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Module 9/Normal approximation & the CLT

The CLT & the normal approximation

For large nn, a Bin(n,p)\mathrm{Bin}(n,p) count behaves like a normal with the same mean and variance. That is the Central Limit Theorem at work — and it lets us estimate binomial probabilities with the standard normal.

Before you start — give these a try

Attempting first primes your brain for the lesson — even if you miss. Nothing is graded or saved; it's just a warm-up.

X∼Bin(100,0.5)X\sim\mathrm{Bin}(100,0.5) (μ=50, σ=5\mu=50,\ \sigma=5). Using the normal approximation with continuity correction, estimate P(X≤60)P(X\le 60).

X∼Bin(100,0.5)X\sim\mathrm{Bin}(100,0.5). Using the normal approximation with continuity correction, estimate P(X≥60)P(X\ge 60).

What you’ll be able to do

  • State the Central Limit Theorem for binomial counts and the normal approximation Bin(n,p)≈N(np, np(1−p))\mathrm{Bin}(n,p)\approx N(np,\,np(1-p)).
  • Standardize a binomial variable to a zz-score.
  • Apply the continuity correction when approximating a discrete probability with the normal.
  • Decide when the normal approximation is appropriate (the np, n(1−p)≥10np,\,n(1-p)\ge 10 rule of thumb).

In your course

· MATH2015 · Linear Algebra & Probability
§14.1 Normal Approximation and the Central Limit Theorem
  • Theorem 14.1CLT for binomial random variables
    lim⁡n→∞P(a≤Sn−npnp(1−p)≤b)=12π∫abe−x2/2dx\lim_{n\to\infty}P\big(a\le \tfrac{S_n-np}{\sqrt{np(1-p)}}\le b\big)=\tfrac{1}{\sqrt{2\pi}}\int_a^b e^{-x^2/2}dx.
  • Remark 14.1Continuity correction
  • Remark 14.2When to use the normal approximation (np, n(1−p)≥10np,\ n(1-p)\ge 10)
  • Example 14.1Bin(1000,0.6)\mathrm{Bin}(1000,0.6) vs N(600,240)N(600,240)
1

The Central Limit Theorem, informally

The Central Limit Theorem (CLT) says: if you add up a large number of independent, identically distributed pieces with finite mean and variance, the sum is approximately normal — whatever the distribution of the individual pieces. A Bin(n,p)\mathrm{Bin}(n,p) count is exactly such a sum: Sn=X1+⋯+XnS_n=X_1+\cdots+X_n where each XiX_i is Bernoulli(p)(p) with mean pp and variance p(1−p)p(1-p).

2

Normal approximation to the binomial

Because SnS_n is a sum of nn i.i.d. Bernoullis, for large nn it is approximately normal with the matching mean and variance:\n\nSn∼Bin(n,p) ≈ N(np, np(1−p)).S_n\sim\mathrm{Bin}(n,p)\ \approx\ N\big(np,\ np(1-p)\big).\n\nMean μ=np\mu=np and variance σ2=np(1−p)\sigma^2=np(1-p).

3

Standardizing

To compare across different nn, standardize: subtract the mean and divide by the standard deviation,\n\nZ=Sn−npnp(1−p).Z=\frac{S_n-np}{\sqrt{np(1-p)}}.\n\nThe standardized variable has mean 00 and variance 11, and the CLT says it converges in distribution to the standard normal N(0,1)N(0,1) as n→∞n\to\infty.

4

Continuity correction

The binomial is discrete but the normal is continuous, so we widen each integer by half a unit. To approximate P(X≤k)P(X\le k) use P ⁣(Z≤k+0.5−npnp(1−p))P\!\big(Z\le \tfrac{k+0.5-np}{\sqrt{np(1-p)}}\big); for an interval [k1,k2][k_1,k_2] use the endpoints k1−0.5k_1-0.5 and k2+0.5k_2+0.5. Skipping the correction noticeably worsens the estimate for small nn.

Theorem 14.1 — CLT for binomial random variables

Fix 0<p<10<p<1 and let Sn∼Bin(n,p)S_n\sim\mathrm{Bin}(n,p). Then for any fixed reals a≤ba\le b,\n\nlim⁡n→∞P ⁣(a≤Sn−npnp(1−p)≤b)=12π∫abe−x2/2 dx.\lim_{n\to\infty}P\!\left(a\le \frac{S_n-np}{\sqrt{np(1-p)}}\le b\right)=\frac{1}{\sqrt{2\pi}}\int_a^b e^{-x^2/2}\,dx.

Intuition. The standardized binomial fills in the standard-normal bell curve as nn grows, so normal-table probabilities approximate binomial ones.
Remark 14.2 — When to use the normal approximation

The normal approximation works well when nn is large and pp is not too close to 00 or 11. A common rule of thumb: both npnp and n(1−p)n(1-p) should be at least 1010.

Intuition. If pp is tiny, the distribution is lopsided and the symmetric bell is a poor fit — that regime calls for the Poisson approximation instead.

Worked examples

Example 1

X∼Bin(1000,0.6)X\sim\mathrm{Bin}(1000,0.6). Find the mean and variance used for its normal approximation.

  1. 1

    Mean: μ=np=1000⋅0.6=600\mu=np=1000\cdot 0.6=600.

  2. 2

    Variance: σ2=np(1−p)=1000⋅0.6⋅0.4=240\sigma^2=np(1-p)=1000\cdot 0.6\cdot 0.4=240.

  3. 3

    So X≈N(600,240)X\approx N(600,240), with σ=240≈15.49\sigma=\sqrt{240}\approx 15.49.

Answer. X≈N(600, 240)X\approx N(600,\,240).
Example 2

X∼Bin(100,0.5)X\sim\mathrm{Bin}(100,0.5). Estimate P(X≤60)P(X\le 60) with the normal approximation and a continuity correction.