Let us approximate a discrete distribution by a standard normal distribution, without using a continuity correction factor. Let $X$ be a random variable with discrete distribution, and $Y$ be a random variable with standard normal distribution. Since we did not use a continuity correction factor, can we say that the $P(X \geq x)$ is always greater than or equal to its approximated probability by the standard normal distribution?
2026-03-28 13:27:30.1774704450
Approximating a discrete probability distribution with a standard normal distribution
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If the discrete random variable $X$ takes integer values, then $$P(X > x)= P(X \ge x+1) = P(X \ge x+.5)$$ The continuity correction would use the third expression when using a continuous distribution as an approximation.
Ordinarily, the approximating continuous distribution would have positive probability in the interval $[x, x+.5].$ In that case using the continuity correction will give you a smaller approximated value.
Example: Suppose $X \sim \mathsf{Binom}(n = 64, p = 1/2)$ and you seek $P(X > 30).$ The exact value is $P(X > 30) = 1 - P(X \le 30) = 0.6460096.$
If you use $P(X^\prime > 30) = 1 - P(X^\prime \le 30)$ as an approximation, where $X^\prime \sim \mathsf{Norm}(\mu = 32, \sigma=4),$ you will get $P(X > 300) \approx 0.6914625.$
But if you use the continuity correction, you will use $P(X^\prime > 30.5) = 1 - P(X^\prime \le 30.5) = 0.6461698.$ Hence, your approximation will be $P(X > 30) \approx 0.6461698.$ This is smaller than the value 0.6914625 without the continuity correction. It is also closer to the exact binomial probability.
Usually in textbook examples you can expect about two decimal places of accuracy from a continuity-corrected normal approximation to a binomial distribution. To four decimal places, the exact value in this example is 0.6460 and the continuity-corrected normal approximation is 0.6462. (Here we get three-place accuracy; approximations are often best when $p \approx 1/2.$)
The figure below shows relevant binomial probabilities (vertical bars) and the approximating normal density curve. Notice that be binomial probability $P(X = 31)$ is approximated by the area under the normal curve above the interval $[30.5, 31.5].$ The uncorrected approximation wrongly includes the vertical strip between $x = 30.0$ and $x=30.5$ under the normal curve.
Note: The values I have shown are from R statistical software. If your normal approximations are obtained by standardization and using a printed normal table, then results will be slightly different because of the rounding entailed in the use of the table.