In Lesson 12 you counted things: the number of made free throws out of 20, the number of defective phones out of 50. Those counts are discrete — you can list the possible values (0, 1, 2, …), and each has its own chunk of probability.
But some quantities can't be listed that way. How long until the next bus? Somewhere between 0 and 15 minutes — but it could be 4 minutes, or 4.7, or 4.71828…, with no gaps. A person's height isn't "67 inches" exactly; it's 67.0000… to as many decimals as your ruler allows.
Quick question to sit with: if a bus is equally likely to arrive at any instant in the next 15 minutes, what is the probability it arrives at exactly 4.000000… minutes — not a hair more, not a hair less?
Hold that thought. The answer is genuinely zero, and understanding why is the key that unlocks this entire lesson.
A discrete random variable takes a countable set of values — you can list them, even if the list is infinite (0, 1, 2, 3, …). We describe it with a probability distribution table assigning a probability to each value.
A continuous random variable can take any value in an interval of the number line. Between any two possible values there are infinitely many others. We cannot make a table — there are too many values. Instead we describe a continuous random variable with a density curve.
A density curve is a smooth curve drawn so that:
That third rule is the heart of the lesson. For a continuous random variable, probability = area. You don't add up probabilities of individual values — you measure the area sitting above an interval.
This leads to a result that surprises students every year:
For a continuous random variable,
P(X = any single exact value) = 0.
Why? The "area" above a single point is the area of a line segment — a rectangle with zero width. Zero width means zero area, which means zero probability. So the bus arriving at exactly 4.0000… minutes has probability 0, even though arriving near 4 minutes is perfectly possible. A practical consequence: for continuous variables, < and ≤ give the same probability, because the single endpoint contributes nothing.
P(X < 5) = P(X ≤ 5) and P(a < X < b) = P(a ≤ X ≤ b)(Contrast this with discrete variables, where the endpoint matters: P(X ≤ 5) includes the chunk at 5, but P(X < 5) does not.)
The simplest density curve is a flat horizontal line over an interval — the uniform distribution. Every value in the interval is equally likely. The curve is a rectangle, so finding probabilities is just finding the area of a rectangle: area = height × width.
If X is uniform on the interval from a to b, the total area must be 1. The width of the whole interval is b − a, so the height must be:
height = 1 / (b − a)
That way the full rectangle has area (b − a) × 1/(b − a) = 1. ✓
Worked example — the bus. A bus is equally likely to arrive at any time in the next 15 minutes, so X = wait time is uniform on [0, 15]. The height is 1/(15 − 0) = 1/15 ≈ 0.0667.
[GRAPH: Uniform density curve for wait time X on [0, 15]. X-axis labeled "Wait time (minutes)" from 0 to 15. Y-axis labeled "Density," with a flat horizontal line at height 1/15 ≈ 0.0667 stretching from x=0 to x=15. The region from x=5 to x=12 is shaded, forming a rectangle of width 7 and height 1/15. Caption: shaded area = P(5 < X < 12).]
What's the probability the wait is between 5 and 12 minutes? Shade the rectangle from 5 to 12:
P(5 < X < 12) = height × width = (1/15) × (12 − 5) = 7/15 ≈ 0.4667And P(X < 5) = (1/15) × 5 = 5/15 ≈ 0.3333. Notice we never used a calculator function — just area = height × width.
The uniform distribution also has a tidy mean and standard deviation (on your formula sheet's spirit, but worth knowing): the mean sits at the center, (a + b)/2, and the standard deviation is (b − a)/√12. For the bus, the mean wait is (0 + 15)/2 = 7.5 minutes.
You met the normal distribution back in Lesson 5 as a model for data. Now reframe it: a normal random variable is simply a continuous random variable whose density curve is the bell curve N(μ, σ). Total area under the bell = 1; probability over an interval = area under the bell over that interval. Same logic as the uniform — just a curvier shape, so we let the calculator measure the area.
TI-84: normalcdf(lower, upper, μ, σ)Use −1E99 for "negative infinity" and 1E99 for "positive infinity." For example, if battery life X is N(500, 50) hours, then P(450 < X < 575) = normalcdf(450, 575, 500, 50) ≈ 0.7745. (Check: z = (450−500)/50 = −1.00 and z = (575−500)/50 = 1.50.) Because the normal is continuous, P(X = 500) = 0, and again < and ≤ are interchangeable.
