Standard Deviation the average

[KingBigness]

Not sure if this is where this question belongs because the course I am doing is predominately calculus based (Applied Physics, University level) but we have a small statistics component for obvious reasons.

Any who, sorry if it's in the wrong spot.

Homework Statement

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 fluid 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?

The Attempt at a Solution

For a) Finding the standard deviation of a set of numbers is pretty trivial but I have no idea how to find the average standard deviation from a standard deviation. Someone said we have to use a standard deviation table? If this is true where can I find such table?
The other questions I'm just not sure where to start.

Any help would be appreciated.

Thanks

 

[HallsofIvy]

If random variable X is normally distributed with mean \mu and standard deviation \sigma, then the average of n trials of X is normally distributed with mean \mu and standard deviation \sigma/\sqrt{n-1}. If you are asked to do this problem you are probably expected to be familliar with that theorem.

 

[Filip Larsen]

For a) you should probably use some knowledge about how independent normal distributed variables sum (hint: their mean and variance sum, see for instance [1]). You do not require access to anything but a calculator with square root for this.

For b) and c) you do, as you say, need access standard deviation table or calculator. I suggest you either check your library or search the net for one. Finding the table is probably going to be easier than using it correctly, so you may want to find a relevant textbook that describes this. Given the problem you've been given I'm a bit puzzled that you not already have access to such textbooks or material.

[1] http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

Later: Forgot to refresh the page before answering, so didn't see HallsofIvy's already answered this. However, I am a bit surprised to see the n-1 version of the theorem mentioned in this situation, but too lazy to go search my stat books for it.

 

[Ray Vickson]

If random variable X is normally distributed with mean \mu and standard deviation \sigma, then the average of n trials of X is normally distributed with mean \mu and standard deviation \sigma/\sqrt{n-1}. If you are asked to do this problem you are probably expected to be familliar with that theorem.

This is incorrect. The average of n trials has standard deviation sigma/sqrt(n) [not sqrt(n-1)]. However, if we *don't know sigma*, but must estimate it from the data at the same time we are estimating the mean, _then_ the so-called sample variance s2 = sum{(x - sum(x)/n)^2: i=1..n} has expected value (n-1)*sigma^2, so an estimate of sigma is sqrt(S2/(n-1)). Note: this is a _biased_ estimate of sigma; the quantity S2/(n-1) is an unbiased estimate of sigma^2, but taking the square root introduces some bias. (Usually, though, we do not bother trying to get unbiased estimates of sigma.)

RGV

 

[KingBigness]

Thank you for your replies everyone.

I have to admit I have been very slack when it has come to the stats lectures and haven't really gone...always considered myself more of a calculus student and stats was stupid. But through this physics course I have realized just how fundamental statistics is so I think it's about time I pull my finger out.

So I am sure we have learned all these theorems I just haven't personally learned them. I got my textbook so I'll have a read through that just wanted a guide on where to start this problem.

Thank you again and I will let you know how everything goes.