Thread: [SOLVED] Normal approximation to binomial

1. [SOLVED] Normal approximation to binomial

I've tried so many things but I never get the correct answer. Here's an example:

For $\displaystyle Y \sim B(20,0.4)$ Find $\displaystyle P(3<Y<13)$ using a normal approximation

This is what I do:

$\displaystyle \mu=np=(20)(0.4)=8$
$\displaystyle \sigma^2=np(1-p)=20(0.4)(0.6)=4.8$

But at this point, I get in a mess.

Thanks in advance for any replies.

2. Originally Posted by Quacky
I've tried so many things but I never get the correct answer. Here's an example:

For $\displaystyle Y \sim B(20,0.4)$ Find $\displaystyle P(3<Y<13)$ using a normal approximation

This is what I do:

$\displaystyle \mu=np=(20)(0.4)=8$
$\displaystyle \sigma^2=np(1-p)=20(0.4)(0.6)=4.8$

But at this point, I get in a mess.

Thanks in advance for any replies.
Remember $\displaystyle \sigma^2=np(1-p)=20(0.4)(0.6)=4.8 \implies \sigma = \sqrt{4.8}\approx 2.2$

Now employ $\displaystyle z = \frac{Y-\mu}{\sigma}$

So $\displaystyle P(3<Y<13) = P\left(\frac{3-8}{2.2}<z<\frac{13-8}{2.2}\right)$

Can you contiune?

3. Thanks for the fast reply, but seemingly not

$\displaystyle =P(-2.282<z<2.282)$
$\displaystyle p(-2.282<z)$
$\displaystyle =p(z>2.282)$
$\displaystyle =0.0113$

$\displaystyle p(z<2.282)$
$\displaystyle =0.9887$

$\displaystyle 0.9887-0.113=0.9774$
Which is incorrect. Am I using the tables incorrectly? Or is there something wrong with my method?

4. I would say $\displaystyle P(-2.82<z<2.82) = 1-2P(z>2.82) = 1 - 2\times 0.0112 = 0.9776$

5. 0.9596 is the correct answer though, so I'm confused.

6. All I can suggest is to use a better approximation for $\displaystyle \sqrt{4.8}\approx 2.1909$ . Otherwise the methodology seems good to me.

7. I suggest you apply a "correction for continuity", so instead of finding
P(3 < Y < 13),
you find
P(2.5 < Y < 13.5),
where Y has a Normal distribution with mean 8 and standard deviation 2.19.