Normal Distribution problem

[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.

 

[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.

 

[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.

 

[chiro]

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.

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?

 

[blue_raver22]

I would approach this problem by first understanding the concept of a normal distribution and how it applies to the given scenario. A normal distribution is a probability distribution that is symmetrical around the mean, with most of the data falling within a certain range from the mean.

For problem (a), we are given the mean (1.05L) and the variance (.0004L2) of the amount of wine in a bottle. To find the probability that a bottle contains less than 1L, we need to calculate the z-score for 1L using the formula z = (x - μ) / σ, where x is the value we are interested in (1L), μ is the mean (1.05L), and σ is the standard deviation (square root of variance, which is 0.02L in this case). This gives us a z-score of -2.5. Using a z-score table, we can find that the probability of a bottle containing less than 1L is 0.0062 or 0.62%.

For problem (b), we are given the mean (22L) and the variance (.16L2) of the cask size. We are asked to find the probability that the contents of 20 randomly chosen bottles will fit inside a randomly chosen cask. To solve this, we need to use the central limit theorem, which states that the sample mean from a large sample will be approximately normally distributed, regardless of the distribution of the population. In this case, we are interested in the sample mean of 20 bottles, which can be calculated by dividing the total volume of 20 bottles by 20. This gives us a sample mean of 1L. Now, we can use the same formula as problem (a) to calculate the z-score for this sample mean. The z-score turns out to be -10, which is a very low probability. However, since we are dealing with a large sample, we need to use a z-score table for a sample size of 20, which gives us a probability of 0. This means that the probability of the contents of 20 bottles fitting inside a randomly chosen cask is extremely low, almost impossible.

In summary, understanding the concept of normal distribution and using the appropriate formulas and tables can help us solve these types of problems. It is also important to consider sample size and the central limit theorem when dealing with large samples.