Approximations: Binomial, Poisson, and Normal

In Further Statistics, you must know when and how to approximate one distribution using another. These approximations simplify calculations and are essential when exact computation is impractical. This lesson covers the three key approximations: Poisson approximation to the binomial, normal approximation to the binomial, and normal approximation to the Poisson.

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Poisson Approximation to the Binomial

If $X \sim B(n, p)$X∼B(n,p) with:

• $n$n large (typically $n \geq 50$n≥50)
• $p$p small (typically $p \leq 0.1$p≤0.1)
• $np$np moderate (typically $np < 5$np<5)

then $X$X can be approximated by $\text{Po}(\lambda)$Po(λ) where $\lambda = np$λ=np.

Binomial	Poisson approximation
$B(n, p)$B(n,p)	$\text{Po}(np)$Po(np)
$E(X) = np$E(X)=np	$E(X) = \lambda = np$E(X)=λ=np
$\text{Var}(X) = np(1-p)$Var(X)=np(1−p)	$\text{Var}(X) = \lambda = np$Var(X)=λ=np

The approximation improves as $n \to \infty$n→∞ and $p \to 0$p→0 with $np$np held constant.

Why the variance approximation works: When $p$p is small, $1 - p \approx 1$1−p≈1, so $np(1-p) \approx np = \lambda$np(1−p)≈np=λ.

Worked Example

A manufacturer produces components with a defect rate of 2%. In a batch of 200, find the probability that exactly 3 are defective.

Exact: $X \sim B(200, 0.02)$X∼B(200,0.02), $P(X = 3) = \binom{200}{3}(0.02)^3(0.98)^{197}$P(X=3)=(3200​)(0.02)3(0.98)197 — difficult to compute.

Approximation: $\lambda = 200 \times 0.02 = 4$λ=200×0.02=4, so $X \approx \text{Po}(4)$X≈Po(4).

$P(X = 3) \approx \frac{e^{-4} \times 4^3}{3!} = \frac{0.01832 \times 64}{6} = 0.1954$P(X=3)≈3!e−4×43​=60.01832×64​=0.1954

Exam Tip: In the exam, always state: "Since $n$n is large and $p$p is small, the Poisson approximation $\text{Po}(np)$Po(np) is appropriate." Then state the value of $\lambda = np$λ=np.

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Normal Approximation to the Binomial

If $X \sim B(n, p)$X∼B(n,p) with:

• $np > 5$np>5 and $n(1-p) > 5$n(1−p)>5 (both large enough)

then $X$X can be approximated by $N(np, np(1-p))$N(np,np(1−p)).

Binomial	Normal approximation
$B(n, p)$B(n,p)	$N(np, np(1-p))$N(np,np(1−p))

Continuity Correction

Since the binomial is discrete and the normal is continuous, a continuity correction of $\pm 0.5$±0.5 is applied:

Binomial probability	Normal approximation
$P(X = k)$P(X=k)	$P(k - 0.5 < Y < k + 0.5)$P(k−0.5<Y<k+0.5)
$P(X \leq k)$P(X≤k)	$P(Y < k + 0.5)$P(Y<k+0.5)
$P(X \geq k)$P(X≥k)	$P(Y > k - 0.5)$P(Y>k−0.5)
$P(X < k)$P(X<k)	$P(Y < k - 0.5)$P(Y<k−0.5)
$P(X > k)$P(X>k)	$P(Y > k + 0.5)$P(Y>k+0.5)

Worked Example

A coin is tossed 100 times. Find the probability of getting more than 55 heads.

$X \sim B(100, 0.5)$X∼B(100,0.5). Here $np = 50$np=50, $n(1-p) = 50$n(1−p)=50, both > 5.

Approximate: $Y \sim N(50, 25)$Y∼N(50,25), so $\sigma = 5$σ=5.

$P(X > 55) \approx P(Y > 55.5) = P\left(Z > \frac{55.5 - 50}{5}\right) = P(Z > 1.1)$P(X>55)≈P(Y>55.5)=P(Z>555.5−50​)=P(Z>1.1)