Please Help :(

[statshead]

I been trying to figure this out for so long, i just dont get it. if anyone can explain the answer it'd be very much appreciated.


A box of 100 machine components is delivered to a factory: experience shows that the components fail 2% of the time. A random sample of
5 components is tested to see if they function correctly.


Which of the following random variables can represent the events "a component functions correctly" and "a component does not function correctly"?

(a) Z~ N (0, 1).
(b) X ~ Bernoulli with Pr (X = 1) = 0.02.
(c) U ~ U (0, 5).
(d) Y ~ Bin (0.02, 5).
(e) V ~ Poisson (0,02).



Which of the following random variables can represent the number of defective components found in the random sample of 5 components?

(a) Z~ N (0, 1).
(b) X ~ Bernoulli with Pr (X = 1) = 0.02.
(c) U ~ U (0, 5).
(d) Y ~ Bin (0.02, 5).
(e) V ~ Poisson (0,02).




A random sample of size 25 is drawn from a population represented by the random variable Z ~N (0, 1): What is the probability distribution of the sample mean Z obtained from this random sample?

thanks!!

[chippy]

Hints for Q's :
1)

a component

can either

function correctly

or

not function correctly

. Therefore do a google search on bernouli random variables.
2) I think when you first read a bit on bernoulli random variables, you can then do a google search on binomial models - to further your understanding.
3) Google "central limit theorem".

[statshead]

Thanks! I'm just reading up on this now!

[statshead]

Ok i've tried to figure it out. is 1 B?.... 2 D?... and 3 Z(mean) ~ N (0, 1/25)?

[Dason]

3) Google "central limit theorem".

Technically since the parent population was normally distributed we don't need to invoke the CLT since we already know the sampling distribution of the mean will be normally distributed. We just need to use expectation and variance rules.

[statshead]

Technically since the parent population was normally distributed we don't need to invoke the CLT since we already know the sampling distribution of the mean will be normally distributed. We just need to use expectation and variance rules.

No you're right, i didn't have to use clt for this. do you know if the answers i came up with are right though? thanks

[Dason]

As long as you're using the variance as the second parameter of the normal distribution then they look good to me.

[statshead]

As long as you're using the variance as the second parameter of the normal distribution then they look good to me.

cheers mate!!!

[chippy]

Technically since the parent population was normally distributed we don't need to invoke the CLT...

You got me...oops...was a bit lazy there. Thanks for pointing that out.