Week 7 · Chapter

Chapter 7 — Binomial & Normal Models

Introduction to Statistics Free sample edition

The module's primary reading · ~40–50 minutes · Objective 4 (apply the basic rules of probability, including conditional probability and independence, and compute and interpret the expected value and standard deviation of a discrete random variable, including binomial settings), Objective 5 (use the normal model and sampling distributions, including the Central Limit Theorem, to reason about variability in individual values and in sample statistics) · Read it first this week; the lecture, the tutorial, the quiz, and the Data Lab all build on it.

In plain language

Many questions in life come down to counting how many times something happens in a set number of tries, such as how many of 10 guessed quiz answers are right. This week you learn the binomial model (a ready-made probability model for counting successes in repeated, independent tries), so you can find those chances without listing every outcome. You will also find the average count and its typical spread. Then you meet the normal curve (a smooth, bell-shaped model), see when it can stand in for a count, and see why it is the working model for many measurements.

Why you already care about this

Picture a 10-question multiple-choice quiz with four options per question. A classmate did not study and plans to guess on every question. Will they pass? Could they score 5 or more by luck? On a 100-question test, would guessing ever get them close to half? These are questions about a count, and chance decides the count.

The same shape of question shows up far from a classroom. A clinic books 12 appointments and wonders how many patients will not show up. A basketball player takes 5 free throws and a coach wonders how many will go in. A factory ships 400 parts and a buyer wonders how many are faulty. In each case there is a fixed number of tries, each try either works or does not, and you want the chance of each possible count. Last week you learned that a random variable is a number chance decides, with a long-run average. This week you meet the most useful random variable for counting.

By the end of this chapter you will be able to:

check the four conditions of a binomial setting (a fixed number of independent tries with the same chance of success), and spot a setting that breaks them;
find a binomial probability by formula for a small case, counting the number of ways;
find exact and running-total binomial probabilities in a spreadsheet, including "at least" chances;
compute and interpret the binomial mean and standard deviation;
decide with the rule of thumb whether a normal curve may stand in for a binomial, and describe what a normal curve is.

Carry this sentence with you: a binomial chance is every arrangement added — not one arrangement.

Where we are

Already covered: Weeks 1 to 4 taught you to describe data. Weeks 5 and 6 taught you to measure chance and to give a random variable a mean and a standard deviation.
This week: the binomial model for counting successes in n independent tries, its mean and standard deviation, its shape, and a first look at the normal curve.
How they connect: each binomial chance is built from Week 5's multiplication and addition rules, and the binomial mean and standard deviation are Week 6's expected value and standard deviation for one special random variable.
Where we're going next: Week 8 reviews the first seven weeks before the midterm and adds z-scores, and Week 9 returns to the normal curve to find areas under it.

Each earlier week feeds this one. Week 2 taught you to read the shape of a distribution and to name the skew by its tail. Week 3 gave you the standard deviation as a typical distance from the mean. Week 5 gave you the complement rule, the addition rule for outcomes that cannot happen together, and the multiplication rule for independent events. Week 6 turned those into a random variable X with an expected value, also called its mean μ, and a standard deviation σ. It also showed you that for a smooth curve, probability is area. This chapter uses every one of those tools.

1. The binomial setting: count the successes in a fixed number of tries

Go back to the classmate who guesses on all 10 questions. Each question is one try. On each try the guess is either right or wrong. A pure guess picks one of four options, so it is right with chance 1 ⁄ 4 = 0.25 on every question. And one guess has no effect on the next. At the end, you count the right answers. That count is what this week is about.

A trial is one repetition of a chance process whose result you record, such as one guessed quiz question or one free throw.
Success and failure are the two labels given to the two possible results of a trial in a binomial setting; a success is the result you are counting, whether or not it is a good thing, and a failure is the other result.
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.
A binomial random variable is the count of successes in the n trials of a binomial setting; it can take the whole-number values 0, 1, 2, and so on up to n.

