- #1

- 35

- 0

I'm doing some homework over the break (!) so I don't have access to my usual lines of help. I've hit a wall:

I assume the "N(12.2, .04)" notation refers to that the distribution of the boxes has mean 12.2 and variance .04. I think 'a' has something to do with the normal distribution, how standard deviations mark percentages of the curve. 'b' I just have no clue on how to proceed.

Also:

What do they mean by triangular?

The last portion of the class I would hazard a guess that I could integrate 'f(w)' to get the CDF 'W', but I'm not sure if that applies to this. The answer is supposed to be, 'mean = 0' and 'variance = 1' but damned if I can get it to come out that way just by manipulating the numbers.

A cereal manufacturer packages cereal in boxes that have 12-ounce label weight. Suppose that the actual distribution of weights is N(12.2, .04).

a) What percentage of the boxes have cereal weighing under 12 ounces?

b) if x-bar is the mean weight of the cereals in n = 4 boxes

selected at random, compute P(x-bar < 12).

I assume the "N(12.2, .04)" notation refers to that the distribution of the boxes has mean 12.2 and variance .04. I think 'a' has something to do with the normal distribution, how standard deviations mark percentages of the curve. 'b' I just have no clue on how to proceed.

Also:

Let W have the triangular PDF f(w) = 2w, 0 < w < 1.

What are the mean and variance of U = sqrt(18) * (W - 2/3)?

What do they mean by triangular?

The last portion of the class I would hazard a guess that I could integrate 'f(w)' to get the CDF 'W', but I'm not sure if that applies to this. The answer is supposed to be, 'mean = 0' and 'variance = 1' but damned if I can get it to come out that way just by manipulating the numbers.

Last edited: