When n is large, the binomial bar chart can settle into a bell curve.
So we replace B(n,p) with a matching normal.
The matching normal has the same mean np and variance npq.
The catch: a discrete bar chart becomes a smooth curve.
So every integer boundary needs a continuity correction of ±0.5.
np, npq, and the ±0.5 patch
Big n → bars become a bell. Replace B(n,p) with N(np, npq), nudge the boundary by 0.5, then standardise with σ = √(npq).
Allowed only if np>5 and nq>5. Then X≈N(np,npq).
Drag n up and watch B(n,p) settle into a bell — that is exactly why the approximation works.
B(20, 0.40)·mean np = 8.0·variance npq = 4.80
Each integer k owns the strip [k−0.5,k+0.5]; shift the boundary to keep the bar you mean, drop the one you don't:
event X < 43
normal cut 42.5
z-score 0.510
43 is excluded, so cut before its bar. The gold bars are the binomial values kept by the event; the gold area is the matching normal estimate.
exact binomial ≈ 0.6967·normal with correction ≈ 0.6951
Check np>5 AND nq>5 (else don't approximate).
State X≈N(np,npq).
σ=npq.
Apply ±0.5: decide if each integer boundary is in or out, keep/drop its bar.
Standardise the corrected boundary(ies), then read Φ.
Phrase
Includes k?
Use
P(X<k)(fewer than)
no
k−0.5
P(X≤k)(at most)
yes
k+0.5
P(X>k)(more than)
no
k+0.5
P(X≥k)(at least)
yes
k−0.5
P(a<X<b)
both out
a+0.5,b−0.5
Worked example
Simplest rung: one clean tail
X∼B(100,0.4): np=40,nq=60 (both > 5), so X≈N(40,24). For P(X<43) the integer 43 is excluded, so cut at 42.5: z=2442.5−40=0.510, and P≈Φ(0.510)=0.695. (9709/52/M/J/24 Q6)
Answer
≈0.695
Worked example
Same setup, the OTHER side
Same X≈N(40,24). For P(X>43), 43 is again excluded but now we keep the bars above it, so cut at 43.5: z=2443.5−40=0.714, and P≈1−Φ(0.714). Notice the same 43 corrects to a different value depending on which bars you keep.
Rare events, and chaining the result into a binomial
np>5 can hold even for a tiny p if n is huge — so the normal route still works. And just like §8.2, the approximated probability can become the p of a small binomial.
Worked example
Rare events: small p, large n
Boxes of 8000 items, 0.2% yellow: X∼B(8000,0.002), so np=16, npq=15.968 (tiny p, but np>5): X≈N(16,15.968). For P(X<20) use 19.5: z=15.96819.5−16=0.876, so P≈Φ(0.876)=0.809.
Answer
≈0.809
Worked example
Approximate, then chain to a binomial
From above, a box has fewer than 20 yellow items with probability p=0.8094. Find the chance that exactly 2 of 3 boxes do. Let Y∼B(3,0.8094): P(Y=2)=(23)(0.8094)2(0.1906)=0.375. Keep the unrounded p through the chain.
Answer
0.375
Worked example
The hardest shape: a 'between' with two corrections
X∼B(80,0.45): np=36,npq=19.8, σ=4.450. P(30<X<40) corrects inward to 30.5 and 39.5: z1=4.4530.5−36=−1.236,z2=4.4539.5−36=0.787, so P=Φ(0.787)+Φ(1.236)−1. (9709/52/O/N/24 Q7)
Answer
≈0.676
Examiner note
A 5-mark ladder — lay out every line: B1 state N(np,npq) (label it variance, not SD); M1 the ±0.5 continuity correction (its own mark, every paper); M1 standardise with σ=npq; M1 a valid Φ area; A1 the final value.
Common mistake
Forgetting the continuity correction (the most-caught error) or correcting the wrong way; on a “between” question both ends must move inward. Also using σ=npq instead of σ2=npq, and skipping the np>5,nq>5 check.
Your turn— tap to reveal the worked answer (9709-style)
A fair coin is tossed 888 times. Estimate P(at most 450 heads).
np=nq=444 (ok), ≈N(444,222). “At most 450” includes 450, so use 450.5: z=222450.5−444=0.436.
Answer
Φ(0.436)≈0.669
Why exactly ±0.5
The discrete bar over integer k spans the continuous strip [k−0.5,k+0.5]. “Fewer than 22” excludes that whole bar, so the cut sits at 21.5; “at most 22” keeps it, so the cut sits at 22.5. The rule is just “does the boundary integer's bar belong in the region?”