We write n (the number of trials) and p (the chance of a success on one trial). We call the count X (the number of successes). For the quiz, n is 10 and p is 0.25, so the chance of a failure on one question is 1 − p = 0.75.

Worked example — the guessing quiz, condition by condition. Check all four conditions out loud before you use the model.

Fixed number of trials. The quiz has n = 10 questions, set before the guessing starts.
Independent trials. A guess on question 3 has no effect on the guess for question 4.
Two outcomes. Each guess is right (a success) or wrong (a failure).
Same chance every time. Each question has four options, so p = 1 ⁄ 4 = 0.25 on every question.

In words: all four conditions hold, so X, the number of correct answers, is a binomial random variable with n = 10 and p = 0.25. That one sentence lets you use every tool in the rest of this chapter.

Worked example — free throws. Suppose a basketball player makes each free throw with chance 0.8, and one shot does not affect the next. The player takes 5 shots.

Fixed number of trials. n = 5 shots.
Independent trials. We are supposing that a miss does not change the next shot.
Two outcomes. Each shot goes in (a success) or misses (a failure).
Same chance every time. p = 0.8 on every shot.

In words: under those assumptions, the number of made shots is binomial with n = 5 and p = 0.8. Caveat: the "independent" and "same chance" conditions are assumptions here. A tired player late in a game might not meet them, and a coach would want data before trusting them.

Misconception check — "success" must mean something good. Students often think a success has to be a happy result. It does not. A success is the result you are counting. If a clinic counts patients who miss their appointment, a missed appointment is the success, even though nobody is glad about it. The cure: decide what X counts first, then call that result the success.

Memory hook for every binomial: fixed n, independent, two outcomes, same p.

2. Check the conditions: catch the setting that breaks them

The binomial model is only as good as its four conditions. When one fails, the formula still produces a number, but the number answers a different question. So the check is not a formality. It is the first step of every binomial problem.

Worked example — a raffle jar, drawn without replacement. A jar holds 10 raffle tickets, and 4 of them are winners. You draw 3 tickets, one at a time, and keep each ticket you draw. Count X, the number of winners you draw.

First draw. 4 of the 10 tickets win, so the chance of a winner is 4 ⁄ 10.
Second draw, after a winner. Now 3 winners sit among 9 tickets, so the chance is 3 ⁄ 9.
Second draw, after a loser. Now 4 winners sit among 9 tickets, so the chance is 4 ⁄ 9.
Compare. The chance changes from draw to draw, and it depends on what came before.

In words: the draws are not independent and p does not stay the same, so X is not binomial. Drawing without replacement from a small group breaks the model.

Two other settings fail for different reasons. A call center that phones customers until the first one answers has no fixed n, because nobody knows in advance how many calls it will take. A survey answer of "agree", "neutral", or "disagree" has three outcomes, not two. You can rescue that one by deciding to count "agree" as the success and everything else as a failure.

Limitation: real data rarely meet the conditions perfectly. If you draw 3 people from a city of 100,000, the chance barely moves from one draw to the next, and the binomial model is a fine working approximation. The trouble comes when the group is small, like 10 tickets in a jar.

Misconception check — using the binomial when the chance changes. Students often see "a fixed number of tries" and reach for the binomial right away. The raffle jar has a fixed number of draws, yet it fails, because each draw changes what is left in the jar. The cure: check all four conditions, and ask out loud, "Does one trial change the chance for the next?"

Pause and reflect — choose one. Pick either of the 2 prompts; they count the same.

(a) Think of a count you could track in your own week, such as the number of on-time buses out of 10 rides. Which of the four conditions do you trust, and which one would you want data to check?
(b) Think of a setting where you draw without replacement from a small group, such as picking 3 people from a club of 8 to lead a project. Explain in your own words why the chance changes from one pick to the next.

3. The binomial formula: count the ways, then multiply

Before you meet the formula, try a small case by hand. Suppose a ride-share driver gives 3 rides, and each rider tips with chance 1 ⁄ 2, independently. What is the chance that exactly 2 of the 3 riders tip?

