Comparing Exact Binomial and CLT Probabilities

Why Compare Exact and Approximate Probabilities?

A binomial count has a discrete distribution: it can take only whole-number values. A normal distribution is continuous, so it assigns probability to intervals rather than to individual points. The Central Limit Theorem explains why a normal curve can approximate the distribution of a suitably well-behaved binomial count, but the approximation is not automatically exact. Comparing it with the binomial calculation shows how much accuracy is gained or lost, and how a continuity correction can help.

For a moderate sample size, an exact probability may be straightforward to calculate by adding binomial probabilities. The normal approximation can be useful when a calculation involves many possible counts or when we want a quick approximation. In either case, the event must be translated correctly: using a continuous boundary at an integer rather than halfway between integers can omit or include substantial area.

Exact Binomial Probabilities and Their Normal Approximation

Definition: If $X$ counts successes in $n$ independent trials, each with the same success probability $p$, then $X\sim\operatorname{Bin}(n,p)$. Its exact probability at an integer $k$, for $k=0,1,\ldots,n$, is $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}. $$ For an event consisting of several possible counts, add the probabilities for those counts. The matching normal approximation uses a random variable $Y\sim N(np,np(1-p))$, whose mean is $np$ and whose variance is $np(1-p)$.

The binomial model requires a fixed number of trials, two possible outcomes per trial, independent outcomes, and a common probability $p$ of success on each trial. The normal model uses the binomial mean and variance; its standard deviation is $\sqrt{np(1-p)}$. A normal approximation is generally more plausible when both $np$ and $n(1-p)$ are sufficiently large. This is a diagnostic, not a guarantee of a particular error size, especially for tail probabilities.

Why the Continuity Correction Uses Half Units

Integer counts occupy separate points on the number line, while a normal curve spreads probability continuously across it. A continuity correction assigns each possible integer count a surrounding interval of width one. For example, the count $k$ is represented by the interval from $k-0.5$ to $k+0.5$. Thus an event that includes the integers $a$ through $b$, inclusive, is approximated by the continuous interval from $a-0.5$ to $b+0.5$.

Continuity-correction rule: For an integer-valued binomial count, approximate $P(X\le k)$ by $P(Y\le k+0.5)$, and approximate $P(X\ge k)$ by $P(Y\ge k-0.5)$. For $a\le X\le b$, use $a-0.5\le Y\le b+0.5$. For a strict inequality, first identify which integer counts satisfy it, then place the corrected boundary halfway between the included and excluded counts.

The rule follows from the unit-width representation. The union of the intervals belonging to counts $a,a+1,\ldots,b$ runs from $a-0.5$ to $b+0.5$. A continuous random variable has probability zero at any single boundary, so whether an endpoint is written with a strict or inclusive inequality does not alter the normal probability.

Worked Example: A Central Range for a Symmetric Binomial Count

Suppose, hypothetically, that 40 independent trials each have success probability $0.5$. Find the exact probability that the success count is from 18 through 22, then compare two normal approximations.

Here $X\sim\operatorname{Bin}(40,0.5)$, so the model conditions are built into the hypothetical setup: there are 40 trials, each has two outcomes, the trials are independent, and each has the same success probability. The expected success and failure counts are both 20, so the normal approximation has a balanced center.

Exact probability. By symmetry, $P(X=19)=P(X=21)$ and $P(X=18)=P(X=22)$. Also, $P(X=20)\approx0.1253707$. Ratios of adjacent binomial probabilities give $P(X=19)=P(X=20)(20/21)$ and $P(X=18)=P(X=20)(20/21)(19/22)$. Therefore,

\begin{aligned} P(18\le X\le22) &=P(X=20)\left[1+2\left(\frac{20}{21}+ \frac{20}{21}\frac{19}{22}\right)\right]\\ &\approx0.1253707(4.5497835)\\ &\approx0.5704095. \end{aligned}

Uncorrected approximation. The matching normal variable has mean $40(0.5)=20$, variance $40(0.5)(0.5)=10$, and standard deviation $\sqrt{10}\approx3.1623$. Using the integers themselves as boundaries gives

P(18\le Y\le22) =P\left(\frac{18-20}{\sqrt{10}}\le Z\le \frac{22-20}{\sqrt{10}}\right) \approx0.4729,

where $Z\sim N(0,1)$. Its absolute error relative to the exact probability is approximately $|0.4729-0.5704095|=0.0975$.

Continuity-corrected approximation. The corrected interval is from 17.5 to 22.5. Thus

P(17.5\le Y\le22.5) =P\left(\frac{-2.5}{\sqrt{10}}\le Z\le \frac{2.5}{\sqrt{10}}\right) \approx0.5708.

The corrected absolute error is approximately $|0.5708-0.5704095|=0.0004$. In this example, the continuity correction greatly improves the approximation by including the normal area corresponding to the endpoint counts.

Comparing Approximation Errors in Tail Events

For a given event, the absolute error of an approximation is the absolute difference between the approximate probability and the exact probability. It measures the size of the discrepancy in probability units. A small absolute error does not establish that every other event will be approximated just as well. In particular, the normal curve may represent a skewed binomial distribution poorly in a tail.

Worked Example: The Probability of No Successes

Imagine 10 independent trials, each with success probability $0.2$. Compare the exact probability of zero successes with normal approximations with and without a continuity correction.

The hypothetical setup satisfies the binomial model conditions. The expected number of successes is $np=2$, and the expected number of failures is $n(1-p)=8$. These counts are small, so a normal approximation deserves caution.

Exact probability. The event is $X=0$, which requires failure on all 10 trials:

P(X=0)=\binom{10}{0}(0.2)^0(0.8)^{10} =0.8^{10}\approx0.1073742.

