• The probability distribution, which tells us which values a variable takes, and how often it takes them.




• The mean of the random variable, which tells us the long-run average value that the random variable takes.




• The standard deviation of the random variable, which tells us a typical (or long-run average) distance between the mean of the random variable and the values it takes.




• Binomial Experiment : Binomial experiments are random experiments that consist of a fixed number of repeated trials, like tossing a coin 10 times, randomly choosing 10 people, rolling a die 5 times, etc. These trials, however, need to be independent in the sense that the outcome in one trial has no effect on the outcome in other trials. In each of these repeated trials there is one outcome that is of interest to us (we call this outcome "success"), and each of the trials is identical in the sense that the probability that the trial will end in a "success" is the same in each of the trials. So for example, if our experiment is tossing a coin 10 times, and we are interested in the outcome "heads" (our "success"), then this will be a binomial experiment, since the 10 trials are independent, and the probability of success is 1/2 in each of the 10 trials. Let's summarize and give more examples.




• To summarize, the requirements for a random experiment to be a binomial experiment are :




1. a fixed number (n) of trials




2. each trial must be independent of the others




3. each trial has just two possible outcomes, called "success" (the outcome of interest) and "failure"




4. there is a constant probability (p) of success for each trial, the complement of which is the probability (1 - p) of failure




5. In binomial random experiments, the number of successes in n trials is random. It can be as low as 0, if all the trials end up in failure, or as high as n, if all n trials end in success.




6. The random variable X that represents the number of successes in those n trials is called binomial, and is determined by the values of n and p. We say, "X is binomial with n = ... and p = ..."




• Example: Random Experiments (Binomial or Not?)




• Let's consider a few random experiments.




• In each of them, we'll decide whether the random variable is binomial. If it is, we'll determine the values for n and p. If it isn't, we'll explain why not.




1. A fair coin is flipped 20 times; X represents the number of heads. X is binomial with n = 20 and p = 0.5.




2. You roll a fair die 50 times; X is the number of times you get a six. X is binomial with n = 50 and p = 1/6.




3. Roll a fair die repeatedly; X is the number of rolls it takes to get a six. X is not binomial, because the number of trials is not fixed.




4. Draw 3 cards at random, one after the other, without replacement, from a set of 4 cards consisting of one club, one diamond, one heart, and one spade; X is the number of diamonds selected. X is not binomial, because the selections are not independent. (The probability (p) of success is not constant, because it is affected by previous selections.)




5. Draw 3 cards at random, one after the other, with replacement, from a set of 4 cards consisting of one club, one diamond, one heart, and one spade; X is the number of diamonds selected. Sampling with replacement ensures independence. X is binomial with n = 3 and p = 1/4.




6. Approximately 1 in every 20 children has a certain disease. Let X be the number of children with the disease out of a random sample of 100 children. Although the children are sampled without replacement, it is assumed that we are sampling from such a vast population that the selections are virtually independent. X is binomial with n = 100 and p = 1/20 = 0.05.




7. The probability of having blood type B is 0.1. Choose 4 people at random; X is the number with blood type B. X is binomial with n = 4 and p = 0.1.




8. A student answers 10 quiz questions completely at random; the first five are true/false, the second five are multiple choice, with four options each. X represents the number of correct answers. X is not binomial, because p changes from 1/2 to 1/4.

• Example 4 above was not binomial because sampling without replacement resulted in dependent selections. In particular, the probability of the second card being a diamond is very dependent on whether or not the first card was a diamond: the probability is 0 if the first card was a diamond, 1/3 if the first card was not a diamond.




• In contrast, Example 5 was binomial because sampling with replacement resulted in independent selections: the probability of any of the 3 cards being a diamond is 1/4 no matter what the previous selections have been.




• On the other hand, when you take a relatively small random sample of subjects from a large population, even though the sampling is without replacement, we can assume independence because the mathematical effect of removing one individual from a very large population on the next selection is negligible. For example, in Example 6, we sampled 100 children out of the population of all children. Even though we sampled the children without replacement, whether one child has the disease or not really has no effect on whether another child has the disease or not.




