Sampling distribution question

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?

Hi, yep it's a binomial approximation to normal, and help would be appreciated.
Thanks

Hi, yep it's a binomial approximation to normal, and help would be appreciated.
Thanks

last try, is your answer to the first part 0.8869?

Sorry. it's 0.0084

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 $N(\mu, \sigma^2)$, then X-bar is distributed as $N(\mu, \sigma^2/n)$ 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