Here's where continuous and discrete meet. A binomial count X (Lesson 12) is discrete — but when n is large, its probability histogram is so finely stepped and so bell-shaped that a normal curve lays right over it. We can then approximate binomial probabilities using areas under that normal curve.
The catch: the bars need to be bell-shaped first. That's the Large Counts condition:
np ≥ 10ANDn(1 − p) ≥ 10
When this holds, the binomial count X is approximately:
X ≈ N( np, √(np(1 − p)) ) — mean μ = np, standard deviation σ = √(np(1−p)).
These are exactly the binomial mean and SD from Lesson 12; we're just feeding them into a normal curve. Then we find binomial probabilities with normalcdf.
Worked example — contrasting approximate vs. exact. A basketball team takes n = 200 three-point shots, each made with probability p = 0.40. Let X = number made. Estimate P(X ≤ 70).
Check Large Counts: np = 200(0.40) = 80 ≥ 10 and n(1−p) = 200(0.60) = 120 ≥ 10. ✓ Approximation is reasonable.
Set up the normal: μ = np = 80, σ = √(200 · 0.40 · 0.60) = √48 ≈ 6.9282.
Approximate: normalcdf(−1E99, 70, 80, 6.9282) ≈ 0.0745
(check z = (70 − 80)/6.9282 ≈ −1.44)Now the exact binomial answer, which your TI-84 can also give directly:
Exact: binomcdf(200, 0.40, 70) ≈ 0.0844The approximation (0.0745) is close to the truth (0.0844), but not identical — the normal is a smooth stand-in for the jagged binomial bars. Bottom line: when your calculator can give the exact binomial answer with binomcdf, prefer it. The normal approximation is a backup for when you only have summary information, for understanding why binomials look bell-shaped, and for problems that explicitly ask you to use it. On the AP exam, read carefully: if a question says "use a normal approximation," do that; otherwise the exact binomial is the gold standard.
A coffee shop's morning wait time (minutes) is uniform on [0, 8]. Find P(X > 6) and P(2 < X < 5).
Strategy: Uniform → area = height × width. Height = 1/(8 − 0) = 1/8 = 0.125.
Solution:
P(X > 6) = (1/8) × (8 − 6) = 2/8 = 0.25
P(2 < X < 5) = (1/8) × (5 − 2) = 3/8 = 0.375Interpretation: There's a 25% chance of waiting more than 6 minutes, and a 37.5% chance of waiting between 2 and 5 minutes. No calculator function needed — geometry does the work.
Adult female heights (inches) are modeled as N(64, 2.5). What proportion of women are taller than 67 inches?
Strategy: Continuous normal → use normalcdf for the area above 67.
Solution:
z = (67 − 64)/2.5 = 1.20
P(X > 67) = normalcdf(67, 1E99, 64, 2.5) ≈ 0.1151Interpretation: About 11.5% of women in this model are taller than 67 inches. (And P(X = 67) = 0, so it doesn't matter whether we wrote > or ≥.)
A vaccine is 80% effective. In a trial of n = 150 people, let X = number for whom it works. Use a normal approximation to estimate P(X ≥ 130).
Strategy: Binomial count, n = 150, p = 0.80. Check Large Counts, build the normal, find the area.
Solution:
Large Counts: np = 150(0.80) = 120 ≥ 10 ✓ n(1−p) = 150(0.20) = 30 ≥ 10 ✓
μ = np = 120 σ = √(150 · 0.80 · 0.20) = √24 ≈ 4.899
z = (130 − 120)/4.899 ≈ 2.04
P(X ≥ 130) ≈ normalcdf(130, 1E99, 120, 4.899) ≈ 0.0206Interpretation: About a 2% chance that 130 or more of the 150 people are protected. (For comparison, the exact answer is 1 − binomcdf(150, 0.80, 129) ≈ 0.0224 — very close, confirming the approximation is solid here. If the question hadn't required the approximation, the exact value would be preferred.)
A factory produces parts with a 2% defect rate. In a batch of n = 40, can we use a normal approximation for the number of defects?
Strategy: Check Large Counts before doing anything.
