Bars shaped like a bell
Toss a fair coin 6 times and count the heads. X ~ , and out of the 64 equally likely sequences, 1, 6, 15, 20, 15, 6 and 1 give 0 to 6 heads. Drawn as bars, these rise to a peak in the middle and fall away evenly on both sides, already close to the shape of a normal curve.
With more trials the binomial gets harder to use. For X ~ B(100, 0.3), is a sum of 26 terms, each with a coefficient like 100C25. A normal curve of the right shape gives the same probability as one area under the curve.
np < 5: the distribution is still skewed toward 0, so the Normal curve is not yet a safe stand-in for it; np = 1.5
Increase n until np ≥ 5
The bars of B(n, 0.15), with a normal curve of the same mean and standard deviation over them. At n = 10, np = 1.5: the bars are crowded against 0 with a tail to the right, and the curve does not follow them. Drag n up past np = 5 and the bars turn nearly symmetric, with the curve on top of them.
Match the mean and the variance
A binomial X ~ B(n, p) has mean np and variance np(1 − p). The normal curve that fits it is the one with the same two numbers: Y ~ N(np, np(1 − p)).
For X ~ B(100, 0.3), the mean is 100 × 0.3 = 30 and the variance is 100 × 0.3 × 0.7 = 21, so X is approximately N(30, 21). The standard deviation is , and that is what a z-score divides by.
For X ~ , the mean is and the variance is , so X is approximately N(12, 9), with standard deviation 3.
The bars of B(100, 0.3) from 17 to 43, each one unit wide, with the gold curve N(30, 21). The curve runs through the tops of the bars: at 30 the bar is 0.0868 high and the curve 0.0871.
When the fit is good
A normal curve is symmetric and runs on forever in both directions. A binomial count cannot go below 0 or above n, and when p is small its bars are crowded against 0. The curve then fits badly, because it puts area on counts that cannot happen.
So the approximation is used only when np > 5 and n(1 − p) > 5: on average more than 5 successes and more than 5 failures, which leaves room for both tails. For B(100, 0.3), np = 30 and n(1 − p) = 70, both well over 5.
For B(20, 0.1), np = 2 and the test fails. The curve N(2, 1.8) puts 0.0312 of its area below −0.5, on negative counts. Its estimate of is 0.3547, against the exact 0.3917.
A large n alone is not enough. B(200, 0.02) has np = 4, so it fails even with 200 trials. B(100, 0.97) fails on the other side: n(1 − p) = 3.
The bars of B(20, 0.1), where np = 2, with the gold curve N(2, 1.8). The bars start at 0, but the curve runs on to the left of 0, and it misses the bars: at 1 the bar is 0.2702 high and the curve only 0.2252, and at 3 the bar is 0.1901 and the curve 0.2252.
Bars have width
Each bar is one unit wide, so the bar for 25 covers 24.5 to 25.5. To approximate for B(100, 0.3), take the area under N(30, 21) to the left of 25.5, the right edge of that bar: , and the area is 0.1631 (0.1635 from the table with z rounded to −0.98). The exact binomial sum is also 0.1631.
This half unit at the edge is the continuity correction. Using 25 itself gives 0.1376, because it leaves out half of the bar for 25.
The usual mistakes
Giving the variance the mean’s value. has mean 12 and variance , not 12. The factor 1 − p is what makes the variance smaller than the mean.
Dropping the p. is the expected number of failures, not the variance; the variance np(1 − p) has both factors.
Dividing by the variance. A z-score for N(30, 21) divides by , not by 21.
Checking only np. B(100, 0.97) has np = 97 but n(1 − p) = 3, so its bars are crowded against 100 and the normal does not fit.
An application
In the application below, a poll of 100 voters is B(100, 0.5), approximated by , and the answers are checked against the exact binomial sums.
Worked example: A Poll of 100 Voters on a Referendum Too Close to Call, Approximated by a Normal Curve
Question In a referendum, exactly half of the voters support a new bridge. A polling company asks 100 voters chosen at random, and X of them support it. (a) Use a normal approximation with a continuity correction, and Φ(1.5) = 0.9332, to find P(X ≥ 58). Compare it with the exact binomial probability, which is 0.0666 to 4 decimal places. (b) Use the same approximation, with Φ(0.1) = 0.5398, to find the probability that exactly 50 of the 100 voters support the bridge, and explain why the continuity correction is essential here.
1.X ∼ B(100, 0.5) has mean np = 50 and variance np(1 − p) = 100 × 0.5 × 0.5 = 25. Both np and n(1 − p) are 50, well above 5, so X is approximately N(50, 52).
Each bar of B(100, 0.5) has width 1, and the normal curve with the same mean 50 and variance 25 runs through their tops. 2.The event X ≥ 58 includes the whole bar for 58, which starts at 57.5. With the continuity correction, P(X ≥ 58) ≈ P(Y > 57.5), where Y ∼ N(50, 25), and z = 57.5 − 505 = 1.5.
The bars for 58 and above start at 57.5, so the curve is read from the corrected edge, z = 1.5. 3.(a) P(X ≥ 58) ≈ 1 − Φ(1.5) = 1 − 0.9332 = 0.0668. The exact binomial sum is 0.0666, so the approximation is out by only 0.0002.
(a) 1 − Φ(1.5) = 0.0668, against the exact binomial sum of 0.0666. 4.Exactly 50 is the single bar from 49.5 to 50.5. Standardize both edges: z = 49.5 − 505 = −0.1 and z = 50.5 − 505 = 0.1.
The single bar for 50 runs from 49.5 to 50.5, from z = −0.1 to z = 0.1. 5.(b) P(X = 50) ≈ Φ(0.1) − Φ(−0.1) = 2 × 0.5398 − 1 = 0.0796, which agrees with the exact value 10050 × 0.5100 = 0.0796. Without the correction the interval would have no width, and a continuous curve gives a probability of 0 to any single value.
(b) 2 × 0.5398 − 1 = 0.0796, the same as the exact 10050 × 0.5100 to 4 decimal places.
Answer: (a) P(X ≥ 58) ≈ 0.0668, against the exact 0.0666; (b) P(X = 50) ≈ 0.0796, the same as the exact value to 4 decimal places; without the correction a single value would have probability 0
Common mistakes
- Using 58 itself as the edge: z = 58 − 505 = 1.6 gives 0.0548, which is too small. The bar for 58 belongs to the event, and it starts at 57.5.
- Dividing by the variance, 25, instead of the standard deviation: z = 7.525 = 0.3. The standard deviation is √25 = 5.