normal approx to binomialWatch

#1
Please can anyone explain (go on ... convince me!) why there's any need to use the Normal Distribution to approximate the Binomial Distribution?
1
1 week ago
#2
(Original post by begbie68)
Please can anyone explain (go on ... convince me!) why there's any need to use the Normal Distribution to approximate the Binomial Distribution?
I don't know the actual answer but it's a nice part of the spec since it introduces students to the Central Limit Theorem (often without students knowing about the CLT) as well as continuity correction. So it could be one of those things in school level maths which is a useful part of the syllabus but doesn't have much practical usage.

I'm thinking though since it's part of the new A Level spec there's probably a good reason why you'd want to approximate binomial->normal. Gregorius is probably the best person for this.
0
1 week ago
#3
(Original post by begbie68)
Please can anyone explain (go on ... convince me!) why there's any need to use the Normal Distribution to approximate the Binomial Distribution?
2 billion people toss a coin 1000 times each. What's the probability there's a total of more than 1000001000000 (i.e. 10^12+10^6) heads?

If that's not a good answer for you, you might need to explain you're thinking a bit.

[I think there are less artificial examples when you look at statistical physics, where you're looking at the aggregate behaviour of trillions of particles. But it's not my area of expertise].
1
1 week ago
#4
(Original post by Notnek)
I'm thinking though since it's part of the new A Level spec there's probably a good reason why you'd want to approximate binomial->normal. Gregorius is probably the best person for this.
DFranklin has it. Doing calculations with large factorials and high powers of probabilities is difficult!
0
1 week ago
#5
(Original post by Notnek)
I don't know the actual answer but it's a nice part of the spec since it introduces students to the Central Limit Theorem (often without students knowing about the CLT) as well as continuity correction. So it could be one of those things in school level maths which is a useful part of the syllabus but doesn't have much practical usage.

I'm thinking though since it's part of the new A Level spec there's probably a good reason why you'd want to approximate binomial->normal. Gregorius is probably the best person for this.
To be honest, a large part of the new spec seems to be simply "let's go back to 'the good old days'" (it's very similar to the spec when I did A-level), so I'm not sure if that plays a part.

However, if it's the main exposure students get to the CLT, then that would justify it by itself.
0
1 week ago
#6
(Original post by DFranklin)
2 billion people toss a coin 1000 times each. What's the probability there's a total of more than 1000001000000 (i.e. 10^12+10^6) heads?

If that's not a good answer for you, you might need to explain you're thinking a bit.

[I think there are less artificial examples when you look at statistical physics, where you're looking at the aggregate behaviour of trillions of particles. But it's not my area of expertise].
Couldn’t a computer tackle this okay without the need for a normal approximation?
0
1 week ago
#7
(Original post by Notnek)
Couldn’t a computer tackle this okay without the need for a normal approximation?
Yes, but non-trivially. Done naively, you'd need to calculate (2x10^12)! which has over 10 trillion digits for example. And then you'd have to sum the result for 1000000 cases.

So such an implementation that didn't want to take literally weeks of CPU time would end up using lots of approximations anyhow.

Edit: of course, you can calculate P(X=k+1) quickly if you know P(X=k), so it's not quite as bad as I portray above. But it's still going to be a lot of CPU time.

Edit 2: a realistic attempt without approximations would store intermediate results at double precision (so 20 digits or so). Since you'd have literally trillions of roundoff errors, it would be sensible to estimate how much these might affect the result. How would you make that estimate? By approximating the accumulated error by a normal distribution. {Mic-drop}
Last edited by DFranklin; 1 week ago
1
1 week ago
#8
Just to offer another practical perspective on this, the largest factorial that the popular Casio fx-991 calculator can compute is 69! (= 1.71 x 10^98).
0
1 week ago
#9
(Original post by old_engineer)
Just to offer another practical perspective on this, the largest factorial that the popular Casio fx-991 calculator can compute is 69! (= 1.71 x 10^98).
You can get a bit of wiggle room on that; I'm guessing it can handle nCr for n > 69 as long as nCr is < 10^100. You can do some cancelling to avoid intermediate results getting too big.

By also cancelling down by the powers of p and (1-p) you can calculate the binomial probability for much larger numbers; the following code works for n = 10^9 (with r=n/2):

Code:
```double bigBinomial(int n, int r, double p)
{
double q = 1 - p;
double P = 1;
int np = r;
int nq = n - r;
for (int i=1;i<=r;i++)
{
P = (P * (n+1-i)) / i;
while ((P > 1 || i == r) && np > 0)
{
P*=p;
np--;
}
while ((P > 1 || i == r) && nq > 0)
{
P*=q;
nq--;
}
}
return P;
}```
Execution time for n = 10^12 would be a couple of hours, and, as I suggested, round-off would be considerable.
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