# [SOLVED] Normal approximation to binomial

• Mar 8th 2010, 12:58 PM
Quacky
[SOLVED] Normal approximation to binomial
I've tried so many things but I never get the correct answer. (Worried) Here's an example:

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

This is what I do:

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

But at this point, I get in a mess.

Thanks in advance for any replies.
• Mar 8th 2010, 01:12 PM
pickslides
Quote:

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

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

This is what I do:

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

But at this point, I get in a mess.

Thanks in advance for any replies.

Remember $\sigma^2=np(1-p)=20(0.4)(0.6)=4.8 \implies \sigma = \sqrt{4.8}\approx 2.2$

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

So $P(3

Can you contiune?
• Mar 8th 2010, 01:27 PM
Quacky
Thanks for the fast reply, but seemingly not (Doh)

$=P(-2.282
$p(-2.282
$=p(z>2.282)$
$=0.0113$

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

$0.9887-0.113=0.9774$
Which is incorrect. Am I using the tables incorrectly? Or is there something wrong with my method?
• Mar 8th 2010, 01:42 PM
pickslides
I would say $P(-2.822.82) = 1 - 2\times 0.0112 = 0.9776$
• Mar 8th 2010, 02:01 PM
Quacky
0.9596 is the correct answer though, so I'm confused.
• Mar 8th 2010, 02:16 PM
pickslides
All I can suggest is to use a better approximation for $\sqrt{4.8}\approx 2.1909$ . Otherwise the methodology seems good to me.
• Mar 10th 2010, 02:03 PM
awkward
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.