List every way it can happen, writing T for a tip and N for no tip. Exactly two tips can land as TTN, TNT, or NTT. That is 3 different arrangements. Each arrangement has the same chance, 1 ⁄ 2 × 1 ⁄ 2 × 1 ⁄ 2 = 1 ⁄ 8, by the multiplication rule for independent events. The three arrangements cannot happen together, so you add them: 1 ⁄ 8 + 1 ⁄ 8 + 1 ⁄ 8 = 3 ⁄ 8. In words, the driver should expect exactly two tippers on about 3 shifts of three rides in 8.

That small case holds the whole idea.

The number of ways is how many different arrangements of k successes among n trials are possible, such as the 10 different questions that could hold the one right answer on a 10-question quiz.
The binomial probability formula is the rule for the chance of exactly k successes in a binomial setting: multiply the number of ways the k successes can land by p once for each of the k successes and by 1 − p once for each of the n − k failures.
The binomial distribution is the list of every possible count of successes in a binomial setting, from 0 to n, together with the probability of each count.

In short, P(X = k) = (number of ways) × (p multiplied k times) × (1 − p multiplied n − k times). Here k (the count you ask about) can be any whole number from 0 to n.

Worked example — zero and one correct guess. Use the guessing quiz, with n = 10 and p = 0.25.

Zero correct: count the ways. There is only 1 way to miss all ten questions.
Zero correct: multiply. All 10 guesses are failures, so P(X = 0) = 1 × 0.25⁰ × 0.75¹⁰. Here 0.75¹⁰ means 0.75 multiplied by itself 10 times, and 0.25⁰ is 1. The binomial table in the next section pins this at 0.0563, about 5.6% of guessers.
One correct: count the ways. The one right answer can be any of the 10 questions, so there are 10 ways.
One correct: multiply. One success and nine failures, so P(X = 1) = 10 × 0.25 × 0.75⁹. The binomial table pins this at 0.1877, about 18.8% of guessers.

In words: about 6 guessers in 100 get nothing right, and about 19 in 100 get exactly one right. Guessing leaves most students with a low score.

For two correct answers, the two right answers can sit on 45 different pairs of questions. Counting larger arrangements by hand gets slow, which is why the next section hands the work to technology.

Misconception check — dropping the number of ways. Students often compute P(X = 1) as 0.25 × 0.75⁹ and stop. That is about 1.9%, the chance of one particular arrangement, such as "right on question 1 and wrong on the other nine". But the right answer could sit on any of the 10 questions. The cure: before you multiply, ask "how many arrangements give this count?" and multiply by that number. Chatbots make this slip too, so watch for it in the Data Lab.

4. Binomial probabilities by technology: one value or a running total

Counting the ways for 3 correct out of 10, or 40 out of 100, is a job for a computer. A spreadsheet will give you any binomial probability in one formula. It will also give you a running total, which is often the chance you really want.

A cumulative probability is the probability that a count is at or below a given value, P(X ≤ k), found by adding the probabilities of every count from 0 up to k.

Here is the whole binomial distribution for the guessing quiz, as a spreadsheet returns it.

Binomial distribution for guessing on a 10-question, four-option quiz: chance of each number of correct answers, rounded to four decimals
Correct answers k	Chance of exactly k
0	0.0563
1	0.1877
2	0.2816
3	0.2503
4	0.1460
5	0.0584
6	0.0162
7	0.0031
8	0.0004
9	0.0000
10	0.0000

Read a row as "the chance of exactly that many correct answers". The eleven rounded values add to 1.0000, as every probability distribution must. The counts 9 and 10 are possible, but their chances are so small that they round to zero.

Worked example — at most two correct. You want P(X ≤ 2), the chance a guesser gets 0, 1, or 2 right.

Pick the rows. The rows for k = 0, 1, and 2 are the counts you want.
Add them. P(X ≤ 2) = 0.0563 + 0.1877 + 0.2816 = 0.5256, rounded to four decimals.

