710 · 0.1030 = 0.0123 + 0.0613 + 0.2001 + 0.3872 + 0.3372 = 0.9981.
This is a lot of work to do each one so we can use technology to find the answer.
Note that Excel and the older TI-84 programs only find the probability below x so you have to use the complement rule.
For the TI-84 calculator shortcut use binompdf(10,0.897,6) + binompdf(10,0.897,7) + binompdf(10,0.897,8) + binompdf(10,0.897,9) + binompdf(10,0.897,10) = 0.9981 or use the complement rule P(X ≥ 6) = 1– P(X≤5) = 1–binomcdf(10,0.897,5) = 0.9981.
On the TI-89, just use the binomcdf with the lower x value as 6 and the upper x value as 10.
In Excel use =1–BINOM.DIST(5,10,0.897,TRUE).
c) A success is to graduate so p = 0.897 and q = 0.103. Find P(X = 10) = 10C10 · 0.89710 · 0.1030 = 0.3372.
On the TI-84 use binompdf(10,0.897,10) = 0.3372.
In Excel =BINOM.DIST(10,10,0.897,FALSE) = 0.3372.
When looking at a person’s eye color, it turns out that only 2% of people in the world have green eyes (not to be confused with hazel colored eyes). Consider a randomly selected group of 20 people.
a) Compute the probability that at most 3 have green eyes.
b) Compute the probability that less than 3 have green eyes.
c) Compute the probability that more than 3 have green eyes.
d) Compute the probability that 3 or more have green eyes.
Solution
a) This fits a binomial experiment. P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 20C0 · 0.020 · 0.8020 + 20C1 · 0.021 · 0.8019 + 20C2 · 0.022 · 0.8018 + 20C3 · 0.023 · 0.8017 = 0.667608 + 0.272493 + 0.05283 + 0.006469 = 0.9994.
On the TI-84 use binomcdf(20,0.02,3) = 0.9994.
In Excel =BINOM.DIST(3,20,0.02,true) = 0.9994.
b) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.667608 + 0.272493 + 0.05283 = 0.9929.
On the TI-84 use binompdf(20,0.02,2) = 0.9929.
In Excel =BINOM.DIST(2,20,0.02,true) = 0.9929.
c) P(X > 3) = 1 – P(X ≤ 3) = 1 – 0.9994 = 0.0006.
On the TI-84 use 1-binompdf(20,0.02,3) = 0.0006.
In Excel =1-BINOM.DIST(3,20,0.02,true) = 0.0006.
d) P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – 0.9929 = 0.0071
On the TI-84 use 1-binompdf(20,0.02,2) = 0.0071.
In Excel =1-BINOM.DIST(2,20,0.02,true) = 0.0071.
As of 2018, the Centers for Disease Control and Prevention (CDC) reported that about 1 in 88 children in the United States have been diagnosed with autism spectrum disorder (ASD). A researcher randomly selects 10 children. Compute the probability that at least 2 children have been diagnosed with ASD.
Solution
\(\mathrm{P}(X \geq 2)=1-\mathrm{P}(X \leq 1)=1-\left({ }_{10} C_{0} \cdot\left(\frac{1}{88}\right)^{0} \cdot\left(\frac{87}{88}\right)^{10-0)}+{ }_{10} C_{1} \cdot\left(\frac{1}{88}\right)^{1} \cdot\left(\frac{87}{88}\right)^{(10-1)}\right)=1-(0.89202+0.102528)=0.0055\)
On the TI-84 use 1-binomcdf(10,1/88,1) = 0.0055.
In Excel =1-BINOM.DIST(1,10,1/88,true) = 0.0055.
Mean, Variance & Standard Deviation of a Binomial Distribution
If you list all possible values of x in a Binomial distribution, you get the Binomial Probability Distribution (pdf). You can then find the mean, the variance, and the standard deviation using the general formulas μ = Σ(xi ∙ P(xi)) and σ2 = ∑(xi2 ∙P(xi)) – μ2 . This, however, would take a lot of work if you had a large value for n. If you know the type of distribution, like binomial, then you can find the mean, variance and standard deviation using easier formulas. They are derived from the general formulas. For a Binomial distribution, μ, the expected number of successes, σ2 , the variance, and σ, the standard deviation for the number of success are given by the formulas, where p is the probability of success and q = 1 – p.
\(\mu=n \cdot \mathrm{p} \quad \sigma^{2}=n \cdot \mathrm{p} \cdot \mathrm{q} \quad \sigma=\sqrt{n \cdot p \cdot q}\)
A random experiment consists of flipping a coin three times. Let X = the number of heads that show up. Compute the mean and standard deviation of X, that is, the mean and standard deviation for the number of heads that show up when a coin is flipped three times.
