# Thread: A very hard Statistic Problem

1. ## A very hard Statistic Problem

A packaging unit is regulated so that it discharges 12 oz. of cereal per box. Due to many factors such as slight variations in machines and variations between uses the amount actually discharged into a box varies slightly and can be shown to be a random variable with mean equal to 12 oz. and standard deviation of 0.25 oz., i.e., m=12 oz. and s=0.25 oz.

(a)Use Chebyshev’s theorem to obtain a lower bound to the probability that the amount discharged into a box will be on the interval from 11.2oz. to 12.8 ml.

(b) Assume that the p.d.f. of the amount discharge is the normal p.d.f. and determine the probability of observing a box with between 11.2 oz. and 12.8 oz. of cereal.

(c) The manager wishes to state that he can guarantee that at least “X’ ounces are in the 99% of the boxes. Find X. Assume that the p.d.f. for the amount discharged is the normal p.d.f. and determine the value of the bottom 1st percentile of the amount discharged in a box. Find c such that P(X<c) =0.01.

(d)We sample box of cereal produced by the machine and observe that in fact we have only 11.0 oz in the box. Suppose we conclude that the machine is “out of control” if we observe a box having amount that is below what we would expect for 1% of the boxes; i.e. b such that. P(X<b) <0.01. Do we consider an 11.0 oz. box as indicating that the process is “out of control”? (You can and should use the answer to (c) above to answer this question.) . Write a sentence.

2. Originally Posted by simple_guy8
A packaging unit is regulated so that it discharges 12 oz. of cereal per box. Due to many factors such as slight variations in machines and variations between uses the amount actually discharged into a box varies slightly and can be shown to be a random variable with mean equal to 12 oz. and standard deviation of 0.25 oz., i.e., m=12 oz. and s=0.25 oz.

(a)Use Chebyshev’s theorem to obtain a lower bound to the probability that the amount discharged into a box will be on the interval from 11.2oz. to 12.8 ml.

(b) Assume that the p.d.f. of the amount discharge is the normal p.d.f. and determine the probability of observing a box with between 11.2 oz. and 12.8 oz. of cereal.

(c) The manager wishes to state that he can guarantee that at least “X’ ounces are in the 99% of the boxes. Find X. Assume that the p.d.f. for the amount discharged is the normal p.d.f. and determine the value of the bottom 1st percentile of the amount discharged in a box. Find c such that P(X<c) =0.01.

(d)We sample box of cereal produced by the machine and observe that in fact we have only 11.0 oz in the box. Suppose we conclude that the machine is “out of control” if we observe a box having amount that is below what we would expect for 1% of the boxes; i.e. b such that. P(X<b) <0.01. Do we consider an 11.0 oz. box as indicating that the process is “out of control”? (You can and should use the answer to (c) above to answer this question.) . Write a sentence.
I assume that (a) is the problem (the rest look to be routine applications of the normal distribution):

$11.2 = 12 - 0.8 = 12 - (3.2)(0.25) = \mu - 3.2 \sigma$.

$12.8 = 12 + 0.8 = 12 + (3.2)(0.25) = \mu + 3.2 \sigma$.

Therefore:

$\Pr(11.2 \leq X \leq 12.8) = \Pr(\mu - 3.2 \sigma \leq X \leq \mu + 3.2 \sigma)$ $= {\color{red}\Pr(-3.2 \sigma \leq X - \mu \leq 3.2 \sigma) = \Pr(|X - \mu| \leq 3.2 \sigma)} \geq 1 - \frac{1}{(3.2)^2}$

using Chebyshev's Theorem.