In words: a little over half of all pure guessers, about 53 in 100, get two or fewer questions right.

Worked example — at least five correct. You want P(X ≥ 5), the chance a guesser scores 50% or better. The rows for 5 through 10 are tiny and easy to mis-add, so use the complement rule from Week 5.

Name the complement. "At least 5" fails exactly when the count is 4 or fewer.
Subtract from 1. P(X ≥ 5) = 1 − P(X ≤ 4).
Let technology do the total. The spreadsheet formula below returns 0.0781, rounded to four decimals.

In words: a pure guesser scores 50% or better on about 8 tries in 100. A teacher who sees a 5 out of 10 cannot be sure the student studied, but guessing alone gets there rarely.

In practice (your tools). In Google Sheets or Excel, =BINOM.DIST(k, n, p, FALSE) returns the chance of exactly k successes, and =BINOM.DIST(k, n, p, TRUE) returns the running total P(X ≤ k). Type =BINOM.DIST(2, 10, 0.25, TRUE) to get P(X ≤ 2) and =1-BINOM.DIST(4, 10, 0.25, TRUE) to get P(X ≥ 5). To build the table above, put the counts 0 to 10 in cells A2 to A12, type =BINOM.DIST(A2, 10, 0.25, FALSE) in B2, and fill it down to B12. Then use Format, then Number, and set four decimals.

Misconception check — the off-by-one complement. Students often write P(X ≥ 5) = 1 − P(X ≤ 5). But a count of exactly 5 belongs to "at least 5", so subtracting it removes a count you wanted. The cure: list the counts in the event you want, then list the counts in its complement, and check that no count sits in both and none is missing. For "at least 5", the complement is "4 or fewer".

5. The binomial mean and standard deviation: the long-run center and the typical miss

In Week 6 you found the mean of a random variable by weighting every value by its chance. For a binomial, that long sum collapses into a short formula. Ten questions with a quarter chance each should give about a quarter of 10 right on average, and that is exactly the result.

The binomial mean is the long-run average count of successes in a binomial setting, μ = np; it need not be a count that can actually happen.
The binomial standard deviation is the typical distance between a binomial count and its mean, σ = √(np(1 − p)), measured in the same units as the count.

Worked example — the guessing quiz's center and spread. Use n = 10 and p = 0.25 again.

Mean. μ = np = 10 × 0.25 = 2.5 correct answers.
Standard deviation. σ = √(np(1 − p)) = √(10 × 0.25 × 0.75) ≈ 1.37 correct answers, rounded to two decimals.

In words: over many guessed quizzes, a guesser averages 2.5 correct answers, and a typical guessed score lands about 1.37 answers away from 2.5. So scores of 1, 2, 3, and 4 are ordinary for a guesser, which matches the big rows in the table.

Worked example — the free throws. For the player with n = 5 shots and p = 0.8, the mean is μ = np = 5 × 0.8 = 4 made shots. In words: over many sets of 5 free throws, the player averages 4 makes.

Misconception check — "expect 2.5 on my quiz". Students often read μ = 2.5 as the score one guesser should get. No quiz can score 2.5; the count is a whole number. The 2.5 is a long-run average over many guessed quizzes, exactly as in Week 6. The cure: say "on average, over many quizzes" every time you report a binomial mean.

Memory hook for the center: the mean is n times p — the typical miss is the square root of n p times 1 minus p.

6. The shape of a binomial: lopsided for small n, a bell for large n

Look down the table in Section 4 again. The chances climb from k = 0 to a peak at k = 2, then shrink, and from k = 6 on they are nearly zero. The pile sits on the left and a long, thin tail runs out to the right. By the rule from Week 2, you name skew by the tail, so this binomial distribution is right-skewed.

