Lecture: Binomial & Normal Models
Lecture script · about 15–20 minutes spoken · Week 7
Opening
Picture a classmate who didn't study for a 10-question quiz. Every question has four options, and they plan to guess on all of them. Could they pass? Could luck carry them to 5 or more right? By the end of this lecture you'll be able to put a number on that, and you won't have to list every possible answer sheet to do it.
Counting successes in repeated tries is everyday work. A clinic manager plans for patients who miss their appointments. A coach tracks a player's free throws. A quality inspector counts faulty parts in a shipment. All of them reach for the same model, the binomial. And when the counts get large, they reach for a smooth curve called the normal curve.
Here's where we are. In Weeks 1 through 4 you learned to describe data you already had. In Week 5 you learned the rules of chance: the complement rule, the addition rule, and the multiplication rule for independent events. In Week 6 you met random variables, numbers that chance decides, each with a mean and a standard deviation. Today we build one special random variable out of those rules. Next week, in the midterm review, you'll add z-scores, and in Week 9 the normal curve gets a full week of its own.
We'll take eight steps: the binomial setting, checking its conditions, the formula, technology, the mean and standard deviation, the shape, when a curve can stand in, and a first look at the normal curve itself. Most of our examples come from that guessing classmate, with stops at a raffle jar, a ride-share driver, a basketball court, and a clinic.
1. The binomial setting: count the successes in a fixed number of tries
Go back to the guessing classmate. Each question is one try. Each guess is right or wrong. A guess picks one of four options, so it's right with chance 1 in 4, which is 0.25, on every question. One guess doesn't affect the next. At the end, you count the right answers.
That picture has a name. A trial is one repetition of a chance process whose result you record, like one guessed question. Each trial has two results, and we call them success and failure. A success is the result you're counting, whether or not it's a good thing. A binomial setting is a chance process with a fixed number of trials n, trials that are independent of one another, exactly two outcomes per trial, and the same probability of success p on every trial. The count of successes is a binomial random variable, and we call it X.
For the quiz, n is 10 and p is 0.25. So the chance a single guess is wrong is 0.75.
Here's a second example from a different place. Suppose a basketball player makes each free throw with chance 0.8, and one shot doesn't affect the next. Over 5 shots, the number of makes is binomial with n of 5 and p of 0.8. Notice the word "suppose". Whether the shots really are independent is an assumption a coach would want to check.
Pause and try this. A clinic counts patients who miss their appointment. In that count, what is a success?
Decide what the count is counting.
Here's the answer. A missed appointment is the success. A success is whatever result you count, even one nobody is happy about.
2. Check the conditions: catch the setting that breaks them
The binomial model is only as good as its four conditions: fixed n, independent trials, two outcomes, and the same p. If one fails, a formula will still hand you a number. It won't answer your question.
Picture a jar of 10 raffle tickets, and 4 of them are winners. You draw 3 tickets and keep each one. On the first draw, the chance of a winner is 4 out of 10. If that first ticket wins, the second draw has 3 winners among 9 tickets. If it loses, there are 4 winners among 9. The chance moves from draw to draw, and it depends on what came before. So the draws aren't independent, p doesn't stay put, and the count of winners is not binomial. This is the classic trap of the week: drawing without replacement from a small group.
Other settings fail in other ways. A call center that phones people until the first one answers has no fixed n. A survey answer of agree, neutral, or disagree has three outcomes, though you can rescue it by counting agree as the success and everything else as a failure.
Here's an honest caveat. Real data are rarely perfect. Draw 3 people from a whole city, and the chance barely moves from one draw to the next, so the binomial is a fine working model. The trouble comes when the group is small, like 10 tickets in a jar.
Pause and try this. In the raffle jar, which two of the four conditions fail?
Ask whether one draw changes the next.
Here's the answer. Independence fails, and so does "the same p every time." Each ticket you keep changes what's left in the jar, so the chance of a winner changes.
3. The binomial formula: count the ways, then multiply
Let's do a tiny case by hand first. A ride-share driver gives 3 rides, and each rider tips with chance one half, independently. What's the chance exactly two riders tip? Write T for a tip and N for none. Two tips can land as tip-tip-none, tip-none-tip, or none-tip-tip. That's 3 arrangements. Each has chance one half times one half times one half, which is 1 out of 8. Add the three arrangements and you get 3 out of 8.