Solution
This experiment follows a binomial distribution; hence, we can use the mean and standard deviation formulas for a binomial. The mean of number of heads is µ = 3·0.5 = 1.5. The standard deviation of X is \(\sigma=\sqrt{n \cdot p \cdot q}=\sqrt{(3 \cdot 0.5 \cdot(1-0.5))}=0.8660\)
When looking at a person’s eye color, it turns out that only 2% of people in the world have green eyes (not to be confused with hazel colored eyes). Consider a randomly selected group of 20 people. Compute the mean, variance and standard deviation.
Solution
Since this is a binomial experiment, then you can use the formula μ = n ∙ p. So μ = 20 · 0.02 = 0.4 people. You would expect on average that out of 20 people, less than 1 would have green eyes. The variance would be σ2 = n ∙ p ∙ q = 20(0.02)(0.98) = 0.392 people2 . Once you have the variance, you just take the square root of the variance to find the standard deviation σ = \(\sqrt0.392\) = 0.6261 people. We would expect on average spread of the distribution to have 0.4 \(\pm\) 0.6261 or 0 to 1 person out of 20 people to have green eyes.
5.05: Hypergeometric Distribution
The hypergeometric distribution is a discrete probability distribution used to find the probability of success when there are two outcomes to each trial, and there are a set number of dependent trials. A hypergeometric probability distribution results from a random experiment that meets all of the following requirements. 12
- The procedure has a fixed sample size n taken from a population N.
- The trials are taken without replacement (dependent).
- Each trial has exactly two outcomes, the number of successes = a and the number of failures = b. Note a + b = N.
If a random experiment satisfies all of the above, the distribution of the random variable X, where X counts the number of successes, is called a hypergeometric distribution, we write X ~ H(n, a, N).
The hypergeometric distribution is P(X = x) = \(\frac{ {}_{a} C_{x} \cdot {}_{b} C_{n-x}}{ {}_{N} C_{n}}\), x = 0, 1, 2, … , n or a, whichever is smaller. Where n is the sample size taken without replacement from the population size N, x is the number of successes out of n that you are trying to find the probability for, a is the number of a successes and b = N – a is the number of failures out of the population N.
A committee of 5 people is to be formed from 15 volunteers and 6 appointed positions. Compute the probability that the committee will consist of 4 randomly selected volunteers.
Solution
There are a total of N = 15 + 6 = 21 people. A sample of n = 5 people are selected without replacement. The success is choosing a volunteer, so a = 15. Of the 15 volunteers, we are asked to find the probability of selecting x = 4. Substitute in each value: P(X = 4) = \(\frac{ {}_{15} C_{4} \cdot { }_{6} C_{1}}{ {}_{21} C_{5}} = 0.4025\).
There is a pattern to the numbers in the formula. The top two numbers on the left of the C’s add up to the bottom number on the left of C, and the top two numbers on the right of the C’s add up to the bottom number on the right of C.
The waiting room for jury selection has 40 people with a college degree and 30 people without. A jury of 12 is selected from this waiting room. What is the probability that exactly 5 of the jury members will have a college degree?
Solution
There are a total of N = 40 + 30 = 70 people. A sample of n = 12 people are selected without replacement. The success is choosing someone with a college degree, so a = 40. Of the 12 jurors we are asked to find the probability of selecting x = 5.
Substitute in each value: P(X = 5) = \(\frac{ {}_{40} C_{5} \cdot { }_{30} C_{7}}{ {}_{70} C_{12}} = 0.1259\).
A wallet contains three $100 bills and five $1 bills. You randomly choose four bills. What is the probability that you will choose exactly two $100 bills?
Solution
There are a total of N = 3 + 5 = 8 bills. A sample of n = 4 bills are selected without replacement. The success is choosing a $100 bill, so a = 3. Of the 3 $100 bills we are asked to find the probability of selecting x = 2. Substitute in each value: P(X = 2) = \(\frac{ {}_{3} C_{2} \cdot { }_{5} C_{2}}{ {}_{8} C_{4}} = 0.4286\).
Note: the calculator does not have the hypergeometric distribution shortcut. The following online calculator will calculate the probability: https://homepage.divms.uiowa.edu/~mbognar/applets/hg.html. The M = a is the number of successes a from our example, or use Excel.