Now imagine the same guessing on a 100-question test. The mean becomes μ = np = 100 × 0.25 = 25 correct answers. The standard deviation becomes σ = √(100 × 0.25 × 0.75) = √18.75 ≈ 4.33 correct answers, rounded to two decimals. If you drew the bars for every count from 0 to 100, you would see a nearly symmetric bell centered at 25. Most of the bars would sit within a few standard deviations of 25, and the bars far from 25 would be almost flat.

Worked example — what changed between 10 and 100 questions. Compare the two quizzes step by step.

Center. The mean moves from 2.5 to 25, ten times as far from zero.
Room on the left. At n = 10, the mean of 2.5 sits only about 2 standard deviations above zero, so the left side of the pile is squashed against zero.
Room on both sides. At n = 100, the mean of 25 sits nearly 6 standard deviations above zero, so the pile can spread evenly both ways.

In words: a binomial looks lopsided when its mean sits close to 0 or close to n, and it looks like a bell when the mean has plenty of room on both sides. A fair coin with p = 0.5 is the exception: its binomial distribution is symmetric for every n, because a success and a failure are equally likely.

Misconception check — naming the skew by the pile. Students often call the 10-question distribution "left-skewed" because the big bars are on the left. Skew is named for the tail, not the pile. The long, thin tail of the guessing quiz runs toward the high counts on the right, so the distribution is right-skewed.

7. When a normal curve can stand in: the rule of thumb

A smooth bell is much easier to work with than a hundred separate bars. So statisticians often let a smooth curve stand in for a binomial distribution. The question is when that shortcut is honest.

The large-counts rule of thumb is the working rule this course uses to decide when a normal curve may approximate a binomial distribution: the expected numbers of successes np and failures n(1 − p) are both at least 10.
A normal approximation uses a normal curve with the same mean and standard deviation as a binomial distribution to stand in for that distribution when the large-counts rule of thumb holds.

It is a rule of thumb, not a law of nature. Some courses use 5 instead of 10. This course uses 10, and no graded question here depends on which number you learned.

Worked example — 10 questions versus 100 questions. Guessing still has p = 0.25.

Ten questions. np = 10 × 0.25 = 2.5 and n(1 − p) = 10 × 0.75 = 7.5.
Ten questions, verdict. 2.5 is below 10, so a normal curve is not an appropriate stand-in. The table's right skew is the reason.
A hundred questions. np = 25 and n(1 − p) = 100 × 0.75 = 75.
A hundred questions, verdict. Both are at least 10, so a normal curve centered at 25 with spread about 4.33 is a reasonable stand-in.

In words: guessing on 10 questions gives an expected 2.5 successes versus the 10 the rule asks for, a 4-fold shortfall, so you stick with the exact binomial. For 100 questions the curve is a fair shortcut.

Worked example — a clinic's missed appointments. Suppose each booked patient misses an appointment with chance 0.2, independently, and a missed appointment is the success being counted.

A small clinic morning. With n = 12 bookings, np = 12 × 0.2 = 2.4, below 10. Use the exact binomial, not a curve.
A large clinic week. With n = 200 bookings, np = 200 × 0.2 = 40 and n(1 − p) = 200 × 0.8 = 160. Both are at least 10, so a normal curve with mean 40 is a reasonable stand-in.

In words: a scheduler planning one morning should use the exact binomial chances, while a manager planning a whole week can use the smooth curve to plan staff.

Misconception check — "a big n is always enough". Students often think any large n makes the normal curve safe. The rule checks both expected counts, not n alone. Suppose a part is faulty with chance 0.01 and a buyer checks n = 500 parts. Then np = 500 × 0.01 = 5, below 10, even though 500 is a big number. With so few expected faults, the distribution is still piled against zero. The cure: compute np and n(1 − p) every time.

Pause and reflect — choose one. Pick either of the 2 prompts; they count the same.

(a) Think of a count in your major or your job where the chance of a success is small, such as rare errors on forms. How large would n need to be before np reaches 10, and what does that tell you about using a curve?
(b) Explain to a friend who missed class why "10 questions" and "100 questions" get different verdicts from the same rule, using the words "expected successes" in your answer.

