Normal Approximation to the Binomial

The normal approximation to the binomial uses a normal curve to estimate binomial probabilities when the number of trials is large. If np ≥ 10 and n(1 − p) ≥ 10, the binomial count X is close to normal, with mean np and standard deviation \sqrt{np(1 - p)}.

Before this lesson: Binomial Distribution, Normal Distribution and Normal Probability Calculations

Why a binomial can look normal

A binomial count X is the number of successes in n independent trials, each with success probability p. With only a few trials, its histogram is often skewed. With many trials, the bars form a smooth, nearly symmetric mound.

That mound is close to a normal curve with the same mean and standard deviation. So you can estimate a binomial probability with a z-score and the z-table, without adding up dozens of binomial terms.

Conditions and formulas

Use the normal approximation only when the large counts condition holds: np ≥ 10 and n(1 − p) ≥ 10. That means at least 10 expected successes and at least 10 expected failures. (Some books use 5 instead of 10.)

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

These are the same mean and standard deviation as the binomial, from expected value rules. Then X is approximately N\left(np, \sqrt{np(1 - p)}\right).

The continuity correction

A binomial count takes only whole numbers, but the normal curve is continuous. The bar for a whole number k covers k − 0.5 to k + 0.5. Moving each boundary by 0.5 to match the bars is called the continuity correction.

AP Statistics usually skips the continuity correction, and many college courses use it. It brings the estimate closer to the exact answer. If your class skips it, use the boundary itself, as in Example 3.

Calculator: for the exact binomial answer, use binomcdf(n, p, k) on a TI-84, or binomialdist(n, p).cdf(k) in Desmos. For the normal estimate, use normalcdf with μ = np and σ = √(np(1 − p)).

Normal approximation examples

Example 1: Check the conditions

Can you use the normal approximation for (a) n = 200 and p = 0.3? (b) n = 40 and p = 0.1?

200(0.3) = 60

200(0.7) = 140

60 \ge 10 \text{ and } 140 \ge 10

40(0.1) = 4

4 \lt 10

Answer: (a) yes; (b) no. With only 4 expected successes, that binomial is skewed right, so a normal curve fits poorly.

Example 2: Find the mean and standard deviation

Suppose 30% of a city’s voters support a new library. A poll takes a random sample of 200 voters. X is the number of supporters in the sample. Find the mean and standard deviation of X.

\mu = 200(0.3) = 60

60(0.7) = 42

\sigma = \sqrt{42} \approx 6.48

Answer: μ = 60 and σ ≈ 6.48, so X is approximately N(60, 6.48). Example 1 showed the conditions hold.

Example 3: Estimate a probability without the correction

Use the poll from Example 2. Estimate the probability that 70 or more voters in the sample support the library. Don’t use the continuity correction.

z = \dfrac{70 - 60}{6.48}

= \dfrac{10}{6.48}

\approx 1.54

P(Z \lt 1.54) = 0.9382

1 - 0.9382 = 0.0618

Answer: about 0.0618. The exact binomial answer, 1 − binomcdf(200, 0.3, 69), is about 0.0728.

Example 4: Use the continuity correction

Repeat Example 3 with the continuity correction.

P(X \ge 70) \approx P(X \gt 69.5)

z = \dfrac{69.5 - 60}{6.48}

= \dfrac{9.5}{6.48}

\approx 1.47

P(Z \lt 1.47) = 0.9292

1 - 0.9292 = 0.0708

Answer: about 0.0708. That is much closer to the exact 0.0728 than the estimate without the correction.

Example 5: Overbooking a flight (word problem)

An airline sells 190 tickets for a flight with 180 seats. Each ticket holder shows up with probability 0.94, independently of the others. Estimate the probability that more than 180 people show up.

190(0.94) = 178.6

190(0.06) = 11.4

178.6(0.06) = 10.716

\sigma = \sqrt{10.716} \approx 3.27

P(X \gt 180) \approx P(X \gt 180.5)

z = \dfrac{180.5 - 178.6}{3.27}

= \dfrac{1.9}{3.27}

\approx 0.58

1 - 0.7190 = 0.2810

Answer: about 0.28, so more than 1 flight in 4 like this would be overbooked. The exact answer, 1 − binomcdf(190, 0.94, 180), is about 0.29.

