B. Probability Distributions · Continuous Distributions and the Normal Distribution

The Normal Approximation to the Binomial and Hypergeometric Distributions

B2.7 B2.8About 75 minutes

Learning goals

I can check whether a normal approximation is appropriate and find μ and σ for it.
I can apply the continuity correction to "at least", "at most", "between" and "exactly" questions.
I can use the normal approximation and the z-table to estimate binomial and hypergeometric probabilities.
I can compare an approximation with the exact answer and explain the difference.

Words to know

normal approximation
Using a normal distribution with the same mean and standard deviation to estimate probabilities of a discrete distribution.

continuity correction
Adding or subtracting 0.5 to move from whole-number values of a discrete X to intervals of a continuous Y. For example, P(X ≥ 60) ≈ P(Y > 59.5).

exact probability
The value calculated directly from the binomial or hypergeometric formula (usually with technology when n is large).

Y
In this lesson, the approximating normal variable: Y ~ N(np, npq).

Why approximate?

A fair coin is tossed 100 times. What is the probability of at least 60 heads? The exact binomial answer is

P(X = 60) + P(X = 61) + ... + P(X = 100),

which is 41 terms like C(100, 60)(0.5)¹⁰⁰. That is a job for a computer. Before computers, and still today for a quick estimate, statisticians use the normal approximation: replace the binomial with the normal curve that has the same mean and standard deviation, and find the area with the z-table. The last lesson showed why this works: for large n the histogram is bell-shaped, and each bar's area matches an area under the curve.

The method

Check np ≥ 5 and nq ≥ 5.
Find μ = np and σ = √(npq). Let Y ~ N(μ, σ²).
Apply the continuity correction (see the table below). The bar for a whole number k runs from k − 0.5 to k + 0.5, so include or exclude whole bars.
Standardize and use the z-table.

Discrete question	Normal approximation	Why
P(X ≥ k)	P(Y > k − 0.5)	include the whole bar for k
P(X > k)	P(Y > k + 0.5)	start with the bar for k + 1
P(X ≤ k)	P(Y < k + 0.5)	include the whole bar for k
P(X < k)	P(Y < k − 0.5)	stop after the bar for k − 1
P(X = k)	P(k − 0.5 < Y < k + 0.5)	just the bar for k
P(a ≤ X ≤ b)	P(a − 0.5 < Y < b + 0.5)	bars a to b

Continuity corrections. Y is the approximating normal variable N(np, npq).

Continuity corrections. The shaded boxes are the whole numbers included. The thick lines show where the interval for the normal curve starts and stops: always at the outside edge of the first and last included bar.

z	P(Z < z)	P(Z < −z) = 1 − P(Z < z)
0.10	0.5398	0.4602
0.55	0.7088	0.2912
0.56	0.7123	0.2877
0.89	0.8133	0.1867
0.92	0.8212	0.1788
0.98	0.8365	0.1635
1.02	0.8461	0.1539
1.07	0.8577	0.1423
1.30	0.9032	0.0968
1.49	0.9319	0.0681
1.59	0.9441	0.0559
1.64	0.9495	0.0505
1.65	0.9505	0.0495
1.75	0.9599	0.0401
1.90	0.9713	0.0287
2.00	0.9772	0.0228
2.01	0.9778	0.0222
2.90	0.9981	0.0019
3.47	0.9997	0.0003

z-table values used in this lesson (taken from the full standard normal table in the lesson Using the z-Table to Find Normal Probabilities). The middle column is the area to the left of z; the last column is the area to the left of −z.

Worked example

Example 1: At least 60 heads in 100 tosses

A fair coin is tossed 100 times. Use the normal approximation to estimate P(X ≥ 60).

Check: np = 50 ≥ 5, nq = 50 ≥ 5. ✓
μ = 50, σ = √(100 × 0.5 × 0.5) = 5.
Continuity correction: P(X ≥ 60) ≈ P(Y > 59.5).
z = (59.5 − 50)/5 = 1.90.
P(Z > 1.90) = 1 − 0.9713 = 0.0287.

About 0.0287. The exact binomial value is 0.0284, so the approximation is excellent. Without the continuity correction (using 60), z = 2.00 and the answer would be 0.0228, noticeably worse.

Binomial(100, 0.5) bars from 48 to 72 heads with the normal curve N(50, 5²). "At least 60" includes the whole bar for 60, which starts at 59.5, so the normal area starts at 59.5.

Worked example

Example 2: Cancellations at a ski chalet

A ski chalet takes 200 bookings for a long weekend. From past records, each booking is cancelled with probability 0.15, independently. Estimate the probability that at most 25 bookings are cancelled.

Check: np = 30, nq = 170. ✓
μ = 30, σ = √(200 × 0.15 × 0.85) = √25.5 ≈ 5.0498.
P(X ≤ 25) ≈ P(Y < 25.5).
z = (25.5 − 30)/5.0498 ≈ −0.89.
P(Z < −0.89) = 1 − 0.8133 = 0.1867.

About 0.1867 (exact: 0.1876).

Worked example

Example 3: Between two values

A student guesses all 40 questions on a test with 4 choices each. Estimate the probability of getting between 12 and 15 correct, inclusive.

p = 0.25. Check: np = 10, nq = 30. ✓
μ = 10, σ = √7.5 ≈ 2.7386.
P(12 ≤ X ≤ 15) ≈ P(11.5 < Y < 15.5).
z₁ = (11.5 − 10)/2.7386 ≈ 0.55; z₂ = (15.5 − 10)/2.7386 ≈ 2.01.
0.9778 − 0.7088 = 0.2690.

About 0.2690 (exact: 0.2586). Here p = 0.25 makes the binomial slightly skewed, so the approximation is a little less accurate.

