# Statistics: Applications of sampling thoery

• SparkimusPrime

#### SparkimusPrime

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:

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.

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A triangular pdf could indicate a binomial distribution (i.e. Pascal's Triangle). I'm not sure though.

I'd wager a guess that it just means the probability increases as w increases according to the line 2w. f(w) = 2w will give you a triangle, after all. Naturally, it still needs to be normalized, though.

N(12.2,.04) means a normal distribution with mean 12.2 and standard deviation (not variance) .04.

#1a) Integrate your distribution between 0 and 12.
#1b) Try using one of the tests you learned.
#2a) Okay.
#2b) You have formulas for this.

#1a) Integrate your distribution between 0 and 12.

Huh? I don't have a formula for this ditribution nor do I know how to make one.

#1b) Try using one of the tests you learned.

Haven't learned any tests that look like that.

#2a) Okay.
#2b) You have formulas for this.

Okay? I have formulas? Huh?
I'm going to email my teacher obviously I'm missing something.

I have checked the original question and it seems you are given the variance so you would calculate the z-score and use the z-table (you can search this on the interet I found http://itrs.scu.edu/psychology/faculty/turdan/stats%20ppts/chapter4.revision.tu.ppt [Broken] with no trouble).

As for the "triangluar" you can get the answers by ignoring this term
"triangular" (If you shade in the area under the pdf it is a triangle but who cares).

the expected value of W is
$$E(W)=\int_0^1 w f(w) = \int_0^1 2w^2= 2/3$$
the variance of W is
$$(\int_0^1 w^2f(w) )- (E(W)^2) = (\int_0^1 2w^3) - (\frac{2^2}{3^2})= \frac{1}{18}$$

you should now realize that you are close to the answer just use the formulas E(aW+b) = aE(W)+ b
and Var(aW+b) = a^2 Var(W)

Hint E(a(W-b)) = a E(W-b) = a ( E(W)-b))
subbing crap in gives
E( sqrt(18) (W-2/3) ) = sqrt18( E(W)-2/3) =0

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1a)
The normal distribution:
http://mathworld.wolfram.com/NormalDistribution.html

Your calculator probably already does it for you, though. On a TI-83+, you'll use the normalcdf() function.

1b)

2b)
Mean:
http://mathworld.wolfram.com/Mean.html

Variance:
http://mathworld.wolfram.com/Variance.html

I'm sorry, but those explanations do nothing for me. We haven't learned T-tests as far as I know.

Damned charming :) -

I have checked the original question and it seems you are given the variance so you would calculate the z-score and use the z-table (you can search this on the interet I found http://itrs.scu.edu/psychology/facu...revision.tu.ppt [Broken] with no trouble).

I don't have power point, thanks though.

the expected value of W is
...
the variance of W is
...

I can't get those pictures (?) to paste, but you get the idea. How did you do that anyway? That's kool.

Anyway, I got the first two answers correctly, those ideas go back several chapters, I have a firm grasp on those. But I'm not sure, exactly how they relate to part 'b'. I've tried doing the integrations the same way, but to no avail. The mean comes out correctly, to zero, but the variance is some strange number (sqrt(2)/6).

you should now realize that you are close to the answer just use the formulas E(aW+b) = aE(W)+ b
and Var(aW+b) = a^2 Var(W)

Hint E(a(W-b)) = a E(W-b) = a ( E(W)-b))
subbing crap in gives
E( sqrt(18) (W-2/3) ) = sqrt18( E(W)-2/3) =0

We've done this stuff, but shouldn't the integral come out to the same solution as the shortcut here?

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$$\sqrt{2}/6 = \sqrt{ \frac{1}{18}}$$ which is the
standard deviation when the variance 1/18, as my integration confirms.
I am not sure what you mean when you say when the integration should be same as the short cut. An alternative method using integration is
$$E(W-2/3) = \int_0^1 ( W- 2/3)2W = 0$$

PS if you want to know how I have done the integrals I have attached a txt file containing exactly what I typed.

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• latex.txt
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