• The convention is to "fudge" the requirement of independence as long as the population is at least 10 times the sample size.




• Rule of Thumb The number (X) of successes in a sample of size n taken without replacement from a population with proportion (p) of successes is approximately binomial with n and p as long as the sample size (n) is at most 10% of the population size (N).




• In symbols, this would be: n ≤ 0.10N. This is the same as saying the population size is greater than or equal to 10 times the sample size. In symbols this is: N ≥ 10n

View Answer



View Answer



View Answer



View Answer



View Answer



View Answer



View Answer



View Answer



View Answer



View Answer



View Answer

• Consider a regular deck of 52 cards, in which there are 13 cards of each suit: hearts, diamonds, clubs and spades. We select 3 cards at random with replacement. Let X be the number of diamond cards we got (out of the 3).




• We have 3 trials here, and they are independent (since the selection is with replacement). The outcome of each trial can be either success (diamond) or failure (not diamond), and the probability of success is 1/4 in each of the trials.




• X, then, is binomial with n = 3 and p = 1/4.




• Let's build the probability distribution of X as we did in the chapter on probability distributions. Recall that we begin with a table in which we:




• record all possible outcomes in 3 selections, where each selection may result in success (a diamond, D) or failure (a non-diamond, N).




• find the value of X that corresponds to each outcome.




• use simple probability principles to find the probability of each outcome.




• 




• With the help of the addition principle, we condense the information in this table to construct the actual probability distribution table:




• 




• In order to establish a general formula for the probability that a binomial random variable X takes any given value x, we will look for patterns in the above distribution. From the way we constructed this probability distribution, we know that, in general:




• 




• Let's start with the second part, the probability that there will be x successes out of 3, where the probability of success is 1/4. Notice that the fractions multiplied in each case are for the probability of x successes (where each success has a probability of p = 1/4) and the remaining (3 - x) failures (where each failure has probability of 1 - p = 3/4).




• 




• So in general:




• 




• Let's move on to talk about the number of possible outcomes with x successes out of three. Here it is harder to see the pattern, so we'll give the following mathematical result.




• Result Consider a random experiment that consists of n trials, each one ending up in either success or failure. The number of possible outcomes in the sample space that have exactly k successes out of n is:




• \( \frac{n!}{k!(n - k)!} \)




• Note that n! is read "n factorial" and is defined to be the product 1 * 2 * 3 * ... * n. 0! is defined to be 1.

• You choose 12 male college students at random and record whether they have any ear piercings (success) or not. There are many possible outcomes to this experiment (actually, 4,096 of them!).




• In how many of the possible outcomes of this experiment are there exactly 8 successes (students who have at least one ear pierced)?




• There is no way that we would start listing all these possible outcomes. The result above comes to our rescue.




• The result says that in an experiment like this, where you repeat a trial n times (in our case, we repeat it n = 12 times, once for each student we choose), the number of possible outcomes with exactly 8 successes (out of 12) is:




• \( \frac{n!}{k!(n - k)!} = \frac{12!}{8!(12 - 8)!} = 495 \)

• Let's go back to our example, in which we have n = 3 trials (selecting 3 cards). We saw that there were 3 possible outcomes with exactly 2 successes out of 3. The result confirms this since:




• \( \frac{n!}{k!(n - k)!} = \frac{3!}{2!(3 - 2)!} = 3 \)




• In general, then




• 




• Putting it all together, we get that the probability distribution of X, which is binomial with n = 3 and p = 1/4 is:




• \( P(X=x) = \frac{3!}{x!(3-x)!}(1/4)^x(3/4)^{3-x} for x = 0, 1, 2, 3... \)




• In general, the number of ways to get x successes (and n - x failures) in n trials is \( \frac{n!}{k!(n - k)!} \). Therefore, the probability of x successes (and n - x failures) in n trials, where the probability of success in each trial is p (and the probability of failure is 1 - p) is equal to the number of outcomes in which there are x successes out of n trials, times the probability of x successes, times the probability of n - x failures:




• \( P(X=x) = \frac{n!}{x!(n-x)!}(p)^x(1-p)^{n-x} for x = 0, 1, 2, 3... \)




• where x may take any value 0, 1, ... , n.