The normal variable has mean $10(0.2)=2$, variance $10(0.2)(0.8)=1.6$, and standard deviation $\sqrt{1.6}\approx1.2649$.

Uncorrected approximation. Treating the event $X=0$ as $Y\le0$ gives

P(Y\le0)=\Phi\left(\frac{0-2}{\sqrt{1.6}}\right) =\Phi(-1.5811)\approx0.05692.

The absolute error is approximately $|0.05692-0.1073742|=0.05045$.

Continuity-corrected approximation. The count zero is represented by values below 0.5, so

P(Y\le0.5)=\Phi\left(\frac{0.5-2}{\sqrt{1.6}}\right) =\Phi(-1.1859)\approx0.11784.

The corrected absolute error is approximately $|0.11784-0.1073742|=0.01047$. It is smaller than the uncorrected error here, but it is not zero. The low expected success count helps explain why the normal curve is not especially reliable for this event.

Worked Example: A Lower-Tail Cumulative Probability

Suppose 12 independent trials each have success probability $0.25$. Compare exact and normal probabilities for at most two successes.

The binomial assumptions hold by construction. Here $np=3$ and $n(1-p)=9$, so the expected success count is small; the normal approximation may not represent the lower tail very accurately.

Exact probability. Add the probabilities for zero, one, and two successes:

\begin{aligned} P(X\le2) &=\binom{12}{0}(0.25)^0(0.75)^{12} +\binom{12}{1}(0.25)(0.75)^{11}\\ &\quad+\binom{12}{2}(0.25)^2(0.75)^{10}\\ &\approx0.0316764+0.1267054+0.2322932\\ &\approx0.3906750. \end{aligned}

The matching normal variable has mean $12(0.25)=3$, variance $12(0.25)(0.75)=2.25$, and standard deviation $1.5$.

Uncorrected approximation. Using the boundary 2 gives

P(Y\le2)=\Phi\left(\frac{2-3}{1.5}\right) =\Phi(-0.6667)\approx0.2525.

The absolute error is approximately $|0.2525-0.3906750|=0.1382$.

Continuity-corrected approximation. The counts included are 0, 1, and 2, so the upper boundary becomes 2.5:

P(Y\le2.5)=\Phi\left(\frac{2.5-3}{1.5}\right) =\Phi(-0.3333)\approx0.36944.

Its absolute error is approximately $|0.36944-0.3906750|=0.02123$. The corrected value is closer to the exact value, but still differs meaningfully. The correction fixes the placement of the discrete boundary; it does not make the binomial distribution itself normal.

A Practical Comparison Workflow

Specify the event and model.
Write the event using integer counts, and check that the fixed-trial, independence, and common-probability assumptions support $X\sim\operatorname{Bin}(n,p)$.

Calculate the exact probability.
Use the binomial probability formula for each included count and add the results, or use an exact binomial cumulative probability calculation.

Build the normal approximation.
Set the mean to $np$, the variance to $np(1-p)$, and the standard deviation to $\sqrt{np(1-p)}$. Assess whether the expected success and failure counts make a normal approximation plausible.

Calculate both approximations.
Use the integer cutoffs for an uncorrected result, and shift each relevant cutoff by 0.5 for a continuity-corrected result. Standardize the resulting boundaries.

Compare in probability units.
Subtract each approximation from the exact probability and take the absolute value to report its absolute error. Keep enough digits during calculations, then round the reported results sensibly.

Interpreting the Comparison

The continuity correction is a geometric adjustment: it better aligns the continuous interval with the integer counts in the event. It often improves a binomial normal approximation, particularly when a probability includes one or more whole counts. However, it is not a theorem that the corrected approximation must be closer for every event. Skewness, small expected counts, and the location and width of the event can still matter.

Calculation	Boundary idea	What it provides
Exact binomial probability	Sum probabilities at included integer counts	The probability under the stated binomial model
Uncorrected normal approximation	Use the integer cutoff directly	A quick continuous approximation without boundary adjustment
Continuity-corrected normal approximation	Move the cutoff halfway toward excluded counts	A normal approximation that accounts for discrete spacing

When the exact calculation is manageable, it is useful as a benchmark for assessing an approximation in that particular problem. When the event lies far into a tail or one expected count is small, report that the normal approximation is questionable rather than presenting its numerical precision as evidence of accuracy.

Common Mistakes

Using the wrong correction direction. For $X\le k$, the upper boundary is $k+0.5$, not $k-0.5$. For $X\ge k$, the lower boundary is $k-0.5$.
Comparing unlike events. The exact and approximate calculations must represent the same set of integer counts. For example, $X=0$ corresponds to the corrected interval ending at 0.5, not to an interval ending at 0.
Assuming a correction guarantees accuracy. The continuity correction addresses discrete boundaries, not every difference between a binomial shape and a normal curve. Check expected counts and the event's location.
Calling the approximation error a probability of error. The absolute error is a numerical difference between two probability values; it is not the probability that the approximation is wrong.

Key takeaway: The exact binomial calculation provides the benchmark when it is feasible. A normal approximation uses mean $np$ and variance $np(1-p)$; a continuity correction places boundaries halfway between integer counts. Compare the resulting probabilities rather than assuming the approximation is exact.

Check Your Understanding

Answer each question using the definitions and methods above.

For $X\sim\operatorname{Bin}(n,p)$, what mean and variance are used for the matching normal approximation?
Write the continuity-corrected normal event corresponding to $X\le7$.
A researcher wants to approximate $P(4\le X\le9)$. What continuous interval should be used with a continuity correction?
Why can the continuity-corrected result still be inaccurate when $np$ is small?
An exact probability is 0.31 and an approximation is 0.28. What is their absolute error, and what does that number describe?