Figure 5-7 may help you decide when to use the “True” for cumulative in Excel. Figure 5-7 A bookstore owner examines 5 books from each lot of 25 to check for missing pages. If they find at least 2 books with missing pages, the entire lot is returned. If, indeed, there are 4 books with missing pages, find the probability that the lot will be returned. Sometimes the hardest part is determining which of the numbers goes in the right place in the formula. The population size is always the largest number, N = 25. A success is what you are trying to find the probability for, which is getting a book with a missing page. The number of successes out of the population is a = 4 (the 4 books with the missing pages). This means b = 25 – 4 = 21 (the number of books without missing pages). Note a + b = N. The sample size is the n = 5 is the number of books the owner examines. The owner will send the lot back if X is at least 2 missing pages. We can stop at X = 4 since there are only 4 books with missing pages. Your stopping point will be the smaller of n or a. P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) = \(\frac{ {}_{4} C_{2} \cdot {}_{21} C_{3}}{ {}_{25} C_{5}} + \frac{ {}_{4} C_{3} \cdot {}_{21} C_{2}}{ {}_{25} C_{5}} + \frac{ {}_{4} C_{4} \cdot {}_{21} C_{1}}{ {}_{25} C_{5}} = 0.1502 + 0.0158 + 0.0004 = 0.1664\). In Excel use the formula: =1- HYPGEOM.DIST(1,5,4,25,TRUE). Figure 5-8 shows the online calculator at https://homepage.divms.uiowa.edu/~mbognar/applets/hg.html.
Figure 5-8
Solution
5.06: Poisson Distribution
The Poisson distribution was named after the French mathematician Siméon Poisson (pronounced pwɑːsɒn, means fish in French).
The Poisson discrete probability distribution finds the probability of an event over some unit of time or space. A Poisson probability distribution may be used when a random experiment meets all of the following requirements.
- Events occur independently.
- The discrete random variable X is the number of occurrences over an interval of time, volume, space, area, etc.
- The mean number of successes μ over time, volume, space, area, etc. is given. (Note some textbooks and calculators use lambda = λ instead of mu = μ for the mean).
The formula for the Poisson distribution is P(X = x) = \(\frac{e^{-\mu} \mu^{x}}{x !}\), where e is a mathematical constant approximately equal to 2.71828, x = 0, 1, 2, … is the number successes that you are trying to find the probability for, μ is the mean number of a success over one interval of time, space, volume, etc.
The value of x has no stopping point since there is no set sample size like the binomial distribution. Note that e is not a variable, it is a constant number. Use the ex button on your calculator.
Mean, Variance & Standard Deviation of a Poisson Distribution
For a Poisson distribution, μ, the expected number of successes, and the variance σ2 are equal to one another.
\(\mu=\sigma^{2} \quad \sigma=\sqrt{\sigma^{2}}\)
Sometimes the question will ask for a probability over a different unit of time, space or area than originally given in the problem. Always change the mean to fit the new units in the question.
This formula will help in correctly rescaling the mean to fit the questions new units: New μ = old μ(\(\frac {\text {new units}}{\text {old units}}\)). The “old” would be the original stated mean and units and the “new” is the units from the question.
TI-84: Press [2nd] [DISTR]. This will get you a menu of probability distributions. Press [ALPHA] B or arrow down to B:poissonpdf( and press [ENTER]. This puts poissonpdf( on the home screen. Enter the values for μ and x with a comma between each. Press [ENTER]. This is the probability density function and will return you the probability of exactly x successes. Press [ALPHA] C or arrow down to C:poissoncdf( and press [ENTER]. This puts poissoncdf( on the home screen. Enter the values for μ and x with a comma between each. Press [ENTER]. This is the cumulative distribution function and will return you the probability of at most x successes.
TI-89: Go to the [Apps] Stat/List Editor, then select F5 [DISTR]. This will get you a menu of probability distributions. Arrow down to Poisson Pdf and press [ENTER]. Enter the values for μ and x into each cell. Press [ENTER]. This is the probability density function and will return you the probability of exactly x successes. Arrow down to Poisson Cdf and press [ENTER]. Enter the values for μ and the lower and upper values of x into each cell. Press [ENTER]. This is the cumulative distribution function and will return you the probability between the lower and upper x-values, inclusive.
Excel: Use the formula =POISSON.DIST(x,mean,false) for P(X = x). Use the formula =POISSON.DIST(x,mean,true) for P(X ≤ x).
There are about 8 million individuals in New York City. Using historical records, the average number of individuals hospitalized for acute myocardial infarction (AMI), i.e. a heart attack, each day is 4.4 individuals. What is the probability that exactly 6 people are hospitalized for AMI in NY City tomorrow?