• Example: Blood Type A




1. The probability of having blood type A is 0.4. Choose 4 people at random and let X be the number with blood type A.




2. X is a binomial random variable with n = 4 and p = 0.4.




3. As a review, let's first find the probability distribution of X the long way: construct an interim table of all possible outcomes in S, the corresponding values of X, and probabilities. Then construct the probability distribution table for X.




4. 




5. As usual, the addition rule lets us combine probabilities for each possible value of X:




6. 




7. Now let's apply the formula for the probability distribution of a binomial random variable, and see that by using it, we get exactly what we got the long way.




8. Recall that the general formula for the probability distribution of a binomial random variable with n trials and probability of success p is:




9. \( P(X = x) = \frac{n!}{x!(n - x)!}p^x(1-p)^{n-x} \) for x = 0, 1, 2, 3, ... , n




10. In our case, X is a binomial random variable with n = 4 and p = 0.4, so its probability distribution is:




11. \( P(X = x) = \frac{4!}{x!(4 - x)!}(0.4)^x(0.6)^{4-x} \) for x = 0, 1, 2, 3, 4




12. Let's use this formula to find P(X = 2) and see that we get exactly what we got before.




13. \( P(X = 2) = \frac{4!}{2!(4 - 2)!}(0.4)^2(0.6)^{4-2} = \frac{1 * 2 * 3 * 4}{(1 * 2)(1 * 2)} (0.4)^2(0.6)^{2} = 0.3456\)




• Scenario: Multiple-Choice Test




1. A multiple choice test has 10 questions, each with 5 possible answers, only one of which is correct. A student who did not study is absolutely clueless, and therefore uses an independent random guess to answer each of the 10 questions. Let X be the number of questions the student gets right.




2. X has a binomial distribution. What is the value of the parameter n? Since the multiple-choice test has 10 questions, n = 10.




3. X has a binomial distribution. What is the value of the parameter p? Report your answer to one decimal place. Since each multiple-choice question has 5 options, the probability of guessing the correct answer is 0.2.




4. What is the probability that the student gets exactly 4 questions right, P(X = 4)? Round your answer to TWO decimal places. Plugging in X = 4 in the probability distribution formula we get P(X = 4) = 0.0881 = 0.09.

• Each student in a group of 15 students is asked to each pick a number from 1 to 20 completely at random. 3 of the 15 happen to pick the number 7 (this is a probability of 0.20). Is this an improbably high proportion to choose a particular number?




• If the selections are truly random, then each number from 1 to 20, including 7, has probability p = 1/20 = .05 of being selected. The number of trials is n = 15. The probability of at least 3 successes in 15 trials, when each trial has probability of success 0.05, can be found by applying the binomial formula.




• To make the notation easier, we will use a shorthand notation for the number of possible outcomes with x successes out of n. \( \frac{n!}{x!(n - x)!} \) will be written as: \( {n}\choose{x} \) .




• P(X ≥ 3) = P(X = 3) + P(X = 4) + .. + P(X = 15)




• \( {15}\choose{3} \)\((0.05)^3(0.95)^{12} + \)\({15}\choose{4}\)\((0.05)^4(0.95)^{11} + ... + \)\({15}\choose{15}\)\((0.05)^{15}(0.95)^{0} \)




• \( = 0.0307 + 0.0049 + 0.0006 + ... = 0.0362 \)




• where all remaining terms after the first 3 are less than 0.0001. The probability of at least 3 out of 15 people picking 7, when choosing at random from the numbers 1 to 20, is only 0.0362. Thus, 3 out of 15 is rather improbably high. People may think they are choosing at random, but in fact they tend to favor certain numbers, like the number 7.