Solution:
np = 40(0.02) = 0.8 → NOT ≥ 10. Condition fails.Interpretation: The normal approximation is not appropriate — the distribution is strongly right-skewed, not bell-shaped, so a normal curve would give a poor estimate. Use the exact binomial (binompdf/binomcdf) instead. Recognizing when not to approximate is exactly what AP graders test.
1. Treating P(X = value) as nonzero for a continuous variable. Students write things like "P(X = 5) = 1/15" for a uniform distribution. For any continuous random variable, the probability of a single exact value is 0 — a line has no width, so no area. Probabilities only attach to intervals. (This also means < and ≤ are interchangeable for continuous variables.)
2. Forgetting the Large Counts check before approximating. Jumping straight to normalcdf(np, …) without verifying np ≥ 10 and n(1−p) ≥ 10 can give a badly wrong answer when the binomial is skewed (see Example 4). On the FRQ, the check is often worth its own point — state both products with numbers, not just "conditions met."
3. Uniform height vs. width confusion. Students mix up the height 1/(b−a) with the width of the interval they're shading. The height is always 1/(b−a) for the whole distribution; the width is the length of the specific interval you want. Probability = height × (your interval's width), not height × total width.
4. Using the wrong σ in the normal approximation. The standard deviation is √(np(1−p)), not np and not √(np). Pull μ and σ straight from the binomial formulas of Lesson 12.
5. Approximating when the exact answer is available. If you have n and p and a calculator, binomcdf gives the exact answer. Use the normal approximation only when asked, or when you lack the raw n and p. Don't throw away accuracy for no reason.
1 (MC). X is uniform on [0, 20]. What is P(X < 8)?
(A) 0.08 (B) 0.40 (C) 0.50 (D) 0.60
2 (MC). For a continuous random variable X, which statement is true?
(A) P(X = 3) can be any value between 0 and 1
(B) P(X = 3) = 0
(C) P(a < X < b) is found by adding the probabilities of each value
(D) The total area under the density curve can exceed 1
3 (MC). X is uniform on [5, 25]. What is the height of the density curve?
(A) 1/20 (B) 1/25 (C) 1/5 (D) 20
4 (MC). A binomial count has n = 60 and p = 0.05. Is a normal approximation appropriate?
(A) Yes, because n is large
(B) Yes, because n(1−p) = 57 ≥ 10
(C) No, because np = 3 < 10
(D) No, because p < 0.5
5 (MC). A binomial count has n = 200, p = 0.30. Using the normal approximation, what are μ and σ?
(A) μ = 60, σ = √60 (B) μ = 60, σ = √42 (C) μ = 60, σ = 42 (D) μ = 0.30, σ = 0.46
6 (MC). Battery life is N(500, 50) hours. P(X < 450) is closest to:
(A) 0.16 (B) 0.34 (C) 0.50 (D) 0.84
7 (MC). For the uniform distribution on [10, 18], P(12 < X < 15) equals:
(A) 0.250 (B) 0.375 (C) 0.500 (D) 0.625
8 (MC). A coin is flipped n = 400 times. Using a normal approximation, P(X ≥ 215) heads is closest to (μ = 200, σ = 10):
(A) 0.067 (B) 0.150 (C) 0.250 (D) 0.500
9 (Short answer). A train arrives uniformly between 0 and 60 minutes after the hour. (a) Find P(X < 10). (b) Find P(20 < X < 45). (c) What is the mean arrival time?
10 (Short answer). Explain in one or two sentences why, for a continuous random variable, P(X ≤ 7) and P(X < 7) give the same value, but for a discrete variable they may not.
11 (Short answer, in context). A region's daily rainfall amount (cm) on rainy days is modeled as N(2.4, 0.8). Find the probability a rainy day produces more than 3.0 cm of rain. Show the z-score and the normalcdf setup.
12 (Short answer, in context). In a county, 36% of registered voters support a ballot measure. A pollster samples n = 250 voters; let X = number who support it. (a) Check the Large Counts condition. (b) Find μ and σ for the normal approximation. (c) Use the approximation to estimate P(X ≥ 100).
13 (MC). Which scenario should be modeled with a continuous random variable rather than a discrete one?