Solution
First, note that the probability question is over the same units of time, one day, which the average is given. It is important that these units always match. The random variable X = the number of individuals hospitalized each day in NY City for an AMI.
We are trying to find P(X = 6) and the mean μ = 4.4 people.
Using the formula P(X = 6) = \(\frac{e^{-4.4} 4.4^{6}}{6 !}\) = 0.1237.
TI-84: poissonpdf(4.4,6) = 0.1237.
Excel =POISSON.DIST(6,4.4,FALSE) = 0.1237.
A bank drive-through has an average of 10 customers every hour. Find the following probabilities.
a) Compute the probability that no customers arrive in an hour.
b) Compute the probability of exactly 2 customers use the drive-through in an hour.
c) Compute the probability that exactly 2 customers use the drive-through in a 30-minute period.
d) Compute the probability that fewer than 2 customers use the drive-through in a half hour.
Solution
a) In this case, X is the number of customers that use the drive-through. We are trying to find the probability that x = 0. The average is given in the problem as μ = 10. This gives P(X = 0) = \(\frac{e^{-10} 10^{0}}{0 !}\) = 4.53999E-5 = 0.0000454.
TI-84: poissonpdf(10,0) = 0.0000454.
Excel: =POISSON.DIST(0,10,FALSE) = 0.0000454.
b) P(X = 2) = \(\frac{e^{-10} 10^{2}}{2 !}\) = 0.0023.
TI-84: P(X = 2) = poissonpdf(10,2) = 0.0023.
Excel: =POISSON.DIST(2,10,FALSE) = 0.0023.
c) The unit of time has changed from 1 hour to 30 minutes, so we need to rescale the mean to fit the new unit of time. Use the following formula to convert your units, New μ = old μ(\(\frac {\text {new units}}{\text {old units}}\)) = 10(\(\frac {\text {30 minutes}}{\text {60 minutes}}\)) = 5.
P(X = 2) = \(\frac{e^{-5} 5^{2}}{2 !}\) = 0.0842.
TI-84: P(X = 2) = poissonpdf(5,2) = 0.0842.
Excel: P(X = 2) =POISSON.DIST(2,5,FALSE) = 0.0842.
Note; always rescale the mean to fit the units of the question, not the other way around.
d) The time-period is half an hour = 30 minutes. The mean from part c would be μ = 5. To find “less than” 2 we would have zero or one customer.
P(X < 2) = P(X = 0) + P(X =1) = \(\frac{e^{-5} 5^{0}}{0 !}\) + \(\frac{e^{-5} 5^{1}}{1 !}\) = = 0.0067 + 0.0337 = 0.0404.
TI-84: P(X = 2) = poissoncdf(5,1) = 0.0404.
Excel: =POISSON.DIST(1,5,TRUE) = 0.0404.
So far, most of the examples for the Poisson distribution were for exactly x successes. If we want to find the probability of accumulation of x values then we would use the cumulative distribution function (cdf) instead of the pdf. As you read through a problem look for some of the following key phrases in Figure 5-9. Once you find the phrase then match up to what sign you would use and then use the table to walk you through the computer or calculator formula.
Figure 5-9
A bank drive-through has an average of 10 customers every hour. Find the following probabilities.
a) Compute the probability of at least four customers arriving in an hour.
b) Compute the probability of at most four customers arriving in an hour.
c) Compute the probability of fewer than four customers arriving in an hour.
d) Compute the probability of more than four customers arriving in an hour.
e) Compute the probability that fewer than 2 customers arrive in a 15-minute period.
Solution
a) Because we do not have a set sample size to stop at like the binomial distribution, you would have to find the probability of 4 or more until your answers are small enough that they are not changing your answer out to at least four decimal places 0.0000. This may take a lot of work; instead, we will use the complement rule. The complement to “at least 4,” is “3 or less.” Find P(X ≥ 4) = 1 – P(X ≤ 3) = 1 – P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = \(1-\left(\frac{e^{-10} 10^{0}}{0 !}+\frac{e^{-10} 10^{1}}{1 !}+\frac{e^{-10} 10^{2}}{2 !}+\frac{e^{-10} 10^{3}}{3 !}\right)\) = 1 – 0.000045 + 0.000454 + 0.00227 + 0.007567 = 1 – 0.0103 = 0.9897. Note that the value of x went from 4 to 3. The complement of strictly less than 4 starts at x = 3.
TI-84: 1 – poissoncdf(10,3) = 0.9897.
