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Why did you close it?

i had to so tht i could post another question, how do i tag them?

Using '@'
Example: @Confucious

thanks

I don't know how to do this either, sorry

Ask Satellite, Myininaya, JamesJ or amistre64

k thanks

Can you give me a rundown of what the continuity correction is?

I've never even heard of continuity correction. What course is this?

Definitely some kind of statistics course.

So if \(n\) increases, continuity correction decreases.

...I'll try this. Thank you so much.

You're welcome.

I have 0.006

Simplifying a bit, we get\[P\left( -.4542 \leq z\leq .0279 \right)\]

Which I get to be about \(P\approx 0.1863\)

I got a different answer for the standard deviation. It's 0.0697.

How did you calculate the standard deviation?

I'm going almost entirely by what that book I linked to is telling me.

According to the book\[\mu_{\hat p} = p=0.45\]Which is what I was using for the mean

oh yh

Which finally gets us that \[P\approx 0.5449\]

And that was just for part a.

Okay. We find the continuity correction again for part (b) right?

yup.

But here's the thing. There's no interval so I don't know if I should add or subtract.

Use the interval \[P\left(.35\leq \hat p\leq 9\times 10^{99}\right)\] Or a similar upper bound.

??? why???

Although you said you have to use a chart and not a calculator. That might complicate things.

I don't get that. How can that number be used. It looks like it's too big to be even written out.

so it can be \[P(0.35\le p-\hat \le 0)\] ?
[I don't know why it came out like that]

where did you get .3398 and -0.1574 from??

I subtracted the continuity correction from the lower interval to get .3398
.3398=.35-0.0102

And \[{.3398-.35 \over 0.648} \approx -0.1574\]

Then you want to find the probability for \(z=-0.16\) and subtract it from 1.

ok thanks a lot for your help