Sampling With The Normal Distribution

[ƒ(x) → ∞]

I have been puzzling over this problem for about a week now and cannot find the answer. In my opinion it is very theoretical, but I know I ma not the best mathematician on here so maybe someone else could look at this.

The mean of a random sample of n observations drawn from an N(\mu,\sigma²) distribution is denoted by \bar{X}.

Given that p(|\bar{X}-\mu|>0.5\sigma)>0.05

(a) Find the smallest vaule of n
(b) With this value of n find p(\bar{X}<\mu+0.1\sigma)

Thank you.

 

[torquil]

You need to derive the probability distribution for \bar{X}. This is possible, since it is a sum of n quantities for which you know the distribution. Tthe distribution for \bar{X} willl depend on n. You can then calculate the probability that |\bar{X}-\mu|&gt;\sigma/2 in the ordinary way. This gives you an inequality for n. Then find the smallest n that satisfies the inequality.

When you know n, you have completely specified the distribution for \bar{X}, and you can do part (b).

So the first step is to find the distribution of a sum of quantities, expressed in terms of their individual distributions.

Torquil

 

[ƒ(x) → ∞]

Do you mean \bar{X}~N(\mu,\sigma²/n)

Where do I go from here if I knew the probability then I think I could manage.

 

[torquil]

Cool, that's the same thing I got. From that you can write the explicit gaussian probability density for e.g. u:=\bar{X}-\mu. Call it e.g. \rho_n(u). It will depend on n of course. It will simply the usual normalied gaussian centered around u=0, with variance \sigma^2/n I think.

Then the probability P(n) that |u|&gt;\sigma/2 is given as a integral that you can solve using the error function:

<br /> P(n) = \int_{-\infty}^{-\sigma/2} \rho_n(u)du + \int^{\infty}_{\sigma/2} \rho_n(u)du<br />

Then find the smallest n such that P(n)>0.05. Then fix this value of n, and just do another simple integral to get the answer to part b.

Agree?

Torquil

 

[ƒ(x) → ∞]

Got it now. It comes out n>16 so at least n=16. I used a different method which I will post now.

 

[ƒ(x) → ∞]

p(\mu-(\sigma/2)\leq\overline{X}\leq\mu+(\sigma/2)>0.95

=p(-\sqrt{n}/2\leqz\leq\sqrt{n}/2)>0.95

=\phi(\sqrt{n}/2) - (1-(\phi(\sqrt{n}/2))>0.95

=2\phi(\sqrt{n}/2)-1>0.95

=2\phi(\sqrt{n}/2)>1.95

=\phi(\sqrt{n}/2)>0.975

=(\sqrt{n}/2)>(\phi^-1)(0.975)

=(\sqrt{n}/2>1.96

=\sqrt{n}>3.92

=n>15.3664

But n is integer
Therefore n>16

The minimum value of n is therefore 16