Statistics word problem! :) http://puu.sh/6Z5pY.png
@jim_thompson5910
This seems to me like a binomial distribution problem.
It is! It's what I'm learning right now but I'm not 100% on it:)
What did you get for the first bullet?
Also, do you have a probability distribution chart for when n = 50 for this?
No i don't, this is all that was given. I just started taking statistics and this is my 2nd assignment haha so please understand i don't understand a lot
No worries. I was just wondering because that would make some of these parts much easier to calculate. The formula we are using is \[\left(\begin{matrix}n \\ x\end{matrix}\right)p^xq ^{n-x}\]
I know this is coming back to me it's something like (0.24)^n*(0.76)^n-x
Right!
So what is n? 50?
so that it would be (0.24)^10*(0.76)^40?
Yes. The variable n represents the number of trials, which in this case is 50. That's right.
I was doing this and I had a problem with something so maybe you can help with that! On the 2nd bullet point it says it at least 10. I understand I have to do the ^0 up to ^10 but it takes so long, is there a shorter way to do it?
(0.24)^10*(0.76)^40 I know that's wrong because I'm getting over one, so what am I doing wrong?
The number .24^10 * .76^40 is not greater than 1
I'm getting 1.083068...*10^(-11) which is definitely less than 1.
Remember that you have to multiply by \[\left(\begin{matrix}50 \\ 10\end{matrix}\right)\]
Oh whoops! That's correct sorry:)
Okay so for #2 of it, it says less than 10. Do I have to do all of them or is there a way to do it without having to multiply them all out?
Without a distribution chart, the fastest way I know to do it is to add up all the probabilities 1 through 9 and subtract that value from 1.
Look at this page and scroll down to n = 50.
Well actually, they don't have our .24 probability of success. Hold on.
Uh oh :P haha
Check that out though. It does all the work for you. Remember that when we are talking about at least 10, we mean >= 10
what should I put for "number of success?"
0.24?
10
At least for bullets 1 - 3
so it would be bullet 1: 0.111255856065322 bullet 2: 0.206550433838689 bullet 3: 0.317806289904011
and bullet 4: 0.11628996701893859
Sorry for slow responses, openstudy doesn't like my computer @RBauer4
bullet 2 should be .7934.... , and bullet 1 and bullet 3 are correct.
At least implies greater than or equal to
OOo i did x>10 not x<10
Wording was weird :) and bullet 4 I should do x=21 and x=45 and just add them
0.0022710274978721500000691282565252491
That doesn't seem right to me haha
Well, if you think about it. You practically have a probability of 1 considering the cumulative probability less than 45
Ooooh right then 0.002271.... would be considered "unusual"
Thus, we are only really calculating the probability when x is strictly greater than 20
So 0.11, 0.7934, 0.317, 0.11 and 0.0027
I think the last bullet would be 0.003769047260403 and I'm not sure for the 4th yet
And I believe bullet four is 0.78968051890090805
I must go now. Good luck in your studies
thanks!
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