Re:Statistics: Applications of sampling thoery

[SparkimusPrime]

Edit:

Moved from statistics -

https://www.physicsforums.com/showthread.php?s=&postid=164120#post164120

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 don't really know how to solve a problem like this, from the replies to the previous thread I know I have no idea of the concepts being discussed here.

Also:

What are the mean and variance of U = sqrt(18) * (w - 2/3)?
0 < w < 1

I get the mean correctly, but the standard deviation does not come out correctly, it integrates to sqrt(2) / 12. Its supposed to be a normal distribution, with mean of 0 and standard deviation of 1, but damned if I know how to integrate the standard deviation correctly:

(this is correct, the notation is standard for the TI-89)

mu = mean = integral( w * sqrt(18) * (w-(2/3)),w,0,1 ) = 0

var = integral(w^2 * sqrt(18) * (w-(2/3)),w,0,1) - mu^2 = sqrt(2) / 12

(this is incorrect, the root of the answer [the standard deviation] is clearly not 1. I think I may in fact be mixing up my equations.)

Peter

 

[M98Ranger]

Hello Peter,

I understand that you are facing some difficulties with your statistics homework and are unsure of the concepts being discussed. I will try my best to help you with the problems you have mentioned.

For the first problem, we are given that the actual distribution of weights of the cereal is N(12.2, .04). This means that the mean weight is 12.2 ounces and the standard deviation is 0.2 ounces.

a) To find the percentage of boxes with cereal weighing under 12 ounces, we need to find the probability that a random box weighs less than 12 ounces. This can be calculated using the normal distribution formula: P(X < 12) = Φ((12-12.2)/0.2) = Φ(-1) = 0.1587 or 15.87%. Therefore, approximately 15.87% of the boxes have cereal weighing under 12 ounces.

b) For this part, we need to find the probability that the mean weight of a sample of 4 boxes is less than 12 ounces. This can be calculated using the central limit theorem, which states that the sample mean follows a normal distribution with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size. So, P(x-bar < 12) = Φ((12-12.2)/(0.2/√4)) = Φ(-1) = 0.1587 or 15.87%. Therefore, there is a 15.87% probability that the mean weight of a sample of 4 boxes is less than 12 ounces.

Moving on to the second problem, we are given U = √18 * (w - 2/3), where 0 < w < 1. We need to find the mean and variance of U.

To find the mean, we can use the formula: E(U) = E(√18 * (w - 2/3)) = √18 * (E(w) - 2/3) = √18 * (0.5 - 2/3) = 0.

To find the variance, we can use the formula: Var(U) = Var(√18 * (w - 2/3)) = (√18)^2 * Var(w) = 18 * Var(w