Common mistakes

Related methods

Exact answers with technology

A calculator finds exact binomial probabilities with binomcdf, so you can check any estimate. The normal approximation still matters because it explains why counts and proportions from large samples are close to normal.

Sample proportions

Divide the count by n to get the sample proportion. Its mean is p, and its standard deviation is \sqrt{\dfrac{p(1 - p)}{n}}. A later lesson, Sampling Distribution of a Sample Proportion, builds on this.

Practice problems

1. For n = 50 and p = 0.4, can you use the normal approximation?
2. For n = 60 and p = 0.05, can you use the normal approximation?
3. For n = 400 and p = 0.25, find the mean and standard deviation of X.
4. Use Problem 3. Estimate P(X ≤ 90) with the continuity correction.
5. A fair coin is flipped 100 times. Estimate the probability of at least 60 heads, with the continuity correction.
6. Write the continuity-corrected normal area for (a) P(X ≥ 45) and (b) P(X < 30).
7. A fair coin is flipped 100 times. Estimate the probability of exactly 50 heads.
8. In a large town, 62% of adults own a pet. In a random sample of 1,000 adults, estimate P(X ≥ 650), with the continuity correction.
9. Why does the approximation need np ≥ 10 and n(1 − p) ≥ 10?
10. A gardener plants 300 seeds, and each grows with probability 0.85. Estimate the probability that fewer than 245 grow, with the continuity correction.

50(0.4) = 20

50(0.6) = 30

60(0.05) = 3

3 \lt 10

\mu = 400(0.25) = 100

100(0.75) = 75

\sigma = \sqrt{75} \approx 8.66

P(X \le 90) \approx P(X \lt 90.5)

z = \dfrac{90.5 - 100}{8.66} = \dfrac{-9.5}{8.66}

\approx -1.10

P(Z \lt -1.10) = 0.1357

\mu = 100(0.5) = 50

100(0.5)(0.5) = 25

\sigma = \sqrt{25} = 5

P(X \ge 60) \approx P(X \gt 59.5)

z = \dfrac{59.5 - 50}{5} = \dfrac{9.5}{5}

= 1.9

1 - 0.9713 = 0.0287

P(X \ge 45) \approx P(X \gt 44.5)

P(X \lt 30) \approx P(X \lt 29.5)

P(X = 50) \approx P(49.5 \lt X \lt 50.5)

z = \dfrac{49.5 - 50}{5} = \dfrac{-0.5}{5}

= -0.1

z = \dfrac{50.5 - 50}{5} = \dfrac{0.5}{5}

= 0.1

0.5398 - 0.4602 = 0.0796

\mu = 1000(0.62) = 620

620(0.38) = 235.6

\sigma = \sqrt{235.6} \approx 15.35

z = \dfrac{649.5 - 620}{15.35} = \dfrac{29.5}{15.35}

\approx 1.92

1 - 0.9726 = 0.0274

\mu = 300(0.85) = 255

255(0.15) = 38.25

\sigma = \sqrt{38.25} \approx 6.18

P(X \lt 245) \approx P(X \lt 244.5)

z = \dfrac{244.5 - 255}{6.18} = \dfrac{-10.5}{6.18}

\approx -1.70

P(Z \lt -1.70) = 0.0446

Normal approximation FAQ

When can you use the normal approximation to the binomial?

When np ≥ 10 and n(1 − p) ≥ 10. Then the binomial distribution is close enough to symmetric for a normal curve.

What is the mean and standard deviation for the normal approximation?

Use the binomial’s own values: μ = np and σ = √(np(1 − p)).

What is the continuity correction?

It moves each boundary by 0.5 so the normal curve covers whole bars of the binomial histogram. For example, P(X ≥ 70) becomes P(X > 69.5).

Do you need the continuity correction on the AP Statistics exam?

AP Statistics usually does not use it. Many college courses do, and it gives an estimate closer to the exact answer.

Summary

• A binomial count X is close to normal when np ≥ 10 and n(1 − p) ≥ 10.
• Use μ = np and σ = √(np(1 − p)).
• The continuity correction moves each boundary by 0.5 to match the bars.
• At least k starts at k − 0.5. More than k starts at k + 0.5.
• Check estimates with binomcdf when you have a calculator.

Previous lesson: Normal Probability Plots

Next lesson: Sampling Distributions

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