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($$\mu$$,$$\sigma$$²) 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 $$|\bar{X}-\mu|>\sigma/2$$ 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($$\mu$$,$$\sigma$$²/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 $$|u|>\sigma/2$$ is given as a integral that you can solve using the error function:

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

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$$\leq$$z$$\leq$$$$\sqrt{n}$$/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