Probability of X > Y in Normal Distributions & Sampling Bottle Weights

[Maybe_Memorie]

Homework Statement

If we were to have two normal distributions, say X and Y, which are the scores of students on a test or something, a is the score of a student from X and b is the score of a student from Y, what is the probability that a randomly selected student from X will score more than a randomly selected student from Y, i.ie what's the probability that a>b, or something like a>b+10?
Both means and SDs given.



Also, if we have a normal distribution, say for weights of bottles of something, and we know that A is the probability that a randomly chosen bottle will weigh less than X, and we then take a sample of say 6 bottles, and we want to find the probability that 2 or less will weigh less than X, is it just 6C0(1-X)^6 + 6C1(X)(1-X)^5 + 6C2(X)^2(1-X)^4
or do I have to standardise again for the population of 6 bottles?


Thanks!

 

[Ray Vickson]

Homework Statement

If we were to have two normal distributions, say X and Y, which are the scores of students on a test or something, a is the score of a student from X and b is the score of a student from Y, what is the probability that a randomly selected student from X will score more than a randomly selected student from Y, i.ie what's the probability that a>b, or something like a>b+10?
Both means and SDs given.



Also, if we have a normal distribution, say for weights of bottles of something, and we know that A is the probability that a randomly chosen bottle will weigh less than X, and we then take a sample of say 6 bottles, and we want to find the probability that 2 or less will weigh less than X, is it just 6C0(1-X)^6 + 6C1(X)(1-X)^5 + 6C2(X)^2(1-X)^4
or do I have to standardise again for the population of 6 bottles?


Thanks!

Is this a homework problem? What have you done on it so far? Where are you having trouble?

RGV

 

[Maybe_Memorie]

Is this a homework problem? What have you done on it so far? Where are you having trouble?

RGV

It's not homework, just me doing revision before going back to university in september.

The first one is a problem I came across in a very old exam paper, and don't know where ti begin.

The second one I just need a yes or no answer.

 

[I like Serena]

What kind of distribution is the difference of 2 independent normal distributions?

 

[Maybe_Memorie]

What kind of distribution is the difference of 2 independent normal distributions?

Normal with mean u1 - u2 and variance V1 + V2

 

[Maybe_Memorie]

Wait, think I get it, find the difference as a normal distribution, and then the probability that this is greater than zero for the probability one is bigger than the other?

 

[I like Serena]

Yep!

 

[I like Serena]

As for the second part of your question.

You have an A which is a probability, and you have an X which is an amount.
You thought up a formula for a probability, which contains X, but does not contain A.
Since X is not a probability, isn't it kind of unlikely that you can use it in a formula where a probability is expected?

 

[Maybe_Memorie]

Sorry! Should have been 6C0(1-A)^6 + 6C1(A)(1-A)^5 + 6C2(A)^2(1-A)^4

 

[I like Serena]

All right.
Then I'll just say yes, although I do not understand what you mean by "standardize for the population of 6 bottles".

 

[Maybe_Memorie]

All right.
Then I'll just say yes, although I do not understand what you mean by "standardize for the population of 6 bottles".

I mean do i treat the six bottles as a new normal population, find the z value for X, and a new probability that a bottle chosen from these will be less than X?

 

[I like Serena]

Ah, now I understand.

Well, did you open the bottles and mix them?
Or did you calculate the mean weight of the 6 bottles and compare that?

 

[Ray Vickson]

If the 6 bottles are drawn independentally (say, from a large population) you just have a coin-tossing type problem. Each bottle, independent of the others, has a probability 'A' of weighing more than some fixed weight w (you called it X, but I prefer to use the convention that capital letters are random variables, while small letters are given numbers). This is like the problem of getting 3 heads in 6 coin tosses, where each toss has Pr{heads} = A.

RGV

 

[Maybe_Memorie]

Ah, now I understand.

Well, did you open the bottles and mix them?
Or did you calculate the mean weight of the 6 bottles and compare that?

I think I'm starting to confuse myself...

But basically, if i take 6 bottles from a population, and in this population the probability that a randomly chosen bottle weighs less than X is A,
the probability that 2 of the 6 bottles weighs less than X is the expression i gave above?

 

[I like Serena]

I think I'm starting to confuse myself...

But basically, if i take 6 bottles from a population, and in this population the probability that a randomly chosen bottle weighs less than X is A,
the probability that 2 of the 6 bottles weighs less than X is the expression i gave above?

I already said yes and I'm not changing my answer now!
As RGV already said, it's basically a coin-tossing problem with a binomial distribution, which is what you wrote down.

 

[Maybe_Memorie]

Thanks to both of you!