Excel: =1-POISSON.DIST(3,10,TRUE) = 0.9897.
b) P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = \(\frac{e^{-10} 10^{0}}{0 !}+\frac{e^{-10} 10^{1}}{1 !}+\frac{e^{-10} 10^{2}}{2 !}+\frac{e^{-10} 10^{3}}{3 !}+\frac{e^{-10} 10^{4}}{4 !}\) = 0.000045 + 0.000454 + 0.00227 + 0.007567 + 0.018917 = 0.0293.
TI-84: poissoncdf(10,4) = 0.0293.
Excel: =POISSON.DIST(4,10,TRUE) = 0.0293.
c) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = \(\frac{e^{-10} 10^{0}}{0 !}+\frac{e^{-10} 10^{1}}{1 !}+\frac{e^{-10} 10^{2}}{2 !}+\frac{e^{-10} 10^{3}}{3 !}\) 0.000045 + 0.000454 + 0.00227 + 0.007567 = 0.0103.
TI-84: poissoncdf(10,3) = 0.0103.
Excel: =POISSON.DIST(3,10,TRUE) = 0.0103.
d) Using part b, P(X > 4) = 1 – P(X ≤ 4) = 1 – 0.0293 = 0.9707.
TI-84: 1 – poissoncdf(10,4) = 0.9707.
Excel: =1-POISSON.DIST(4,10,TRUE) = 0.9707.
e) The unit of time has changed from 1 hour to 15 minutes, so we need to rescale the mean to fit the new unit of time. Use the following formula to convert your units, New μ = old μ(\(\frac {\text {new units}}{\text {old units}}\)) = 10(\(\frac {\text {15 minutes}}{\text {60 minutes}}\)) = 2.5.
P(X < 2) = P(X = 0) + P(X = 1) = \(\frac{e^{-2.5} 2.5^{0}}{0 !}\) + \(\frac{e^{-2.5} 2.5^{1}}{1 !}\) = 0.0821 + 0.2052 = 0.2873.
TI-84: poissoncdf(2.5,1) = 0.2873.
Excel: =POISSON.DIST(1,2.5,TRUE) = 0.2873.
The last ever dolphin message was misinterpreted as a surprisingly sophisticated attempt to do a double--‐ backwards – somersault through a hoop whilst whistling the "Star Sprangled Banner," but in fact the message was this: So long and thanks for all the fish.
(Adams, 2002)
5.07: Chapter 5 Exercises
1. Determine if the following tables are valid discrete probability distributions. If they are not state why.
a)
| x | –5 | –2.5 | 0 | 2.5 | 5 |
| P(X = x) | 0.15 | 0.25 | 0.32 | 0.18 | 0.1 |
b)
| x | 0 | 1 | 2 | 3 | 4 |
| P(X = x) | 0.111 | 0.214 | 0.312 | 0.163 | 0.159 |
c)
| x | 0 | 1 | 2 | 3 | 4 |
| P(X = x) | 0.2 | -0.3 | 0.5 | 0.4 | 0.2 |
2. The random variable X = the number of vehicles owned.
| x | 0 | 1 | 2 | 3 | 4 |
| P(X = x) | 0.1 | 0.35 | 0.25 | 0.2 | 0.1 |
a) Compute the probability that a person owns at least 2 vehicles.
b) Compute the P(X > 2).
c) Compute the probability that a person owns less than 2 vehicles.
d) Compute the expected number of vehicles owned.
e) Compute the standard deviation of the number of vehicles owned.
f) Compute σ2 .
3. The following discrete probability distribution represents the amount of money won for a raffle game.
| x | –5 | –2.5 | 0 | 2.5 | 5 |
| P(X = x) | 0.15 | 0.25 | 0.32 | 0.18 | 0.1 |
a) Compute μ.
b) Compute σ
4. Keke's Kookies sells mini cookies in packs of 5 and has determined a probability distribution for the number of cookies that they sell in a given day.
| x = #sold | 0 | 5 | 10 | 15 | 20 |
| P(X = x) | 0.22 | 0.38 | 0.14 | ..? | 0.07 |
a) What is the probability of selling 15 mini cookies in a given day?
b) Find the expected number of mini cookies sold in a day using the discrete probability distribution.
c) Find the variance of the number of mini cookies sold in a day using the discrete probability distribution.
5. The bookstore also offers a chemistry textbook for $159 and a book supplement for $41. From experience, they know about 25% of chemistry students just buy the textbook while 60% buy both the textbook and supplement. Compute the standard deviation of the bookstore revenue.
