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([tex]\mu[/tex],[tex]\sigma[/tex]²) distribution is denoted by [tex]\bar{X}[/tex].

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

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

Thank you.

 

[torquil]

You need to derive the probability distribution for [tex]\bar{X}[/tex]. This is possible, since it is a sum of n quantities for which you know the distribution. Tthe distribution for [tex]\bar{X}[/tex] willl depend on n. You can then calculate the probability that [tex]|\bar{X}-\mu|>\sigma/2[/tex] 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 [tex]\bar{X}[/tex], 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 [tex]\bar{X}[/tex]~N([tex]\mu[/tex],[tex]\sigma[/tex]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. [tex]u:=\bar{X}-\mu[/tex]. Call it e.g. [tex]\rho_n(u)[/tex]. It will depend on n of course. It will simply the usual normalied gaussian centered around u=0, with variance [tex]\sigma^2/n[/tex] I think.

Then the probability [tex]P(n)[/tex] that [tex]|u|>\sigma/2[/tex] is given as a integral that you can solve using the error function:

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

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

=p(-[tex]\sqrt{n}[/tex]/2[tex]\leq[/tex]z[tex]\leq[/tex][tex]\sqrt{n}[/tex]/2)>0.95

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

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

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

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

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

=([tex]\sqrt{n}[/tex]/2>1.96

=[tex]\sqrt{n}[/tex]>3.92

=n>15.3664

But n is integer
Therefore n>16

The minimum value of n is therefore 16