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# Binomial approximation watch

1. Hey, I'm getting an answer close to the one in the book, but its not right!!!
X~B(100,0.1) using an approximation find P (X>15)
The answer is 0.0487

I used a normal approximation, although I think poisson could also be used, and also did a half continuity correction... could someone have a go please

thank youuuu xxx
2. As far as I can remember, you can't use a normal approximation since the "p" value is not close to 0.5, therefore the distribution is not very normal. Poisson should do the trick.
3. As far as I can remember, you can't use a normal approximation since the "p" value is not close to 0.5
I thought p < 0.5 ... I'll try it with poisson
4. oooo yey, poisson works, thank you !!
Just wondering : the rule I have for using normal as an approx to binomial are:
n is large
np > 5
n(1-p) > 5
I think I might be missing something major here - but doesnt that work for this qu?
5. Both approximations are valid. For the normal you should get 0.0336.
6. (Original post by Baal_k)
Both approximations are valid. For the normal you should get 0.0336.
I always thought that unless the p value of the binomial distribution is not 0.5 if you were to plot a graph it would not be as accurate if you approximate it to a normal.
7. For the normal you should get 0.0336
yeah thats what I got , I suppose they just wanted me to use poisson!
Thanks all your help
8. For the normal you should get 0.0336
yeah thats what I got , I suppose they just wanted me to use poisson!
Thanks all your help
9. (Original post by welwyn)
I always thought that unless the p value of the binomial distribution is not 0.5 if you were to plot a graph it would not be as accurate if you approximate it to a normal.
Normal can be used when n is large, np > 5, n(1-p)>5

Poisson can be used when n is large and p is small.

I will calculate the actual value when I find my calculator
10. When X~B(100,0.1), P(X>15) = 0.0399 Both answers give a good approximation, neither is better.
11. How the hell can u standardize it? I mean
z=(15.5-100)/(0.1)^.5=-267.2 which is quite great. The value cannot b read from the table. If u did possion then how?
12. (Original post by driving_seat)
How the hell can u standardize it? I mean
z=(15.5-100)/(0.1)^.5=-267.2 which is quite great. The value cannot b read from the table. If u did possion then how?
X~Bin(100, 0.1)
Using Possion approximation
X~Po(10)
P(X>15)
= 1 - P(X<=15)
= 1 - e-10( 1+ 10 + 10^2/2! + 10^3/3! + ..... + 10^15/15!)
= 0.0487

X~Bin(100, 0.1)
Using Normal approximation
X~N(10, 9)
P(X>15)
= P(Z>[15.5-10]/sqrt9)
= P(Z>1.833)
= 0.0334

I think normal approximation should be used because p is not really small.

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