• Past studies have shown that 90% of the booked passengers actually arrive for a flight. Suppose that a small shuttle plane has 45 seats. We will assume that passengers arrive independently of each other. (This assumption is not really accurate, since not all people travel alone, but we'll use it for the purposes of our experiment).




• Many times airlines "overbook" flights. This means that the airline sells more tickets than there are seats on the plane. This is due to the fact that sometimes passengers don't show up, and the plane must be flown with empty seats. However, if they do overbook, they run the risk of having more passengers than seats. So, some passengers may be unhappy. They also have the extra expense of putting those passengers on another flight and possibly supplying lodging.




• With these risks in mind, the airline decides to sell more than 45 tickets. If they wish to keep the probability of having more than 45 passengers show up to get on the flight to less than 0.05, how many tickets should they sell?




• This is a binomial random variable that represents the number of passengers that show up for the flight. It has p = 0.90, and n to be determined.




• Suppose the airline sells 50 tickets. Now we have n = 50 and p = 0.90. We want to know P(X > 45), which is 1 - P(X ≤ 45) = 1 - 0.57 or 0.43. Obviously, all the details of this calculation were not shown, since a statistical technology package was used to calculate the answer. This is certainly more than 0.05, so the airline must sell fewer seats.




• If we reduce the number of tickets sold, we should be able to reduce this probability. We have calculated the probabilities in the following table:




• # tickets sold	P(X > 45)
  50	0.43
  49	0.26
  48	0.13
  47	0.04
  46	0.008




• From this table, we can see that by selling 47 tickets, the airline can reduce the probability that it will have more passengers show up than there are seats to less than 5%.

• In vitro fertilization is becoming more and more common these days. Suppose each embryo that is implanted has a 20% chance of resulting in a pregnancy that results in delivering a baby. Also, assume that each embryo's chance of surviving and resulting in a baby is independent of the others.




• It is an expensive procedure, so we want to do it only once. We wish to try to find the optimum number of embryos to implant so that the probability of at least 1 child being born is high, but the probability of more than 2 children being born is low. In other words, we want a baby, and we're willing to have twins, but we don't want triplets, quadruplets, etc.




• Note that unlike the airline flight example, where we needed to control only one probability, in this case there are two probabilities that we wish to control.




• We will let X represent the number of implanted embryos resulting in a baby. It is a binomial random variable with n = number of implanted embryos and p = 0.20 (the probability that an implanted embryo results in a baby).




• It is customary to implant between n = 1 and n = 7 embryos. We have provided a table that contains the two probabilities mentioned in the previous question, for values of n ranging from 1 to 7.




• # embryos	P(X ≥ 1)	P(X > 2)
  1	0.20	0
  2	0.36	0
  3	0.488	0.008
  4	0.590	0.027
  5	0.672	0.058
  6	0.738	0.099
  7	0.790	0.148




• Using the table of probabilities provided and our conditions that we want P(X ≥ 1) to be high, while keeping P(X > 2) low, decide how many embryos you'd implant and why.




• Even though the probability of having one child keeps increasing, so does the probability of having more than two children. We think the optimum number is five embryos. You have almost a 70% chance of a baby, while keeping the probability of triplets, quadruplets, etc. to well under 10%.

View Answer



View Answer



View Answer



View Answer

• Example: Blood Type B—Mean




1. Overall, the proportion of people with blood type B is 0.1. In other words, roughly 10% of the population has blood type B.




2. Suppose we sample 120 people at random. On average, how many would you expect to have blood type B?




3. The answer, 12, seems obvious; automatically, you'd multiply the number of people, 120, by the probability of blood type B, 0.1. This suggests the general formula for finding the mean of a binomial random variable:




4. Claim : If X is binomial with parameters n and p, then




5. \( \mu_x = n*p \)




6. Although the formula for mean is quite intuitive, it is not at all obvious what the variance and standard deviation should be. It turns out that:




7. Claim : If X is binomial with parameters n and p, then




8. \( \sigma^2_x = np(1-p) ; \sigma_x = \sqrt{np(1-p)} \)




9. For those who are interested, read below to see how these formulas were derived.

How are formulas for the mean and standard deviation of a binomial random variable derived?