Worked example

Example 4: A hypergeometric situation

A school has 1200 students, and 480 of them (40%) walk to school. A random sample of 60 different students is chosen (without replacement). Estimate the probability that at least 30 of them walk.

This is hypergeometric: N = 1200, r = 480, n = 60. Since 60 is only 5% of 1200, it behaves almost like a binomial with p = 0.4.
μ = n(r/N) = 60(0.4) = 24.
σ² = 60(0.4)(0.6)(1200 − 60)/(1200 − 1) = 14.4 × 1140/1199 ≈ 13.6914, so σ ≈ 3.7002.
P(X ≥ 30) ≈ P(Y > 29.5).
z = (29.5 − 24)/3.7002 ≈ 1.49. P(Z > 1.49) = 1 − 0.9319 = 0.0681.

About 0.0681 (exact hypergeometric: 0.0695).

Worked example

Example 5: An "exactly" question

Estimate the probability of exactly 50 heads in 100 tosses of a fair coin.

P(X = 50) ≈ P(49.5 < Y < 50.5), with μ = 50, σ = 5.
z = ±0.5/5 = ±0.10.
0.5398 − 0.4602 = 0.0796.

About 0.0796 (exact: 0.0796). Without the continuity correction you would get P(Y = 50) = 0, which is clearly wrong.

Is the approximation good enough?

With np and nq both well above 5, answers are usually within about 0.01 of the exact value, and often much closer. Today, spreadsheets compute exact binomial values (=BINOM.DIST(k, n, p, TRUE) gives P(X ≤ k)), so use the approximation to estimate, to check technology, and to understand why so many counts follow a bell curve.

Just for fun

Toss a fair coin 100 times and the chance of exactly 50 heads is only about 8%. So "getting exactly half" is the single most likely result, and it still fails more than 9 times out of 10.

Hands-on activity

Approximate, then check

You need: a calculator, the z-table values in this lesson, and (optional) a spreadsheet.

Choose a real yes-no situation at home, for example: "A dropped buttered toast lands butter-side down with probability 0.5" or "a kernel of popcorn fails to pop with probability 0.1". (If you test popcorn, a grown-up should run the microwave or stove.)
Pick n (for example 200 kernels) and a question such as "at most 25 unpopped".
Check np ≥ 5 and nq ≥ 5, find μ and σ, use the continuity correction and the z-table.
If you have a spreadsheet, compute the exact value with =BINOM.DIST(25, 200, 0.1, TRUE) and compare.
Optional: pop a bag, count the unpopped kernels and see whether your count was likely or unlikely under your model.

What to look for: your normal estimate should be within about 0.01 of the exact value. For popcorn: μ = 20, σ ≈ 4.24, P(X ≤ 25) ≈ P(Z < 1.30) ≈ 0.90.

Your turn

Write the continuity-corrected version of P(X ≥ 45).

Write the continuity-corrected version of P(X < 30).

Write the continuity-corrected version of P(X = 18).

A basketball player makes 80% of free throws. In 50 attempts, estimate the probability of at least 45 baskets.

90% of a certain kind of seed germinates. 400 are planted. Estimate the probability that fewer than 350 germinate.

In a large school board, 30% of students bike to school. In a random sample of 150 students, estimate P(40 ≤ X ≤ 50).

Why can't we use the normal approximation for Binomial(20, 0.1)?

More practice: warm-up

X ~ Binomial(100, 0.3). Find μ and σ.

Continuity-correct P(X ≤ 12).

Continuity-correct P(X > 12).

Continuity-correct P(10 ≤ X ≤ 20).

What is the first thing to check before using a normal approximation?

More practice: on your own

A hotel has 200 rooms and accepts 220 reservations. Each guest independently shows up with probability 0.9. Estimate the probability that more than 200 guests show up.

2% of LED bulbs are defective. In a shipment of 1000, estimate the probability of at most 15 defective bulbs.

A lake has 2000 fish, 500 of which are tagged. A researcher catches 100 different fish. Estimate the probability that at most 20 are tagged.

For 100 tosses of a fair coin, compare P(X ≥ 60) with and without the continuity correction against the exact value 0.0284.

A student guesses on all 40 four-choice questions. Estimate the probability of passing (at least 20 correct).

Challenge

For the hotel above (220 reservations, show-up probability 0.9), how many rooms would the hotel need so that the probability of having enough rooms is at least 0.95?

A coin is tossed 100 times and gives 65 heads. Use the normal approximation to find P(X ≥ 65) for a fair coin. Is the coin likely fair?

Explain why the normal approximation without a continuity correction always gives P(X = k) = 0, and why the correction fixes this.

Check your understanding

Quiz

Which continuity correction matches P(X ≤ 40)?

Which continuity correction matches P(X > 40)?

X ~ Binomial(100, 0.4). Estimate P(X ≥ 45).

For which situation is a normal approximation NOT appropriate?

What are μ and σ of the normal curve used to approximate Binomial(400, 0.5)?

Why is the continuity correction used?

For parents and teachers

Big idea (B2.7, B2.8): a binomial (or hypergeometric) probability with large n can be estimated with the normal curve of the same mean and SD. The continuity correction (±0.5) turns whole-number counts into intervals. The steps are: check np, nq ≥ 5; find μ, σ; correct; standardize; table.

How to help: Ask your teen to draw the little row of boxes (as in the diagram) for each question and mark where the interval starts and stops. If you have a spreadsheet, check answers with =BINOM.DIST(k, n, p, TRUE).

Common difficulties:

Applying the 0.5 in the wrong direction. Drawing the boxes fixes this.
Using npq instead of √(npq) for σ.
Forgetting to check np ≥ 5 and nq ≥ 5.
For hypergeometric problems, forgetting the factor (N − n)/(N − 1) in the variance (it is often close to 1).