Continuity Correction

A bar is one unit wide

A binomial count takes whole-number values only. In its bar chart each bar is one unit wide and stands on its value, so the bar for X = 7 runs from 6.5 to 7.5, and its area is P(X = 7).

The bars for 6, 7 and 8 sit side by side with no gaps: 5.5 to 6.5, 6.5 to 7.5 and 7.5 to 8.5. Every point of the axis belongs to exactly one bar.

The curve has no bars

A normal curve is continuous: it gives a probability to an interval, and the probability of any single value is 0. So P(X = 7) cannot be read as the area at 7 itself. It is the area over the bar for 7, from 6.5 to 7.5.

Every event about the count is turned into an interval of bar edges before the curve is used. Moving a boundary by half a unit to the edge of a bar is called the continuity correction.

Keep the bar or drop it

Ask whether the event includes the bar at its boundary. $X\ge 7$X ≥ 7 includes 7, so it keeps the whole bar for 7, and the area starts at that bar’s left edge: $P\left(X\ge 7\right)\approx P\left(Y>6.5\right)$P(X ≥ 7) ≈ P(Y > 6.5).

X > 7 does not include 7. It starts at 8, so it drops the bar for 7, and the area starts at that bar’s right edge: $P\left(X>7\right)\approx P\left(Y>7.5\right)$P(X > 7) ≈ P(Y > 7.5).

The other side works the same way. $X\le 7$X ≤ 7 keeps the bar for 7, so $P\left(X\le 7\right)\approx P\left(Y<7.5\right)$P(X ≤ 7) ≈ P(Y < 7.5); X < 7 drops it, so $P\left(X<7\right)\approx P\left(Y<6.5\right)$P(X < 7) ≈ P(Y < 6.5). And X = 7 is that one bar: $P\left(X=7\right)\approx P\left(6.5<Y<7.5\right)$P(X = 7) ≈ P(6.5 < Y < 7.5).

An example

X ~ B(100, 0.3) has mean 30 and variance 21, and both np = 30 and n(1 − p) = 70 are greater than 5, so Y ~ N(30, 21) approximates it. Its standard deviation is $\sqrt{21}=4.583$√21 = 4.583.

$P\left(X\le 25\right)$P(X ≤ 25) keeps the bar for 25, so the boundary is 25.5: $z=\frac{25.5-30}{4.583}=-0.98$z = (25.5 − 30)/4.583 = −0.98, and the area to its left is 0.1631 (0.1635 from the table with z rounded to −0.98). The exact binomial sum is 0.1631.

Without the correction, the boundary 25 cuts the bar for 25 down the middle: $z=\frac{25-30}{4.583}=-1.09$z = (25 − 30)/4.583 = −1.09, and the area is 0.1376, too small by about half that bar.

The bars of B(100, 0.3) from 19 to 33 under the gold curve N(30, 21). The shaded area under the curve stops at 25.5, the right edge of the bar for 25, so it takes in that bar whole.

More events

$P\left(X\ge 35\right)$P(X ≥ 35) keeps the bar for 35, so the boundary is its left edge, 34.5: $z=\frac{34.5-30}{4.583}=0.98$z = (34.5 − 30)/4.583 = 0.98, and P(Y > 34.5) = 0.1631. The exact value is 0.1629.

P(X = 30) is the single bar from 29.5 to 30.5: P(29.5 < Y < 30.5) = 0.0869, against the exact 0.0868. Without the correction the interval would have no width and the curve would give 0.

$P\left(25\le X\le 35\right)$P(25 ≤ X ≤ 35) keeps the bars at both ends, so it runs from 24.5 to 35.5: 0.7699, against the exact 0.7704.

P(X = 30) on the curve: the shaded slice from 29.5 to 30.5 sits over the bar for 30, and its area, 0.0869, is close to the bar’s 0.0868.

The usual mistakes

Using the whole number as the boundary. $P\left(X\le 25\right)$P(X ≤ 25) with 25 as the edge cuts the bar for 25 in half and gives 0.1376 instead of 0.1631.

Moving the wrong way. P(X > 7) does not include 7, so its edge is 7.5; using 6.5 sweeps in the bar for 7. $P\left(X\ge 7\right)$P(X ≥ 7) includes 7, so its edge is 6.5; using 7.5 throws that bar out.

Rewriting a strict inequality first, then correcting again. P(X < 7) is $P\left(X\le 6\right)$P(X ≤ 6), and either form gives the edge 6.5. Correcting $X\le 6$X ≤ 6 to 6.5 and then moving another half unit gives 6.

Correcting a variable that was continuous to begin with. A mass or a time has no bars, so a normal variable is used as it is.