A normal approximation on an interval

Topic: Functions of random variables and limit theorems · more questions on this topic
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Let X\sim\operatorname{Bin}(100,1/4). Approximate \mathbb{P}(20\le X\le 30) by a normal probability. Use \Phi(1.15)\approx 0.875.

Worked answer (click to show)

The mean is 25 and the variance is 100\cdot(1/4)\cdot(3/4)=75/4, so the standard deviation is \sqrt{75/4}\approx 4.330. With Y\sim\operatorname{N}(25,75/4),

\mathbb{P}(20\le Y\le 30)=\mathbb{P}(-1.155\le Z\le 1.155)\approx 2\Phi(1.15)-1\approx 0.75,

where Z is standard normal. The exact binomial probability is about 0.797; the normal approximation here is a little low because no continuity correction was used.