# Normal Distribution problem

1. Dec 14, 2011

### MiamiThrice

Hello all,

While studying for exam I came across this practise problem which is giving me some trouble.

The amount , A, of wine in a bottle ~ N(1.05L, .0004L2).

a)The bottle is labelled as containing 1L. What is the probability a bottle contains less than 1L?
b) Casks are available which have a volume, V, which is N(22L,.16L2). What is the probability the contents of 20 randomly chosen bottles will fit inside a randomly chosen cask?

For (a), I simply did P(z < 1-1.05 / 0.02), and the correct answer being 0.0062.

However I am unsure how to approach (b). At first I thought i could just use the mean of the cask size (22L) and do P(Z<= 21-22 / 0.4), but it didnt work out.

Any help is appreciated thanks.

2. Dec 14, 2011

### chiro

Hey MiamiThrice.

You have to be a little careful with this question because you can only have a positive amount of wine (in terms of litres) and the normal distribution is defined over the entire real line which includes all negative values, so just keep that in mind.

For the second one, you are interested in 20 bottles being less than or equal to the free room in the cask. A 95% interval for the cask can be calculated.

From the above you can calculate the distribution for 20 wine bottles (each bottle is independent) using rules for adding normal distributions together, and then calculate the probability that corresponds with your cask confidence interval.

3. Dec 14, 2011

### MiamiThrice

Hey Chiro,

I'm not completely sure what you mean.

I have the following formula:
ƩXi ~ N(nμ, nδ2).

Should I use that to sum of the 20 wines? (n = 20, μ and δ given)

I'm unsure how to use this together with the normal distrubition for the volume of the cask.

4. Dec 15, 2011

### chiro

Yes that is the right idea.

You have a distribution for 20 bottles of wine using formula above.

Now you need to calculate the probability that lies within your casket interval.

Your casket interval is given to you, but you need to specify a confidence interval. Most staistical applications use 95%, so you need to calculate the 95% interval for the casket which is (a,b) [a being the lower bound, be being the upper bound] where P(a < Y < b) = 0.95 where the probability refers to casket distribution.

Now given (a,b) you need to find out P(a < X < b) where X is the distribution for the 20 wines in litres.

Does this make sense to you?