• For those who are interested, these formulas may be derived with the aid of the rule for variance of a sum of random variables.




• We think of our binomial X with n and p as the sum of n identical "Bernoulli" random variables X1 through Xn,




• each representing one of the n trials. Each of the Xi has just two possibilities: success (X = 1, with probability p), or failure (X = 0, with probability 1 - p), as shown in the probability distribution table below.




• 




• The mean of each Xi is 0(1 - p) + 1(p) = p, and the variance is




• 




• Because variance is additive, if we sum up n such random variables to form the binomial random variable X = X1 + ... + Xn, its variance will be the sum of n variances p(1 - p), or np(1 - p).

Comment : The binomial mean and variance are special cases of our general formulas for the mean and variance of any random variable.

• \( \mu_x = np ; \sigma^2_x = np(1-p) ; \sigma = \sqrt{np(1-p)} \)

Example: Blood Type B—Standard Deviation

• Suppose we sample 120 people at random. The number with blood type B should be about 12, give or take how many? In other words, what is the standard deviation of the number X who have blood type B?




• Since n = 120 and p = .1,




• \( \sigma^2_x = 120(0.1)(1-0.1) = 10.8; \sigma_x = \sqrt{10.8} = 3.3 \)




• In a random sample of 120 people, we should expect there to be about 12 with blood type B, give or take about 3.3.

A Gallup Poll in May 2004 estimated that roughly 70% of U.S. adults are in favor of the death penalty for a person convicted of murder. A random sample of 750 U.S. adults is chosen. Let X be the number of adults (out of 750) who favor the death penalty. The distribution of X is binomial. What are the values of n and p?

• Each sampled adult is a trial, so n = 750. In this case "success" (our outcome of interest) is "favoring the death penalty for convicted murderers." The Gallup Poll estimates that p = 0.7.

View Answer

• Using the figure below we will see that for different values of n and p, binomial distributions can be symmetric, skewed right or skewed left.




• 




• What is the shape of the binomial distributions for p = 0.5 and the different values of n? Does the general shape change as n increases from 10 to 60? When p=0.5,the distribution is symmetric for any value of n from 10 to 60.




• With the value of p to 0.2, and set n all the way to the left to n=10. What is the shape of the distribution? The distribution is not symmetric. Most of it is on the left, with a long tail to the right. (skewed right)




• As the value of n increases with p constant what would happen to the shape of the distribution? The distribution shape becomes roughly symmetric when n is large with most of the data (high bars) in the middle and very little data (low bars) in the extremes.




• With the value of p to 0.8, and set n all the way to the left to n=10. What is the shape of the distribution? The distribution is not symmetric. Most of it is on the right, with a tail to the left. (skewed left)




• Increase the value of n with p constant. As the value of n increases, what would happen to the shape of the distribution? The distribution shape becomes roughly symmetric when n is large with most of the data (high bars) in the middle and very little data (low bars) in the extremes.

• In order to shift our focus from discrete to continuous random variables, let us first consider the probability histogram below for the shoe size of adult males. Let X represent these shoe sizes. Thus, X is a discrete random variable, since shoe sizes can only take whole and half number values, nothing in between.




• 




• Recall that in all of the previous probability histograms we've seen, the X-values were whole numbers. Thus, the width of each bar was 1. The height of each bar was the same as the probability for its corresponding X-value.




• Due to the principle that states the sum of probabilities of all possible outcomes in the sample space must be 1, the heights of all the rectangles in the histogram must sum to 1. This meant that the area was also 1.




• This histogram uses half-sizes. We wish to keep the area = 1, but we still want the horizontal scale to represent half-sizes. Therefore, we must adjust the vertical scale of the histogram.




