5.5 - Approximating a binomial distribution

5.5 - Approximating a binomial distribution

A binomial probability may involve many terms. Learn when a normal curve is a suitable approximation, use half-integer boundaries to match a discrete count to a continuous area, and check that the final approximation describes the intended event.

When approximation is suitable

Suppose X∼B(100,0.4)X\sim\mathrm B(100,0.4). How can you estimate P(X≤42)\mathrm P(X\le42) without adding 4343 binomial terms? A normal curve can approximate the distribution, but matching just the mean is not enough: its spread and the event boundary matter too.

When both the expected numbers of successes and failures are sufficiently large, the binomial bars form a roughly symmetric bell shape. Very small expected counts on either side can produce pronounced skew, so a normal curve is less suitable.

Normal approximation to a binomial distribution

For X∼B(n,p)X\sim\mathrm B(n,p), put q=1−pq=1-p. Use a normal approximation when

np>5andnq>5.np>5\quad\text{and}\quad nq>5.

Match the mean and variance with

Y∼N(np,npq).Y\sim\mathrm N\bigl(np,npq\bigr).

The binomial mean and variance are in the formula booklet. The approximation conditions and the choice to use this normal model are not: learn them.

These are suitability conditions, not a promise of an exact result or a fixed maximum error. A large nn alone is insufficient if pp is very close to 00 or 11. The underlying experiment must first satisfy the binomial assumptions.

Worked example 1: a suitable approximation

Question. For X∼B(100,0.4)X\sim\mathrm B(100,0.4), justify a normal approximation and state its parameters.

Plan. Check both expected counts, then use the binomial mean and variance.

np=100(0.4)=40>5nq=100(0.6)=60>5npq=100(0.4)(0.6)=24\begin{aligned} np&=100(0.4)=40>5\\ nq&=100(0.6)=60>5\\ npq&=100(0.4)(0.6)=24 \end{aligned}

Answer. Both conditions hold, so use Y∼N(40,24)Y\sim\mathrm N(40,24), with standard deviation 24\sqrt{24}.

Check. The proposed normal mean 4040 equals E(X)\mathrm E(X), and its variance 2424 equals Var⁡(X)\operatorname{Var}(X). The second parameter is not the standard deviation.

Keep 100100 trials and change only the success probability. A large trial count does not rescue a small expected number of failures.

Worked example 2: one condition fails

Question. Is a normal approximation suitable for W∼B(100,0.96)W\sim\mathrm B(100,0.96)?

Plan. Check both npnp and nqnq, rather than just the larger expected count.

np=96>5,nq=100(0.04)=4<5.np=96>5,\qquad nq=100(0.04)=4<5.

Answer. No: the expected number of failures is too small for the stated normal-approximation conditions.

Check. The mean is close to the maximum possible count 100100, so a symmetric bell curve would poorly reflect the short upper side and longer lower side. This agrees with the failed numerical condition.

At equality, such as np=5np=5, the stated condition np>5np>5 is not met. Give both numerical checks when asked to justify the approximation.

Watch out

Do not write N(np,npq)\mathrm N(np,\sqrt{npq}). Normal notation takes the variance as its second parameter. Use npq\sqrt{npq} only when standardising.

Test both conditions in this choice.

Continuity correction

A binomial variable puts probability at integers; a normal variable spreads it continuously. Represent each integer count by a unit-width bar centred on that integer. The bar for 4242 occupies 41.541.5 to 42.542.5, so including all of count 4242 means going to the right edge, 42.542.5.

Continuity correction

A continuity correction replaces an integer event boundary by the appropriate half-integer edge when using a continuous normal model for a discrete count.

Write the allowed integer values first. This is safer than memorising “add 0.50.5” because the direction depends on the event.

Worked example 3: including the boundary count

Question. For X∼B(100,0.4)X\sim\mathrm B(100,0.4), write the continuity-corrected normal event approximating P(X≤42)\mathrm P(X\le42).

Plan. Include the whole unit bar centred at 4242, so use its upper edge.

The allowed counts end at 4242. Their bars extend up to 42.542.5, giving

P(X≤42)≈P(Y<42.5),Y∼N(40,24).\mathrm P(X\le42)\approx\mathrm P(Y<42.5), \qquad Y\sim\mathrm N(40,24).

Answer. Use the normal area below 42.542.5.

Check. The bar from 41.541.5 to 42.542.5 is included in full. Stopping at 4242 would include only half of the boundary bar.

Change only ≤\le to <<. The largest allowed integer becomes 4141, so the corrected boundary moves by a whole unit.

Worked example 4: excluding the boundary count

Question. For the same binomial model, write the continuity-corrected normal event approximating P(X<42)\mathrm P(X<42).

Plan. Translate X<42X<42 to X≤41X\le41, then use the upper edge of the bar for 4141.

P(X<42)=P(X≤41)≈P(Y<41.5).\mathrm P(X<42)=\mathrm P(X\le41) \approx\mathrm P(Y<41.5).

Answer. Use the normal area below 41.541.5.

Check. The bar for 4242, from 41.541.5 to 42.542.5, is excluded. This matches the strict inequality in the original discrete event.

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