That tiny case holds the whole idea. The number of ways is how many different arrangements of k successes among n trials are possible. The binomial probability formula says: multiply the number of ways the k successes can land by p once for each success and by 1 minus p once for each failure.
Now the quiz. To get zero right, there's only 1 way: miss all ten. So you multiply 0.75 by itself 10 times. That comes out to 0.0563, about 5.6 percent of guessers. To get exactly one right, the right answer could sit on any of the 10 questions, so there are 10 ways. Multiply 10, times 0.25, times 0.75 nine times over. That's 0.1877, about 18.8 percent. For two right, there are 45 different pairs of questions, and the chance is about 28.2 percent.
Put all those counts together with their chances and you have the binomial distribution: the list of every possible count of successes, from 0 to n, together with the probability of each count.
Pause and try this. A student works out the chance of exactly one right guess and gets about 1.9 percent. What step did they skip?
Think about where the one right answer could be.
Here's the answer. They dropped the number of ways. About 1.9 percent is the chance of one particular arrangement. The right answer could be on any of 10 questions, so multiply by 10 to get about 18.8 percent.
4. Binomial probabilities by technology: one value or a running total
Counting the ways for 40 right out of 100 by hand would take all day. That's a job for a spreadsheet. In Google Sheets or Excel, the BINOM.DIST function takes four things: the count, n, p, and a true-or-false switch. Set the switch to false and you get the chance of exactly that count. Set it to true and you get a running total.
That running total has a name. A cumulative probability is the probability that a count is at or below a given value, found by adding the probabilities of every count from 0 up to that value.
So, what's the chance a guesser gets 2 or fewer right? Add the chances for 0, 1, and 2: 0.0563, plus 0.1877, plus 0.2816. That's 0.5256. A little over half of all guessers, about 53 in 100, get two or fewer right. The spreadsheet's running total for 2 gives you the same number in one step.
Now the question from the opening: 5 or more right. The chances for 5 through 10 are tiny and easy to mis-add. So use the complement rule from Week 5. "At least 5" fails exactly when the count is 4 or fewer. Take 1 minus the running total at 4, and the spreadsheet hands back 0.0781. A pure guesser scores 50 percent or better on about 8 tries in 100. So our classmate could get lucky, but it's not a plan.
Pause and try this. To find the chance of at least 5, someone takes 1 minus the running total at 5. What's wrong?
Ask which group a score of exactly 5 belongs to.
Here's the answer. A score of exactly 5 counts as "at least 5," so it can't be subtracted away. The complement of "at least 5" is "4 or fewer," so you subtract the running total at 4.
5. The binomial mean and standard deviation: the long-run center and the typical miss
Ten questions, each right a quarter of the time. How many right should a guesser average? A quarter of 10, which is 2.5. That's the whole formula. The binomial mean is the long-run average count of successes, n times p, and it need not be a count that can actually happen.
Spread works the same way, with one more step. The binomial standard deviation is the typical distance between a binomial count and its mean: multiply n, p, and 1 minus p, then take the square root. For the quiz, that's the square root of 10 times 0.25 times 0.75, about 1.37 correct answers.
So here's the sentence that matters. Over many guessed quizzes, a guesser averages 2.5 right, and a typical score lands about 1.37 answers away from that. Scores of 1, 2, 3, and 4 are ordinary for a guesser.
The free-throw shooter works the same way. Five shots at 0.8 each gives a mean of 4 makes. Over many sets of 5 free throws, the player averages 4.
Pause and try this. A friend says, "The mean is 2.5, so a guesser should expect to get 2.5 right on my quiz." What would you tell them?
Think about what one quiz can score.
Here's the answer. No single quiz can score 2.5, because the count is a whole number. The 2.5 is a long-run average over many guessed quizzes, as in Week 6.
6. The shape of a binomial: lopsided for small n, a bell for large n
Let me describe the 10-question distribution in words. The chances climb from 0 right to a peak at 2 right, then shrink. From 6 right on, they're nearly zero, and 9 and 10 round to zero. So the pile sits near the low end, and a long, thin tail runs toward the high counts. By the rule from Week 2, you name skew by the tail, so this is right-skewed.
