Difficulty: Introductory | Prerequisites: Binomial distribution, Normal distribution basics (pnorm, qnorm)
The Binomial distribution counts successes in a fixed number of independent trials, but computing exact Binomial probabilities gets unwieldy for large n. The Normal approximation lets you reuse pnorm and qnorm to get close answers, provided certain conditions are met. Knowing when the approximation is valid (and when it breaks down) is a core skill for this course and a reliable exam topic.
When n is large enough that both np ≥ 5 and n(1−p) ≥ 5, you can approximate a Binomial(n, p) distribution with a Normal distribution whose mean is np and standard deviation is √(np(1−p)). Apply a continuity correction (±0.5) for better accuracy. If either condition fails, the approximation is poor and you should use exact Binomial calculations instead.
Binomial distribution
The distribution of the number of successes in n independent trials, each with success probability p. Written X ~ Bin(n, p). Think of it as the "how many times does this happen out of n tries" distribution.
Normal approximation to the Binomial
The technique of treating Bin(n, p) as approximately N(np, np(1−p)) when n is large enough. This lets you use pnorm instead of summing many individual Binomial probabilities. Think of it as swapping the staircase-shaped Binomial histogram for a smooth bell curve that sits on top of it.
Conditions for the approximation (np rule)
Both np ≥ 5 and n(1−p) ≥ 5 must hold. These ensure the Binomial distribution is symmetric enough for the Normal curve to fit it well. In simple terms, you need enough expected successes and enough expected failures for the bell shape to be a reasonable match.
Continuity correction
An adjustment of ±0.5 when converting a discrete Binomial probability to a continuous Normal one. P(X ≤ 27) becomes P(X ≤ 27.5) under the Normal, and P(X ≤ 1) becomes P(X ≤ 1.5). Think of it as giving each integer its own half-unit of width under the smooth curve.
dbinom / sum(dbinom(...))
R functions for exact Binomial probabilities. dbinom(k, n, p) gives P(X = k). sum(dbinom(0:k, n, p)) gives P(X ≤ k). These are the "ground truth" you compare the Normal approximation against.
Before using the Normal approximation, compute np and n(1−p). Both must be at least 5.
Example where conditions FAIL (Dr. Keaton, n = 100, p = 0.01):
np = 100 × 0.01 = 1 (less than 5).
n(1−p) = 100 × 0.99 = 99 (greater than 5).
Because np < 5, the Normal approximation is not appropriate here. The Binomial is too skewed.
Example where conditions PASS (Dr. Craig, n = 100, p = 0.30):
np = 100 × 0.30 = 30 (greater than 5).
n(1−p) = 100 × 0.70 = 70 (greater than 5).
Both conditions met, so the Normal approximation is appropriate.
Once the conditions are satisfied, use:
Mean = np
SD = √(np(1−p))
For Dr. Craig's example: mean = 30, SD = √(100 × 0.30 × 0.70) = √21 ≈ 4.583.
For P(X ≤ k), compute pnorm(k + 0.5, mean, SD).
Dr. Keaton (conditions NOT met, for comparison):
Normal approximation for P(X ≤ 1): pnorm(1.5, mean = 1, sd = sqrt(0.99)) = 0.6923.
Exact Binomial: sum(dbinom(0:1, 100, 0.01)) = 0.7358.
The approximation is off by about 0.04, a noticeable gap. This is what happens when np < 5.
Dr. Craig (conditions met):
Normal approximation for P(X ≤ 27): pnorm(27.5, mean = 30, sd = sqrt(21)) = 0.2927.
Exact Binomial: sum(dbinom(0:27, 100, 0.30)) = 0.2964.
The approximation is off by only about 0.004, much closer. This is the payoff of meeting both conditions.
Compare the Normal approximation result to the exact Binomial result. When the conditions are met, the two should be close. When they are not met, expect a meaningful gap, and prefer the exact calculation.
