# Homework Help: Help with statistics please,

1. Sep 11, 2011

### forty-twwo

To be honest I have no real idea how to do this. I know of the probability density function, and the cumulative probability function, but using them is another difficulty i'm trying to overcome.

My questions are,

Intravenous fluid bags are filled by an automated filling machine. Assume that the
fill volumes of the bags are independent, normal random variables with a standard
deviation of 0.08 fluid ounces.

(a) What is the standard deviation of the average fill volume of 20 bags?
(b) If the mean fill volume of the machine is 6.16 fluid ounces, what is the
probability that the average fill volume of 20 bags is below 5.95 °uid ounces?
(c) What should the mean fill volume equal in order that the probability that the
average fill volume of 20 bags is below 6 ounces is 0.001?

Second question is,

A continuous random variable X has the probability density function f(x) given
below.
0 x < or equal to 0
f(x) = 4x3 0 < x < or equal to 1
0 x > 1

(a) What is the cumulative distribution function of the random variable X?
(b) What are the mean and median of the random variable X?
(c) What is the variance of the random variable X?
(d) What is the probability that the random variable takes a value greater than
its mean?

Any help or tips so I can figure these out would be greatly appreciated. Thanks,

Last edited: Sep 11, 2011
2. Sep 11, 2011

### HallsofIvy

Well, what formulas and equations do you know? In particular, what is the standard deviation of an average of n trials from a normal distribution?

What have you tried on these?

3. Sep 11, 2011

### forty-twwo

I believe I just solved all of the first question. For Part A I used the equation of the standard deviation divided by the square root of the number or trials, so 0.08 / sqrt of 20.
Then for Part B I used the z score formula of (X-μ)/σ, then used a z-table I found online. For Part C I did a similar thing but basically used the equation and table in reverse.

Question 2 is still a bit of a mystery for now though,

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