Now imagine guessing on a 100-question test. The mean becomes 100 times 0.25, which is 25. The standard deviation becomes the square root of 18.75, about 4.33. Picture a bar for every score from 0 to 100. The tallest bars sit around 25, and they fall away almost evenly on both sides. It's a nearly symmetric bell.
What changed? At 10 questions, the mean of 2.5 sits only about 2 standard deviations above zero, so the pile is squashed against the floor. At 100 questions, the mean of 25 sits nearly 6 standard deviations above zero, so the pile has room to spread both ways.
Pause and try this. The 10-question pile is on the left. Is the distribution left-skewed or right-skewed?
Look for the tail, not the pile.
Here's the answer. Right-skewed. The long, thin tail runs toward the high scores on the right, and skew is named for the tail.
7. When a normal curve can stand in: the rule of thumb
A smooth bell is much easier to work with than 101 separate bars. But when is it honest to let a curve stand in? This course uses one working rule. The large-counts rule of thumb says a normal curve may approximate a binomial distribution when the expected number of successes, n times p, and the expected number of failures, n times 1 minus p, are both at least 10. Using the curve that way is called a normal approximation. Some courses use 5 instead of 10. It's a rule of thumb, not a law of nature.
Check the 10-question quiz. Expected successes: 10 times 0.25 is 2.5. Expected failures: 10 times 0.75 is 7.5. Since 2.5 is below 10, a curve is not an appropriate stand-in. That right skew is the reason. Now the 100-question test: 25 expected successes and 75 expected failures. Both clear 10, so a curve centered at 25 with a spread of about 4.33 is a reasonable stand-in.
The clinic shows the same split. Suppose each booked patient misses with chance 0.2. One morning with 12 bookings expects only 2.4 misses, below 10, so the scheduler uses the exact binomial. A week with 200 bookings expects 40 misses and 160 shows, so the manager can plan with the curve.
Pause and try this. A buyer checks 500 parts, and suppose each is faulty with chance 0.01. Is 500 big enough for the curve?
Compute the expected number of faulty parts.
Here's the answer. No. The expected count is 500 times 0.01, which is 5, below 10. A big n isn't enough on its own; the rule checks both expected counts.
8. A first look at the normal curve: the working model for many measurements
Picture those 101 bars for the 100-question test, and imagine sanding off their corners. What's left is a smooth hill: highest in the middle, falling away evenly on both sides, with tails that thin out but never quite touch zero.
That hill is a normal curve: a smooth, symmetric, bell-shaped model for a quantitative variable, set by its mean at the center and its standard deviation for the spread. The total area under it is 1, and the area over a range of values is the probability of landing in that range. You met "probability is area" in Week 6, and it carries straight over.
Two numbers settle everything. The mean places the peak. The standard deviation sets the width: small means tall and narrow, large means low and wide. For the guessing test, the matching curve peaks at 25 with a spread of about 4.33, and a score of 21 sits at the same height as a score of 29.
Many measurements get the same model. Suppose the coffee bags from one packing machine follow a normal model with mean 340 grams and standard deviation 3 grams. The curve peaks at 340 grams, and bags far from it in either direction are rare. Field biologists often model a body mass within one penguin species this way, too. But notice the word "suppose". A normal curve is a model you assume, never a fact about the data. In Week 9 you'll learn to check whether it's reasonable.
Pause and try this. Two normal curves share a mean of 340 grams. One has a standard deviation of 3 grams and the other a larger one. Which curve is taller and narrower?
Remember which number sets the width.
Here's the answer. The curve with the standard deviation of 3 grams. A smaller standard deviation squeezes the same total area into a narrower, taller bell.
Closing
Let's pull it together in the order we built it. A binomial counts successes in a fixed number of independent trials with two outcomes and the same chance each time. Check those conditions out loud, and watch for drawing without replacement from a small group. A binomial chance is the number of ways times the chance of one arrangement. A spreadsheet's BINOM.DIST gives one value or a running total, and the complement handles "at least". The mean is n times p, a long-run average. Small binomials lean to one side, and big ones look like bells. A curve can stand in when both expected counts reach 10. And a normal curve is a model set by its mean and standard deviation, where area is probability.
Next in this module come the Readings & Resources page, the lecture tutorial, the practice set, and the quiz, then the discussion, the assignment, and the Data Lab. Next week, in Week 8, Midterm Review & Exam, you'll pull the first seven weeks together and meet z-scores, the measuring stick the normal curve is built on.
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