[SOLVED] Normal approximation to binomial

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March 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 $P(3<Y<13)$ 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.

The correct answer is 0.9596

Thanks in advance for any replies.
March 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 $P(3<Y<13)$ 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.

The correct answer is 0.9596

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

Can you contiune?
March 8th 2010, 01:27 PM
Quacky

Thanks for the fast reply, but seemingly not (Doh)

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

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

$0.9887-0.113=0.9774$
Which is incorrect. Am I using the tables incorrectly? Or is there something wrong with my method?
March 8th 2010, 01:42 PM
pickslides

I would say $P(-2.82<z<2.82) = 1-2P(z>2.82) = 1 - 2\times 0.0112 = 0.9776$
March 8th 2010, 02:01 PM
Quacky

0.9596 is the correct answer though, so I'm confused.
March 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.
March 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.
This leads to
P(-2.51 < Z < 2.51)
where Z has a Normal(0,1) distribution, yielding a probability of 0.988.

(Yes, I know, that's not the answer in the book. Sometimes the book has the wrong answer.)
March 10th 2010, 04:06 PM
Quacky

Hm. Seeing as we're all getting around that, I suppose it must be right. Thanks again.