# Normal approx binomial question.

1. Sep 27, 2008

### Shawj02

Ok, the question says

Binomial with n=40. p=0.6 use normal approximation and determine..
a) value at 14.
b)value lest then 12.

So I thought I had this down and packed but the answer in the back of the book tells me otherwise. Anyways the following is my working.

Y~Nor(13.5<x<14.5)
Mean=np=24
S.D = sqrt(np(1-p)) = 3.094

larger prob area
Z=(x-mean)/S.D

=(14.5 - 24)/3.094
=-3.07
(Tables book, 3.07 = 0.9989)
So (1-0.9989)

smaller prob area
Z=(x-mean)/S.D

=(13.5 - 24)/3.094
=-3.39
(Tables book, 3.39 = 0.9997)
So (1-0.9997)

larger prob area - smaller prob area
total prob =(1-0.9989) - (1-0.9997)
=8*10^-4

But the actual answer is meant to be 0.049?

B) I do pretty much the same thing But after the whole Z=(x-mean)/S.D (x=11.5)
end up with Z=-4.034 Which isnt on the tables book. So I cant figgure out the final prob.

Thanks!

2. Sep 30, 2008

$$\sigma \approx 3.0984$$
Regardless of this, your work for the first portion looks good. My calculations (in a spreadsheet and in statistical software) show the normal approximation to equal $$0.0007$$, certainly not $$0.049$$ - are you sure the solution you are looking at is for this particular problem? I also used the binomial distribution to work out $$\Pr(X = 14)$$ for this case, and I get $$0.0008$$, so the approximation is reasonable.
The $$Z$$-value you get for the second problem is correct. The fact that it is below the smallest value in the normal table means that (to the decimal accuracy of the table) the corresponding probability is zero. That matches up rather well with the number from the binomial distribution, so all looks good.
Note: the reason these answers are so small is easy to see: both the cutoffs (12, 14) are several standard deviations from the mean. The normal distribution used for approximation is symmetric (every normal distribution is), and for the given values of $$n$$ and $$p$$ the binomial distribution is also quite symmetric: there just isn't much probability that far out in the tails.