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# Sampling distribution question watch

1. The random variable X has a B(40. 0.3) distribution. The mean of a random sample of n observations of X is denoted by x̄. Find(a) P(x̄≥13) when n is 49,(b) the smallest value of n for which P(x̄≥13)<0.001.
I've done part a and for b got upto (13)-(1 divided by 2n)-(2) all divided by (square root 8.4 divided by square root n)=3.090
How would i obtain a value for n?
2. for more clarity here's a picture
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3. isnt and

(just checking I have this right before I go any further)
4. Wouldn't the variance be 8.4 over n, since mean for binomial approximated to normal is npq, so 40*0.3*0.7=8.4
5. (Original post by gooner1010)
Wouldn't the mean be 8.4 over n, since mean for binomial approximated to normal is npq, so 40*0.3*0.7=8.4
hmm as I feared, I cant remember how to do this sorry

Edit: Is this S2? a binomial to normal approximation question? ill revise it and get back to you if you want?
6. (Original post by DylanJ42)
hmm as I feared, I cant remember how to do this sorry

Edit: Is this S2? a binomial to normal approximation question? ill revise it and get back to you if you want?
Hi, yep it's a binomial approximation to normal, and help would be appreciated.
Thanks
7. (Original post by gooner1010)
Hi, yep it's a binomial approximation to normal, and help would be appreciated.
Thanks
sorry I honestly dont have a clue how to do this. Hopefully someone else can help you
8. (Original post by gooner1010)
Hi, yep it's a binomial approximation to normal, and help would be appreciated.
Thanks

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9. (Original post by DylanJ42)

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Sorry. it's 0.0084
10. (Original post by gooner1010)
Sorry. it's 0.0084
ive got no idea then, sorry

iirc SeanFM knows his stats, maybe he can salvage this
11. (Original post by gooner1010)
The random variable X has a B(40. 0.3) distribution. The mean of a random sample of n observations of X is denoted by x̄. Find(a) P(x̄≥13) when n is 49,(b) the smallest value of n for which P(x̄≥13)<0.001.
I've done part a and for b got upto (13)-(1 divided by 2n)-(2) all divided by (square root 8.4 divided by square root n)=3.090
How would i obtain a value for n?
So this is a normal approximation to the binomial type question. If you have a random variable X distributed as B(n,p) then it is approximated by a variable distributed as N(np, np(1-p)) (where the second parameter for the normal is given as the variance rather that standard deviation).

So here, X is approximated by N(40 * 0.3, 40 * 0.3 * 0.7) = N(12, 8.4)

If X is distributed as , then X-bar is distributed as where n is the number of observations. So here we have the mean distributed as N(12, 0.1714286). So part (a) follows by looking at tables or using software.

If n is unknown, then you need to work in reverse, try a few values of n and see what values you get for the probability - after a while you should be able to pinpoint the n you want.
12. (Original post by Gregorius)
So this is a normal approximation to the binomial type question. If you have a random variable X distributed as B(n,p) then it is approximated by a variable distributed as N(np, np(1-p)) (where the second parameter for the normal is given as the variance rather that standard deviation).

So here, X is approximated by N(40 * 0.3, 40 * 0.3 * 0.7) = N(12, 8.4)

If X is distributed as , then X-bar is distributed as where n is the number of observations. So here we have the mean distributed as N(12, 0.1714286). So part (a) follows by looking at tables or using software.

If n is unknown, then you need to work in reverse, try a few values of n and see what values you get for the probability - after a while you should be able to pinpoint the n you want.
The book has the answer 82, however I seem to be getting 81
13. (Original post by gooner1010)
The book has the answer 82, however I seem to be getting 81
I agree with your value of 81. I'm using the following R code

N <- 81
mm <- 12
ss <- (40 * 0.3 * 0.7)/N
1 - pnorm(13, mm, sqrt(ss))

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