Find the probability of getting 20 heads in 40 flips of a fair coin

Question

Find the probability of getting $20$ heads in $40$ flips of a fair coin

I did this problem with the binomial distribution and got my probability as $0.12537$.

However, I am being asked to do this using normal approximation to the binomial distribution with continuity correction.

So i'm using the formula:

$$ Z=\frac{x-np}{\sqrt{npq}}$$

where $x=20,n=40,p=\frac{1}{2},q=\frac{1}{2}$

I have:

$$P(x=20)=P(19.5<x<20.5)=P(\frac{-.5}{\sqrt{10}} < Z <\frac{.5}{\sqrt{10}}) $$

However, I'm not sure what to do after that. This is the first time I worked with a normal approximation case that involves exactly, it usually involves less than or greater than.

Any help or pointers would be appreciated.

Answer 1

So far, so good.

Once you have got the limits for z, proceed as usual, look up a z-table.

Answer 2

Formally your approach is correct.

However, I would rather use the local limit theorem: $$ P(X = 20) \approx \frac{1}{\sqrt{2 \pi npq}}\,e^{-\frac{(20 -np)^2}{2npq}} = \frac{1}{\sqrt{20\pi}}\approx 0.1261. $$ This gives a worse approximation that the one you wrote (which gives about $0.1257$), but it does not use any special functions.