8. A first look at the normal curve: the working model for many measurements

Picture the bars for the 100-question test again, and then imagine sanding off their corners. What is left is a smooth hill: highest in the middle, falling away evenly on both sides, with tails that thin out but never quite reach zero. That hill is the shape this course leans on for the rest of the term.

A normal curve is a smooth, symmetric, bell-shaped model for a quantitative variable, set by its mean μ at the center and its standard deviation σ 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.

Two numbers settle everything about a normal curve. The mean μ places the peak. The standard deviation σ sets the width: a small σ gives a tall, narrow bell, and a large σ gives a low, wide one. You met the idea that probability is area under a smooth curve in Week 6, and it carries over here unchanged.

Worked example — the guessing curve. For the 100-question test, the matching normal curve has μ = 25 and σ ≈ 4.33.

Peak. The curve is highest at 25 correct answers.
Symmetry. It falls away the same way above and below 25, so a score of 21 and a score of 29 sit at the same height.
Area. The share of guessers who score between two values is the area under the curve between them. Week 9 shows you how to find those areas.

In words: one smooth curve with two numbers summarizes the chances for all 101 possible scores.

Many real measurements are modeled the same way. Suppose the fill weights of coffee bags from one packing machine follow a normal model with mean 340 g and standard deviation 3 g. Then the curve peaks at 340 g, and bags far from 340 g in either direction are rare. Field biologists often model a measurement such as body mass within one penguin species with a normal curve too. In each case the curve is a model you assume, and a good analyst says so.

Honesty note: notice the word "suppose" in the coffee example. A normal curve is a model, and real data are never exactly normal. In Week 9 you will learn to check whether a normal model is reasonable for a set of data before you lean on it.

Misconception check — "the data are normal". Students often say a set of measurements "is normal" as if that were a fact about the world. It is a modeling choice. The cure: write "suppose these values follow a normal model with mean … and standard deviation …", and be ready to check the assumption against the data.

Memory hook for the curve: two numbers draw the bell — the mean places it, the standard deviation spreads it.

Key terms

trial — One repetition of a chance process whose result you record, such as one guessed quiz question or one free throw. (first used: 1. The binomial setting: count the successes in a fixed number of tries)
success and failure — The two labels given to the two possible results of a trial in a binomial setting; a success is the result you are counting, whether or not it is a good thing, and a failure is the other result. (first used: 1. The binomial setting: count the successes in a fixed number of tries)
binomial setting — 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. (first used: 1. The binomial setting: count the successes in a fixed number of tries)
binomial random variable — The count of successes in the n trials of a binomial setting; it can take the whole-number values 0, 1, 2, and so on up to n. (first used: 1. The binomial setting: count the successes in a fixed number of tries)
number of ways — How many different arrangements of k successes among n trials are possible, such as the 10 different questions that could hold the one right answer on a 10-question quiz. (first used: 3. The binomial formula: count the ways, then multiply)
binomial probability formula — The rule for the chance of exactly k successes in a binomial setting: multiply the number of ways the k successes can land by p once for each of the k successes and by 1 − p once for each of the n − k failures. (first used: 3. The binomial formula: count the ways, then multiply)
binomial distribution — The list of every possible count of successes in a binomial setting, from 0 to n, together with the probability of each count. (first used: 3. The binomial formula: count the ways, then multiply)
cumulative probability — The probability that a count is at or below a given value, P(X ≤ k), found by adding the probabilities of every count from 0 up to k. (first used: 4. Binomial probabilities by technology: one value or a running total)
binomial mean — The long-run average count of successes in a binomial setting, μ = np; it need not be a count that can actually happen. (first used: 5. The binomial mean and standard deviation: the long-run center and the typical miss)
binomial standard deviation — The typical distance between a binomial count and its mean, σ = √(np(1 − p)), measured in the same units as the count. (first used: 5. The binomial mean and standard deviation: the long-run center and the typical miss)
large-counts rule of thumb — The working rule this course uses to decide when a normal curve may approximate a binomial distribution: the expected numbers of successes np and failures n(1 − p) are both at least 10. (first used: 7. When a normal curve can stand in: the rule of thumb)
normal approximation — Using a normal curve with the same mean and standard deviation as a binomial distribution to stand in for that distribution when the large-counts rule of thumb holds. (first used: 7. When a normal curve can stand in: the rule of thumb)
normal curve — A smooth, symmetric, bell-shaped model for a quantitative variable, set by its mean μ at the center and its standard deviation σ 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. (first used: 8. A first look at the normal curve: the working model for many measurements)