6. A $100,000 life insurance policy for a 50-year-old woman has an annual cost of $335. The probability that a 50- year-old woman will die is 0.003118. What is the expected value of the policy for the woman’s estate?
7. An LG Dishwasher, which costs $1000, has a 24% chance of needing to be replaced in the first 2 years of purchase. If the company has to replace the dishwasher within the two-year extended warranty, it will cost the company $112.10 to replace the dishwasher.
a) Fill out the probability distribution for the value of the extended warranty from the perspective of the company.
| x | ||
| P(X =x) |
b) What is the expected value of the extended warranty?
c) Write a sentence interpreting the expected value of the warranty.
8. The Oregon lottery has a game called Pick 4 where a player pays $1 and picks a four-digit number. If the four numbers come up in the order you picked, then you win $2000.
a) Fill out the probability distribution for a player's winnings
| x | ||
| P(X =x) |
b) What are your expected winnings?
c) Write a sentence interpreting the expected winnings.
9. The following table represents the probability of the number of pets owned by a college student.
| x | 0 | 1 | 2 | 3 |
| P(X=x) | 0.46 | 0.35 | 0.12 | 0.07 |
a) Is this a valid discrete probability distribution? Explain your answer.
b) Find the mean number of pets owned.
c) Find the standard deviation of the number of cars owned.
d) Find σ2.
10. Suppose a random variable, X, arises from a geometric distribution. If p = 0.13, compute the P(X = 4).
11. Approximately 10% of all people are left-handed. You randomly sample people until you get someone who is left-handed. What is the probability that the 4th person selected will be the left-handed person?
12. An actress has a probability of getting offered a job after a tryout of 0.12. She plans to keep trying out for new jobs until she is offered a job. Assume outcomes of tryouts are independent. Compute the probability she will need to attend more than 7 tryouts.
13. A fair coin is flipped until a head is shown. What is the probability that head shows on the 6th flip?
14. When you post a picture on social media, it seems like your friends randomly "like" the picture. Independent of the quality or the humor of the photo, there seems to be an 18% chance of the picture being "liked" for any given picture. Let X represent the number of pictures you post until one is "liked." (X represents the photo number that is actually liked.) Compute P(X = 15).
15. Suppose a random variable, x, arises from a binomial experiment. If n = 14, and p = 0.13, find the following probabilities.
a) P(X = 3)
b) P(X ≤ 3)
c) P(X < 3)
d) P(X > 3)
e) P(X ≥ 3)
16. Suppose a random variable, X, arises from a binomial experiment. If n = 25, and p = 0.85, find the following probabilities.
a) P(X = 15)
b) P(X ≤ 15)
c) P(X < 15)
d) P(X > 15)
e) P(X ≥ 15)
17. Suppose a random variable, X, arises from a binomial experiment. If n = 14, and p = 0.13, compute the standard deviation.
18. Suppose a random variable, X, arises from a binomial experiment. If n = 25, and p = 0.85, compute the variance.
19. A fair coin is flipped 30 times.
a) What is the probability of getting exactly 15 heads?
b) What is the probability of getting 15 or more heads?
c) What is the probability of getting at most 15 heads?
d) How many times would you expect to get heads?
e) What is the standard deviation of the number of heads?
20. Approximately 10% of all people are left-handed. Out of a random sample of 15 people, find the following.
a) What is the probability that 4 of them are left-handed?
b) What is the probability that less than 4 of them are left-handed?
c) What is the probability that at most 4 of them are left-handed?
d) What is the probability that at least 4 of them are left-handed?
e) What is the probability that more than 4 of them are left-handed?
f) Compute μ.
g) Compute σ.
h) Compute σ2.
21. Approximately 8% of all people have blue eyes. Out of a random sample of 20 people, find the following.
a) What is the probability that 2 of them have blue eyes?
b) What is the probability that at most 2 of them have blue eyes?
c) What is the probability that less than 2 of them have blue eyes?
d) What is the probability that at least 2 of them have blue eyes?
e) What is the probability that more than 2 of them have blue eyes?
f) Compute μ.
g) Compute σ.
h) Compute σ2.
22. An unprepared student takes a 10 question TRUE/FALSE quiz and ended up guessing each answer.
a) What is the probability that the student got 7 questions correct?
b) What is the probability that the student got 7 or more questions correct?
23. A local county has an unemployment rate of 7.3%. A random sample of 20 employable people are picked at random from the county and are asked if they are employed. The distribution is a binomial. Round answers to 4 decimal places.
a) Find the probability that exactly 3 in the sample are unemployed.
b) Find the probability that there are fewer than 4 in the sample are unemployed.
c) Find the probability that there are more than 2 in the sample are unemployed.
d) Find the probability that there are at most 4 in the sample are unemployed.
