# Ask Uncle Colin: A Poor Approximation

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!