Binomial Distribution

🎰 Binomial Distributions

Part 1 of 7 — The Binomial Setting

BINS Criteria

A random variable $X$ is binomial if it counts the number of successes and:

Binomial Distribution: $X \sim B(n, p)$

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ and $k = 0, 1, 2, \ldots, n$

Mean and Standard Deviation

$\mu = np \qquad \sigma = \sqrt{np(1-p)}$

Example: 20 free throws, $p = 0.8$

• $\mu = 20(0.8) = 16$
• $\sigma = \sqrt{20(0.8)(0.2)} = \sqrt{3.2} \approx 1.789$

Concept Check 🎯

Binomial Basics 🧮

A basketball player makes 75% of free throws. She shoots 12.

1) Expected number of makes?

2) Standard deviation? (round to 2 decimal places)

3) $P(X = 12)$? (round to 4 decimal places)

🔢 Binomial Calculations

Part 2 of 7 — Computing Binomial Probabilities

Using the Formula

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

Cumulative Probabilities

• $P(X \leq k)$: use binomcdf(n, p, k) on calculator
• $P(X \geq k) = 1 - P(X \leq k-1)$
• $P(a \leq X \leq b) = P(X \leq b) - P(X \leq a-1)$

Worked Example

$X \sim B(5, 0.4)$. Find $P(X \geq 3)$.

$P(X \geq 3) = P(3) + P(4) + P(5)$

$= \binom{5}{3}(0.4)^3(0.6)^2 + \binom{5}{4}(0.4)^4(0.6)^1 + \binom{5}{5}(0.4)^5$

$= 10(0.064)(0.36) + 5(0.0256)(0.6) + 1(0.01024)$

$= 0.2304 + 0.0768 + 0.01024 = 0.3174$

Concept Check 🎯

Binomial Calculation 🧮

$X \sim B(4, 0.5)$.

1) $P(X = 2) = ?$ (as a fraction, e.g., 3/8)

2) $P(X \leq 1) = ?$ (as a fraction)

3) $P(X \geq 3) = ?$ (as a fraction)

🎯 Binomial Mean and Standard Deviation in Context

Part 3 of 7 — What $np$ and $\sqrt{np(1-p)}$ Tell You

The Two Parameters

If $X \sim B(n, p)$ counts successes in $n$ independent trials:

$\mu_X = np \qquad \sigma_X = \sqrt{np(1-p)}$

Both formulas are on the AP formula sheet. The work the exam rewards is interpreting them in context.

Worked Example

A store finds that 35% of customers use a coupon. Let $X$ = the number of coupon users among 40 randomly selected customers.

• $\mu_X = 40(0.35) = 14$ customers
• $\sigma_X = \sqrt{40(0.35)(0.65)} = \sqrt{9.1} \approx 3.02$ customers

Interpreting the mean: If we took many random samples of 40 customers, the average number of coupon users would be about 14.

Interpreting the standard deviation: The number of coupon users in a sample of 40 typically varies by about 3.02 customers from the mean of 14.

⚠️ Both values are counts of customers, not proportions or percents, so give them in units of the count.

Concept Check 🎯

Is a Result Surprising?

The mean and standard deviation let you judge whether an observed count is unusual.

In the coupon example, suppose a sample of 40 customers has 22 coupon users:

$\frac{22 - 14}{3.02} \approx 2.65$

That count is about 2.65 standard deviations above the mean. A count that far above $np$ would be unusual if the coupon rate really were 35%, so it gives some evidence that the rate for these customers is higher than 35%.

Guessing on a Quiz 🧮

A student guesses on all 20 questions of a multiple-choice quiz; each question has 4 choices. Let $X$ = the number correct.

1) Mean number correct?

2) Standard deviation of $X$? (round to 2 decimals)

3) How many standard deviations above the mean is a score of 10? (round to 2 decimals)

📈 Simulation and Cumulative Binomial Probabilities

Part 4 of 7 — Tables of $P(X \leq k)$ and Estimates from Simulation

Cumulative Probability Distribution

A cumulative probability distribution gives $P(X \leq k)$ for every value $k$. For $X \sim B(5, 0.4)$:

Each cumulative value adds the next exact probability: $0.3370 + 0.3456 = 0.6826$.