• As is, the total area of the histogram rectangles would be 0.50 times the sum of the probabilities, since the width of each bar is 0.50. Thus, the area is 0.50(1) = 0.50. If we double the vertical scale, the area will double and be 1, just like we want.




• This means we are changing the vertical scale from "Probability" to "Probability per half size." The shape and the horizontal scale remain unchanged.




• 




• Now we can tell the probability of shoe size taking a value in any interval, just by finding the area of the rectangles over that interval. For instance, the area of the rectangles up to and including 9 shows the probability of having a shoe size less than or equal to 9.




• 




• Recall that for a discrete random variable like shoe size, the probability is affected by whether we want strict inequality or not. For example, the area—and corresponding probability—is reduced if we only consider shoe sizes strictly less than 9:




• 




• Transition to Continuous Random Variables




1. Now we are going to be making the transition from discrete to continuous random variables. Recall that continuous random variables represent measurements and can take on any value within an interval.




2. For our shoe size example, this would mean measuring shoe sizes in smaller units, such as tenths, or hundredths. As the number of intervals increases, the width of the bars becomes narrower and narrower, and the graph approaches a smooth curve.




3. To illustrate this, the following graphs represent two steps in this process of narrowing the widths of the intervals. Specifically, the interval widths are .25 and .10.




4. 

The shaded area in the probability histogram below represents the probability of a male's shoe size being:

• 

View Answer

• Now consider another random variable X = foot length of adult males. Unlike shoe size, this variable is not limited to distinct, separate values, because foot lengths can take any value over a continuous range of possibilities, so we cannot present this variable with a probability histogram or a table.




• The probability distribution of foot length (or any other continuous random variable) can be represented by a smooth curve called a probability density curve.




• 




• Like the modified probability histogram above, the total area under the density curve equals 1, and the curve represents probabilities by area.




• The probability that X gets values in any interval is represented by the area above this interval and below the density curve. In our foot length example, if our interval of interest is between 10 and 12 (marked in red below), and we would like to know P(10 < X < 12), the probability that a randomly chosen male has a foot length anywhere between 10 and 12 inches, we'll have to find the area above our interval of interest (10,12) and below our density curve, shaded in blue:




• 




• If, for example, we are interested in P(X < 9), the probability that a randomly chosen male has a foot length of less than 9 inches, we'll have to find the area shaded in blue below:




• 




• Comments




1. We have seen that for a discrete random variable like shoe size, whether we have a strict inequality or not does matter when solving for probabilities. In contrast, for a continuous random variable like foot length, the probability of a foot length of less than or equal to 9 will be the same as the probability of a foot length of strictly less than 9. In other words,




2. \( P(X < 9) = P(X \leq 9) \)




3. Visually, in terms of our density curve, the area under the curve up to and including a certain point is the same as the area up to and excluding the point, because there is no area over a single point. Conceptually, because a continuous random variable has infinitely many possible values, technically the probability of any single value occurring is zero!




4. It should be clear now why the total area under any probability density curve must be 1. The total area under the curve represents P(X gets a value in the interval of its possible values). Clearly, according to the rules of probability this must be 1, or always true.




5. Density curves, like probability histograms, may have any shape imaginable as long as the total area underneath the curve is 1.




• Let's Summarize :




1. The probability distribution of a continuous random variable is represented by a probability density curve.




2. The probability that X gets a value in any interval of interest is the area above this interval and below the density curve.




3. 




4. Now that we see how probabilities are found for continuous random variables, we understand why it is more complicated than finding probabilities in the discrete case. As anyone who has studied calculus can attest, finding the area under a curve can be difficult. The general approach is to use integrals. For those of you who did study calculus, the following should be familiar....




5. \( P(a \leq X \leq b) = (\text{area between a and b and below the density curve}) = \int_{a}^{b} f(x) dx\) , where f(x) represents the density curve.

Previous

Next