Summary

A binomial setting has a fixed number of trials, independent trials, two outcomes, and the same chance of success every time.
Drawing without replacement from a small group breaks the model, because the chance changes from draw to draw.
A binomial chance is the number of ways times the chance of one arrangement.
A spreadsheet's BINOM.DIST gives one exact chance or a running total; use the complement for "at least".
The binomial mean is n times p, a long-run average that need not be a possible count.
A binomial with few expected successes is lopsided; with plenty of room on both sides it looks like a bell.
A normal curve may stand in for a binomial when the expected successes and failures are both at least 10.
A normal curve is a model set by its mean and standard deviation, where area is probability, and you state it as an assumption.

Check your understanding

Work these before looking at the answers — they are ungraded, and the answers (with reasoning) are at the bottom of the page.

For each setting, say whether the count is binomial, and name a condition that fails if it is not. (a) An inspector tests 15 light bulbs from a huge production run, where each bulb is faulty with chance 0.02, and counts the faulty bulbs. (b) A coach picks 3 players from a team of 6, of whom 2 are first-years, and counts the first-years picked. (c) A barista counts customers until the first one orders tea.
A commuter takes two buses each morning. Suppose each bus is late with chance 0.3, independently. Find the chance that exactly one of the two buses is late.
Suppose an online shop has 400 orders and each order is returned with chance 0.1, independently. Find the mean and standard deviation of the number of returns, and say whether a normal curve could stand in.
A gardener plants 30 seeds, and suppose each seed fails to sprout with chance 0.05, independently. Count the seeds that fail. Would a normal curve be a good stand-in for this count? Explain.

Answers (worked reasoning — read after attempting)

(a) Binomial: n = 15 is fixed, a huge run makes the bulbs effectively independent with the same chance, and each bulb is faulty or not, so n = 15 and p = 0.02. (b) Not binomial: the coach picks without replacement from only 6 players, so the chance of a first-year changes after each pick, from 2 ⁄ 6 on the first pick to 1 ⁄ 5 or 2 ⁄ 5 on the second. (c) Not binomial: there is no fixed number of trials, because nobody knows in advance how many customers it will take.
There are 2 ways to get exactly one late bus: the first is late and the second is not, or the other way around. Each way has chance 0.3 × 0.7 = 0.21. So the chance is 2 × 0.3 × 0.7 = 0.42. In words: on about 42 mornings in 100, exactly one bus is late. If you forgot the 2 ways, you would get 0.21, half the true answer.
The mean is μ = np = 400 × 0.1 = 40 returns. The standard deviation is σ = √(400 × 0.1 × 0.9) = √36 = 6 returns. Check the rule of thumb: np = 40 and n(1 − p) = 400 × 0.9 = 360, both at least 10, so a normal curve with mean 40 and standard deviation 6 is a reasonable stand-in. In words: the shop should plan for about 40 returns, give or take about 6.
No. The expected number of failures is np = 30 × 0.05 = 1.5, far below 10. The count will usually be 0, 1, 2, or 3, so the distribution is piled against zero with a tail to the right. Use the exact binomial chances instead of a curve.

Next up in this module: the Readings & Resources page, the lecture tutorial, the practice set, and the quiz — and next week the midterm review adds z-scores, the measuring stick that Week 9's normal model is built on.

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