Conditions: np ≥ 5 and n(1−p) ≥ 5
Approximation parameters: mean = np, SD = √(np(1−p))
Left-tail with continuity correction: P(X ≤ k) ≈ pnorm(k + 0.5, np, √(np(1−p)))
Exact Binomial (R): P(X ≤ k) = sum(dbinom(0:k, size = n, prob = p))
Students often skip the condition check and jump straight to pnorm. The conditions (np ≥ 5 and n(1−p) ≥ 5) are not optional. If they fail, the answer from pnorm will be noticeably wrong.
Students sometimes check only one condition (np ≥ 5) and forget n(1−p) ≥ 5. Both must hold.
Forgetting the continuity correction is common. Without it, the approximation is systematically less accurate. P(X ≤ 27) becomes pnorm(27.5, ...), not pnorm(27, ...).
Students confuse the standard deviation formula. It is √(np(1−p)), not √(np) or np(1−p). The square root is easy to drop.
⚠️ Expect a question that gives you n and p and asks whether the Normal approximation is appropriate. The answer is entirely determined by checking np ≥ 5 and n(1−p) ≥ 5. Show your working for both.
⚠️ A classic exam format: compute the approximation, compute the exact answer, then explain why they agree or disagree. The explanation ties back to the conditions.
⚠️ The continuity correction (+0.5 for "at most", −0.5 for "at least") is a frequent source of lost marks. Know which direction to adjust.
⚠️ When p is very small (like 0.01) and n is moderate (like 100), np can easily fall below 5. This is the textbook scenario for a poor approximation, and exams test whether you recognise it.
True or false: If n = 50 and p = 0.08, the Normal approximation to the Binomial is appropriate.
Fill in the blank: The mean of the Normal approximation to Bin(100, 0.30) is ___ and the SD is ___.
True or false: P(X ≤ 27) for a Binomial is approximated using pnorm(27, np, SD) with no adjustment.
Fill in the blank: When np < 5, the Binomial distribution is too ___ for the Normal curve to fit well.
True or false: The exact Binomial and Normal approximation always give the same answer.
Answers: 1. False (np = 4, which is less than 5). 2. 30; √21 ≈ 4.58. 3. False (continuity correction: use 27.5). 4. Skewed. 5. False (they are close when conditions are met, but never identical).
Q: n = 100, p = 0.01. Is the Normal approximation to the Binomial appropriate here? Explain.
A: No. np = 1, which is less than 5. Even though n(1−p) = 99 is well above 5, both conditions must hold. The approximation will be inaccurate.
Q: n = 100, p = 0.30. Approximate P(X ≤ 27) using the Normal distribution.
A: Mean = 30, SD = √(21) ≈ 4.583. With continuity correction: pnorm(27.5, 30, 4.583) ≈ 0.2927.
Q: For the same scenario (n = 100, p = 0.30), what is the exact Binomial P(X ≤ 27)?
A: sum(dbinom(0:27, 100, 0.30)) ≈ 0.2964.
Q: How does the approximation quality in Q2 compare to a case where n = 100 and p = 0.01?
A: It is much better. When p = 0.30, both conditions (np ≥ 5 and n(1−p) ≥ 5) are satisfied, so the Normal curve fits the Binomial histogram well. When p = 0.01, np = 1, the Binomial is heavily skewed, and the Normal approximation overshoots or undershoots noticeably.
Q: What continuity correction would you apply to approximate P(X ≥ 35) for X ~ Bin(100, 0.30)?
A: Use 1 − pnorm(34.5, 30, √21). The correction shifts the boundary down by 0.5 because "at least 35" includes 35 itself.
The Normal approximation to the Binomial is a special case of the Central Limit Theorem (CLT), which says that sums of many independent random variables tend toward a Normal distribution regardless of the original distribution's shape. Understanding why np ≥ 5 matters here prepares you for the CLT's sample-size requirements later. This topic also connects directly to the companion study notes on Normal distribution probabilities, where pnorm and qnorm are introduced.
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