Even More Probability Help?

Question

The weight of a certain type of brick has an expectation of 1.12kg with a standard deviation of .03kg.

Find: expectation and variance given 25 randomly selected bricks, and number of bricks necessary to have a standard deviation of .005 or less.

I know expectation is 1.12kg, and I know two formulas for variance, but neither of those are applicable here. What's the formula to use?

And for the other part of the question, I understand why there's a certain number, but short of trial and error I can't think of a way to actually calculate it. Tips?

I couldn't really follow the first answerer, but my main question is how do you use Variance=E(X^2)-E(X)^2 in this kind of equation? I don't understand how to find the first term.

I also don't really understand where you get the .0009 in the second part.

Answer 1

let X be the r.v. of "(whatever the r.v is above)"

Var (X)
= E(X2) - [E(X)]2
= square of standard deviation

i assume these are the equations you know right?

when given a sample size, such as 25 from above, you need to use sample mean.

do you know if the distribution above is a normal distribution, if not, you'd need to state "By CLT, (X bar, the X with the horizontal line on top)~N (1.12, 0.0009/25) approximately". This is so as the sample variance is that of the variance of X divided by the sample size.

as the the standard deviation of 0.005 or less, let the number of bricks required be n

since standard deviation is sq root of variance, it is √(0.0009/n).
let this be √(0.0009/n) ≤ 0.005,
and solve for n.

i believe n ≥ 36

hope this helped! :)