24. About 1% of the population has a particular genetic mutation. Find the standard deviation for the number of people with the genetic mutation in a group of 100 randomly selected people from the population.
25. You really struggle remembering to bring your lunch to work. Each day seems to be independent as to whether you remember to bring your lunch or not. The chance that you forget your lunch each day is 25.6%. Consider the next 48 days. Let X be the number of days that you forget your lunch out of the 48 days. Compute P(10 ≤ X ≤ 14).
26. A flu vaccine has a 90% effective rate. If a random sample of 200 people are given the vaccine, what is the probability that at most 180 people did not get the flu?
27. The Lee family had 6 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Smith family had at least 4 girls?
28. If a seed is planted, it has a 70% chance of growing into a healthy plant. If 142 randomly selected seeds are planted, answer the following.
a) What is the probability that exactly 100 of them grow into a healthy plant?
b) What is the probability that less than 100 of them grow into a healthy plant?
c) What is the probability that more than 100 of them grow into a healthy plant?
d) What is the probability that exactly 103 of them grow into a healthy plant?
e) What is the probability that at least 103 of them grow into a healthy plant?
f) What is the probability that at most 103 of them grow into a healthy plant?
29. A manufacturing machine has a 6% defect rate. An inspector chooses 4 items at random.
a) What is the probability that at least one will have a defect?
b) What is the probability that exactly two will have a defect?
c) What is the probability that less than two will have a defect?
d) What is the probability that more than one will have a defect?
30. A large fast-food restaurant is having a promotional game where game pieces can be found on various products. Customers can win food or cash prizes. According to the company, the probability of winning a prize (large or small) with any eligible purchase is 0.162. Consider your next 33 purchases that produce a game piece. Calculate the following:
a) What is the probability that you win 5 prizes?
b) What is the probability that you win more than 8 prizes?
c) What is the probability that you win between 3 and 7 (inclusive) prizes? d) What is the probability that you win 3 prizes or fewer?
31. A small regional carrier accepted 20 reservations for a particular flight with 17 seats. 15 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 60% chance, independently of each other.
a) Find the probability that overbooking occurs.
b) Find the probability that the flight has empty seats.
32. A poll is given, showing 72% are in favor of a new building project. Let X be the number of people who favor the new building project when 37 people are chosen at random. What is the probability that between 10 and 16 (including 10 and 16) people out of 37 favor the new building project?
33. A committee of 5 people is to be formed from 10 students and 7 parents. Compute the probability that the committee will consist of exactly 3 students and 2 parents.
34. In endurance horse racing, people over the age of 15 are called seniors and people 15 years or younger are called juniors. An endurance horse race consists of 25 seniors and 5 juniors. What is the probability that the top three finishers were:
a) All seniors.
b) All juniors.
c) 2 seniors and one junior.
d) 1 senior and two juniors.
35. A bag contains 9 strawberry Starbursts and 21 other flavored Starbursts. 5 Starbursts are chosen randomly without replacement. Find the probability that 3 of the Starbursts drawn are strawberry.
36. A jury selection room has 12 people that are married and 20 people that are not married to choose from. What is the probability that in a jury of 12 randomly selected people, exactly 3 of them would be married?
37. A pharmaceutical company receives large shipments of ibuprofen tablets and uses this acceptance sampling plan: randomly select and test 25 tablets, then accept the whole batch if there is at most one that doesn't meet the required specifications. If a particular shipment of 100 ibuprofen tablets actually has 5 tablets that have defects, what is the probability that this whole shipment will be accepted?
38. In a shipment of 24 keyboards, there are 3 that are defective (a success is a defective keyboard). A random sample of 4 keyboards are selected. The shipment will be returned if one or more of the 4 keyboards in the sample is defective. What is the probability that the shipment will be returned?
39. A writer makes on average one typographical error every page. The writer has landed a 3-page article in an important magazine. If the magazine editor finds any typographical errors, they probably will not ask the writer for any more material. What is the probability that the reporter made no typographical errors for the 3-page article?
40. A coffee shop serves an average of 75 customers per hour during the morning rush, which follows a Poisson distribution. Find the following.
a) Compute the probability that 80 customers arrive in an hour during tomorrow’s morning rush.
b) Compute the probability that less than 60 customers arrive in an hour during tomorrow’s morning rush.
c) Compute the probability that more than 60 customers arrive in an hour during tomorrow’s morning rush.