(A) The number of heads in 10 coin flips
(B) The number of defective items in a shipment of 100
(C) The exact time (in seconds) a runner takes to finish a 100 m dash
(D) The number of students absent today
14 (MC). For a uniform distribution on [0, 4], a student computes P(1 < X < 3) = (1/4) × 4 = 1. What did they do wrong?
(A) Used the wrong height
(B) Multiplied by the total width (4) instead of the interval width (2)
(C) Forgot the distribution is continuous
(D) Nothing — the answer is correct
15 (Short answer, in context). A vaccine is 90% effective. In a group of n = 300, let X = number protected. (a) Verify Large Counts. (b) Find μ and σ. (c) Use the normal approximation to estimate P(X ≤ 265), and state whether the exact binomial would be preferred here.
Statistical Practice 3 — Analyze Data
A county election official reports that 36% of registered voters in the county support a proposed ballot measure. A local newspaper takes a random sample of n = 250 registered voters and records X, the number in the sample who support the measure.
(a) Explain why X can be modeled as a binomial random variable, and state the mean and standard deviation of X. (3 points)
(b) Verify that the conditions for using a normal approximation to the distribution of X are met. (2 points)
(c) Use a normal approximation to estimate the probability that at least 100 of the 250 sampled voters support the measure. Show your work, including a z-score. (3 points)
(d) The newspaper's statistician computes the exact binomial probability and reports
P(X ≥ 100) ≈ 0.1059, which differs slightly from your answer in part (c). Explain why the two values differ and which one is more accurate. (2 points)
(a) X is binomial because the four conditions of a binomial setting hold: (B)inary — each voter either supports the measure or does not; (I)ndependent — voters are sampled at random, and the sample (250) is small relative to the county's population of voters, so we can treat trials as independent; (N)umber of trials is fixed at n = 250; (S)uccess probability is the same, p = 0.36, for each voter.
μ = np = 250(0.36) = 90
σ = √(np(1 − p)) = √(250 · 0.36 · 0.64) = √57.6 ≈ 7.5895So X has mean 90 supporters and standard deviation about 7.59 supporters.
(b) Large Counts condition:
np = 250(0.36) = 90 ≥ 10 ✓
n(1 − p) = 250(0.64) = 160 ≥ 10 ✓Both expected counts are at least 10, so the distribution of X is approximately normal and the normal approximation is appropriate.
(c) Model X as approximately N(90, 7.5895).
z = (100 − 90) / 7.5895 ≈ 1.32
P(X ≥ 100) ≈ normalcdf(100, 1E99, 90, 7.5895) ≈ 0.0938There is approximately a 0.094 probability that at least 100 of the 250 sampled voters support the measure.
(d) The normal curve is a continuous, smooth approximation to the discrete binomial distribution, so the area it gives (0.0938) is close to — but not exactly equal to — the true binomial probability (0.1059). The exact binomial value (0.1059) is more accurate, because X is genuinely a discrete count and the binomial calculation reflects its actual distribution; the normal approximation introduces small error by smoothing the discrete bars into a continuous curve. When n and p are known and a calculator is available, the exact binomial is preferred.
| Part | Point | Earned for |
|---|---|---|
| (a) | 1 | Identifies binomial setting / states the conditions (BINS or equivalent) hold in context |
| (a) | 1 | Correct mean: μ = np = 90 |
| (a) | 1 | Correct SD: σ = √(np(1−p)) = √57.6 ≈ 7.59 |
| (b) | 1 | States Large Counts condition: np ≥ 10 and n(1−p) ≥ 10 |
| (b) | 1 | Shows both values with numbers (90 and 160) and concludes condition is met |
| (c) | 1 | Correct z-score: z = (100 − 90)/7.59 ≈ 1.32 (or correct normalcdf bounds) |
| (c) | 1 | Correct probability: ≈ 0.094 (accept 0.093–0.095) |
| (c) | 1 | Answer stated as a probability in context (about the number of supporters) |
| (d) | 1 | Explains the difference: normal is a continuous approximation to a discrete binomial |
| (d) | 1 | Identifies the exact binomial (0.1059) as more accurate, with reason |
Where students lose points:
σ = np or σ = √(np) instead of √(np(1−p)).normalcdf number with no z-score and no setup shown can cost the work point; also, an answer with no context (just "0.094") misses the interpretation point.1. (B) 0.40. Uniform on [0,20]: height 1/20, P(X<8) = (1/20)(8) = 0.40. (A) multiplies height by itself / mis-reads as 8/100; (C) assumes the midpoint; (D) uses width 12.
