Dear Uncle Colin,

I am trying to work out a probability using a binomial distribution and I’m getting a very different answer when I use a normal distribution:

Binomially:

$X \sim B(380, 0.3)$

$P(X \ge 152) = 1 - P(X \le 151)$

$\dots = 1 - 0.9999789688$

$\dots \approx 0.00002103$

Normally:

$Y \sim N(114, 79.8)$

$P(Y \ge 151.5) \approx 0.00001347$

That’s almost a third off! What am I doing wrong?

Did I Somehow Calculate Rotten Estimate? Perhaps A Normal Confuses Y

Hi, DISCREPANCY, and thanks for your message!

I think there are two sources of the error here, but it’s hard for me to be sure. They are:

• The binomial distribution is cut off at zero
• It’s an approximation, and the error has to go somewhere

I suspect the first one is more important, so let’s start there.

## The zero cut-off

The normal distribution is perfectly symmetrical about its mean – there is a non-zero (although small) probability that $P(Y \lt 0)$. So, the normal distribution probabilities in the domain where the binomial distribution makes sense are systematically smaller than the binomial ones.

In most places, the difference is so small that it makes next to no difference. However, out in the tails, a tiny absolute error becomes a significant relative error.

## The error has to go somewhere

The other thing is that the normal and binomial distributions are good approximations to each other – but they’re not precisely equal. Again, that good absolute agreement everywhere can translate into a wild relative error.

So I can’t give a definitive answer; this is the best I’ve got. I hope it helps!

- Uncle Colin

• Thanks to @mrsouthernmaths for the problem.
• Edited 2022-06-18 to fix formatting and a typo – thanks to Teymour for letting me know!
• Edited 2022-07-03 to fix a mistranscribed digit – thanks for checking, Adam!