41. Suppose a random variable, x, follows a Poisson distribution. Let μ = 2.5 every minute, find the following probabilities.
a) P(X = 5) over a minute.
b) P(X < 5) over a minute.
c) P(X ≤ 5) over a minute.
d) P(X > 5) over a minute.
e) P(X ≥ 5) over a minute.
f) P(X = 125) over an hour.
g) P(X ≥ 125) over an hour.
42. The PSU computer help line receives, on average, 14 calls per hour asking for assistance. What is the probability that the company will receive more than 20 calls per hour?
43. There are on average 5 old-growth Sitka Spruce trees per 1/8 of an acre in a local forest.
a) Compute the probability that there are exactly 30 Sitka Spruce trees in 1 acre.
b) Compute the probability that there are more than 8 Sitka Spruce trees in a ¼ acre.
44. The number of rescue calls received by Pacific Northwest Search & Rescue follows a Poisson distribution with an average of 2.83 rescues every eight hours.
a) What is the probability that the squad will have exactly 4 calls in two hours?
b) What is the probability that the company will receive 2 calls in a 12-minute period?
c) What is the probability that the squad will have at most 2 calls in an hour?
45. Suppose a random variable, x, follows a Poisson distribution. Let μ = 3 every day; compute the P(X ≤ 12) over a week.
- Answer to Odd Numbered Exercises
-
1) a) Yes b) No c) No
3) a) -$0.425 b) $2.9592
5) $69.283
7) a) x -112.1 887.9 P(X = x) 0.76 0.24 b) $127.90 c) For many of these extended warranties bought by customers, they can expect to gain 127.9 dollars per warranty on average.
9) a) Yes, Σ P(x) = 1 and 0 ≤ P(x) ≤ 1 b) 0.8 c) 0.9055 d) 0.82
11) 0.0729
13) 0.0156
15) a) 0.1728 b) 0.9021 c) 0.7292 d) 0.0979 e) 0.2708
17) 1.2583
19) a) 0.1445 b) 0.5722 c) 0.5722 d) 15 e) 2.7386
21) a) 0.2711 b) 0.7879 c) 0.5169 d) 0.4831 e) 0.2121 f) 1.6 g) 1.2133 h) 1.472
23) a) 0.1222 b) 0.9464 c) 0.1759 d) 0.9873
25) 0.5923
27) 0.3438
29) a) 0.2193 b) 0.0191 c) 0.9801 d) 0.0199
31) a) 0.6826 b) 0.087
33) 0.4072
35) 0.1238
37) 0.6328
39) 0.0498
41) a) 0.0668 b) 0.8912 c) 0.958 d) 0.042 e) 0.1088 f) 0.0039 g) 0.9835
43) a) 0.0185 b) 0.6672
45) 0.0245
5.08: Chapter 5 Formulas
|
Discrete Distribution Table: 0 ≤ P(xi) ≤ 1 ∑ P(xi) = 1 |
Discrete Distribution Mean: μ = Σ(xi ∙ P(xi)) |
| Discrete Distribution Variance: σ2 = ∑(xi2 ∙P(xi)) – μ2 | Discrete Distribution Standard Deviation: σ = \(\sqrt {\sigma ^{2}}\) |
| Geometric Distribution: P(X = x) = p ∙ q (x – 1) , x = 1, 2, 3, … |
Geometric Distribution Mean: μ = \(\frac {1}{p}\) Variance: σ2 = \(\frac {1−p}{p ^{2}}\) Standard Deviation: σ = \(\sqrt \frac {1-p}{p ^{2}}\) |
| Binomial Distribution: P(X = x) = nCx·px ·q(n-x) , x = 0, 1, 2, … , n |
Binomial Distribution Mean: μ = n ∙ p Variance: σ 2 = n ∙ p ∙ q Standard Deviation: σ = \(\sqrt n \cdot p \cdot q\) |
| Hypergeometric Distribution: P(X = x) = \(\frac{a C_{x} \cdot {}_b C_{n-x}}{ _{N} C_{n}}\) |
p = P(success) q = P(failure) = 1 – p n = sample size N = population size |
| Unit Change for Poisson Distribution: New μ = old μ(\(\frac{\text { new units }}{\text { old units }}\)) | Poisson Distribution: P(X = x) = \(\frac{e^{-\mu} \mu^{x}}{x !}\) |
From Mostly Harmless Statistics by Rachel L. Webb, adapted from LibreTexts. Licensed CC BY-SA 4.0. XYZ Homework OER web edition, adapted with 7 verified corrections (see errata).