2. (B) P(X = 3) = 0. A single point has zero width, hence zero area, hence zero probability. (A) confuses continuous with discrete; (C) describes a discrete distribution; (D) violates rule that total area = 1.
3. (A) 1/20. Height = 1/(b−a) = 1/(25−5) = 1/20. (B) uses 1/b; (C) uses 1/a; (D) inverts the formula.
4. (C) No, because np = 3 < 10. np = 60(0.05) = 3, which fails Large Counts. (A) "n large" alone isn't the condition; (B) checks only one product; (D) p < 0.5 is irrelevant to the condition.
5. (B) μ = 60, σ = √42. μ = np = 60; σ = √(np(1−p)) = √(200·0.3·0.7) = √42 ≈ 6.48. (A) uses √(np); (C) uses np(1−p) without the root; (D) uses p instead of np.
6. (A) 0.16. z = (450−500)/50 = −1.00; normalcdf(−1E99, 450, 500, 50) ≈ 0.1587 ≈ 0.16. (B) misreads the tail; (C) is the center; (D) is the complement.
7. (B) 0.375. Height 1/(18−10) = 1/8; width 15−12 = 3; area = 3/8 = 0.375. (A) uses width 2; (C) uses width 4; (D) uses width 5.
8. (A) 0.067. z = (215−200)/10 = 1.50; normalcdf(215, 1E99, 200, 10) ≈ 0.0668 ≈ 0.067. (The exact binomial is ≈ 0.0735.) (B)–(D) misplace the tail area.
9. Uniform on [0,60], height 1/60. (a) P(X<10) = 10/60 ≈ 0.1667. (b) P(20= (0+60)/2 = 30 minutes.
10. For a continuous variable, P(X = 7) = 0 (a point has no area), so dropping or including the endpoint 7 changes nothing: P(X ≤ 7) = P(X < 7). For a discrete variable, the value 7 carries a nonzero chunk of probability, so P(X ≤ 7) includes it while P(X < 7) does not — the two differ by P(X = 7).
11. z = (3.0 − 2.4)/0.8 = 0.75; P(X > 3.0) = normalcdf(3.0, 1E99, 2.4, 0.8) ≈ 0.2266. About a 22.7% chance a rainy day exceeds 3.0 cm.
12. (a) np = 250(0.36) = 90 ≥ 10 and n(1−p) = 250(0.64) = 160 ≥ 10 ✓. (b) μ = 90, σ = √(250·0.36·0.64) = √57.6 ≈ 7.59. (c) z = (100−90)/7.59 ≈ 1.32; P(X ≥ 100) = normalcdf(100, 1E99, 90, 7.59) ≈ 0.094. (Exact ≈ 0.106.)
13. (C) the exact finish time in seconds is continuous (any value in an interval). (A), (B), (D) are all counts — discrete.
14. (B) They multiplied height 1/4 by the total width 4 instead of the interval width 3 − 1 = 2. Correct: P(1
15. (a) np = 300(0.90) = 270 ≥ 10 and n(1−p) = 300(0.10) = 30 ≥ 10 ✓. (b) μ = 270, σ = √(300·0.9·0.1) = √27 ≈ 5.196. (c) z = (265−270)/5.196 ≈ −0.96; P(X ≤ 265) = normalcdf(−1E99, 265, 270, 5.196) ≈ 0.168. Yes — since n and p are known, the exact binomial (binomcdf(300, 0.9, 265) ≈ 0.191) would be preferred; the approximation is a bit off here because, with n(1−p) = 30, the distribution is mildly skewed.
FRQ: See the rubric above — 10 points total: (a) 3, (b) 2, (c) 3, (d) 2.
StatsIQ · Lesson 13 of 30 · Unit 2: Probability, Random Variables, and Probability Distributions · Phase 2: Probability
This lesson is aligned to the revised 2026–27 AP® Statistics Course and Exam Description (first exam May 2027). AP® is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.
Accuracy review: All density-curve areas, uniform-rectangle probabilities, z-scores, normalcdf outputs, and exact binomcdf comparisons in this lesson were independently recomputed and verified. Reviewed for statistical accuracy by a retired actuary.