Reading a Cumulative Table

Example: $P(X \geq 3) = 1 - P(X \leq 2) = 1 - 0.6826 = 0.3174$, which matches the sum of $P(3) + P(4) + P(5)$ from Part 2.

Concept Check 🎯

Estimating Binomial Probabilities by Simulation

When you cannot (or do not want to) use the formula, simulate the random process.

Simulating $X \sim B(5, 0.4)$ with random digits:

1. Let digits 0–3 represent a success (4 of the 10 digits, so $p = 0.4$) and 4–9 a failure.
2. Read 5 digits and count the successes: that is one value of $X$.
3. Repeat many times and record each count.
4. Estimate a probability with the relative frequency, e.g. (number of repetitions with $X \geq 3$) ÷ (total repetitions).

Example: In 50 repetitions, 17 had $X \geq 3$, so the estimate is $17/50 = 0.34$ (the exact value is 0.3174).

🔑 Law of large numbers: as the number of repetitions grows, the simulated relative frequency tends to get closer to the true probability.

Simulation Check 🎲

Cumulative Practice 🧮

$X \sim B(8, 0.25)$.

1) $P(X \leq 1)$? (round to 4 decimals)

2) $P(X \geq 2)$? (round to 4 decimals)

3) A simulation of 400 repetitions gives 252 with $X \geq 2$. What is the simulated estimate of $P(X \geq 2)$? (2 decimals)

🔄 Is It Binomial? Checking the Conditions

Part 5 of 7 — Justifying a Binomial Model

Check Each Condition in Context

The 10% Condition

Sampling without replacement makes trials slightly dependent. If the sample is no more than 10% of the population ($n \leq 0.10N$), the trials are close enough to independent to use a binomial model.

Example: Choosing 5 students from a class of 20 is 25% of the class, so the selections are NOT approximately independent and $X$ is not binomial. Choosing 50 adults from a city of 200,000 is fine.

🔑 On the AP exam, name the condition and tie it to the context: "There are $n = 15$ independent rolls, each roll is a 4 or not a 4, and $P(4) = 1/6$ on every roll."

Concept Check 🎯

From Counts to Proportions

A binomial $X$ counts successes. Dividing by $n$ gives the sample proportion $\hat{p} = X/n$, which you will use for inference in Unit 3.

Example: $X \sim B(200, 0.3)$ has $\mu_X = 60$ and $\sigma_X = \sqrt{42} \approx 6.48$. For $\hat{p} = X/200$: mean $0.3$ and SD $6.48/200 \approx 0.032$.

Counts vs. Proportions 📐

Checking and Computing 🧮

In a large city, 18% of adults are left-handed. A random sample of 60 adults is taken; $X$ = number who are left-handed.

1) Mean of $X$?

2) Standard deviation of $X$? (round to 2 decimals)

3) Mean of the sample proportion $\hat{p}$?

🏆 Problem-Solving Workshop

Part 6 of 7 — AP-Style Practice

AP Free-Response Strategy

When the AP exam gives a binomial scenario:

1. State the distribution and parameters: “$X \sim B(n, p)$” with $X$ defined in context
2. Verify conditions (BINS, plus the 10% condition when sampling without replacement)
3. Calculate using the formula or calculator (binompdf / binomcdf), showing the setup
4. Interpret in context with proper probability language

🔑 Calculator syntax alone does not earn credit: write $P(X \leq 2)$ with $n$ and $p$ identified.

Concept Check 🎯

Mixed Practice 🧮

10% of products are defective. A random sample of 20 is inspected.

1) $P(\text{exactly 2 defective}) = ?$ (round to 4 decimals)

2) Expected number defective?

3) $P(\text{at most 1 defective}) = ?$ (round to 4 decimals)

📝 Review & Applications

Part 7 of 7 — Comprehensive Review

Key Formulas

Common AP Mistakes

• Calling a variable binomial when the number of trials is not fixed
• Forgetting $\binom{n}{k}$ in the binomial formula
• Writing $P(X \geq k) = 1 - P(X \leq k)$, which drops $X = k$
• Leaving out the square root in $\sigma_X$
• Interpreting the mean as the count that "will happen" instead of a long-run average

Concept Check 🎯

Final Challenge 🧮

$X \sim B(100, 0.3)$

1) $\mu_X = ?$

2) $\sigma_X = ?$ (round to 2 decimals)

3) How many standard deviations above the mean is $X = 40$? (round to 2 decimals)