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Combinations of Random Variables Awkward Variance Questions

M~N (80,400)

W~N (70,225)

what IS THE probability that the average weight of 4 randomly selected men is less than 70?

Okay so Z=(W1+W2+W3+W4)/4

So for the variance do I multiply by 1/16 I guess so...

But in edexcel EX4a they do a similar thing but don't include the coefficient being squared in the variance

Thoughts?
Reply 1
Original post by L'Evil Wolf


But in edexcel EX4a they do a similar thing but don't include the coefficient being squared in the variance

Thoughts?


1. Which module?

2. Show us. Why do you think they haven't included the coefficient being squared?

3. It's probably because you have to add the variances up, so: 225 + 225 + 225 + 225, and then divided that by 16. Which is 4×22516=2254\frac{4 \times 225}{16} = \frac{225}{4}. But I can't say that for sure because you haven't shown anything.

In the future, please try to provide some sort of link or picture or anything instead of just paraphrasing.
(edited 8 years ago)
Reply 2
S3.

imgur is playing up it corresponds to the bottom of the page 3 s3 guide - assuming you possess it.



It is the part where you deduce whether the operation is Var(12X) ir 12^2(Var X)

Original post by Zacken
1. Which module?

2. Show us. Why do you think they haven't included the coefficient being squared?

3. It's probably because you have to add the variances up, so: 225 + 225 + 225 + 225, and then divided that by 16. Which is 4×22516=2254\frac{4 \times 225}{16} = \frac{225}{4}. But I can't say that for sure because you haven't shown anything.

In the future, please try to provide some sort of link or picture or anything instead of just paraphrasing.
Reply 3
Original post by L'Evil Wolf
S3.

imgur is playing up it corresponds to the bottom of the page 3 s3 guide - assuming you possess it.

It is the part where you deduce whether the operation is Var(12X) ir 12^2(Var X)


You've given me two different questions now. Which is it? The weight of men or bottles?
Reply 4
Original post by Zacken
You've given me two different questions now. Which is it? The weight of men or bottles?


I Get How The Operation Works With The Men, Although I Am Unsure Of The Operation On The Bottles. Could You Please Tell Me That When:

I Have 12 Bottles, How do you distinguish between doing 12Var(X) and Var(12X) you explained the men question, thanks.
Reply 5
Original post by L'Evil Wolf
I Get How The Operation Works With The Men


You could have said that instead of just ignoring my post. :tongue:

Although I Am Unsure Of The Operation On The Bottles. Could You Please Tell Me That When:

I Have 12 Bottles, How do you distinguish between doing 12Var(X) and Var(12X) you explained the men question, thanks.


It's 12 bottles, so B1+B2++B12B_1 + B_2 + \cdots + B_{12} Hence you would do:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\text{Var} \left(\frac{B_1 + B_2 + \cdots + B_{12}}{12}\right) = \frac{1}{12^2}\text{Var}(B_1 + \cdots + B_{12}) = \frac{12}{12^2} \text{Var}(B) = \frac{\text{Var} B}{12}\end{equation*}

Reply 6
Original post by Zacken
You could have said that instead of just ignoring my post. :tongue:



It's 12 bottles, so B1+B2++B12B_1 + B_2 + \cdots + B_{12} Hence you would do:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\text{Var} \left(\frac{B_1 + B_2 + \cdots + B_{12}}{12}\right) = \frac{1}{12^2}\text{Var}(B_1 + \cdots + B_{12}) = \frac{12}{12^2} \text{Var}(B) = \frac{\text{Var} B}{12}\end{equation*}



Perfect explanation, thank you. It is a shame edexcel did not do what you did.

I apologise, I did not intend to do that.

Lastly do you know of any learning material on this topic. There is a plethora of questions availiable, however scarce learning material I would say.

I suppose with the madas and edexcel questions is enough, but it would be better to have good background :P
Reply 7
Original post by L'Evil Wolf
Lastly do you know of any learning material on this topic. There is a plethora of questions availiable, however scarce learning material I would say.


I thought the textbook was more than enough. I don't know of anything else.
Reply 8
Original post by Zacken
I thought the textbook was more than enough. I don't know of anything else.


Okay thanks for your help. It will have to do.
Reply 9
Original post by L'Evil Wolf
Okay thanks for your help. It will have to do.


No worries.
Original post by Zacken
No worries.


I really see what you have done, you modelled it as the variable being multiplied as in 12*(all the bottles) where I was doing bottle1+bottle2+...+bottle12.

Where the addition is what edexcel were referring too.

So the multiplication of the variable as in Var(3X) is what you would do when you have 9(VarX1) - so then 3(Var(X1)) = Var (X1+X2+X3).

Your explanation was very useful, I imagine it will be of great use in the future.

Thanks again.
I would rep you but I have ran out.
Reply 11
Original post by L'Evil Wolf
I really see what you have done, you modelled it as the variable being multiplied as in 12*(all the bottles) where I was doing bottle1+bottle2+...+bottle12.


No, I'm adding as well. It's just that when you add two random variables, you add their variances. But instead of me adding all 12 variances, I just multiplied by 12 instead.
Original post by Zacken
No, I'm adding as well. It's just that when you add two random variables, you add their variances. But instead of me adding all 12 variances, I just multiplied by 12 instead.


Thanks.
Reply 13
Original post by L'Evil Wolf
Thanks.


No worries.
Original post by Zacken
You could have said that instead of just ignoring my post. :tongue:



It's 12 bottles, so B1+B2++B[b]12[/b]B_1 + B_2 + \cdots + B_{[b]12[/b]} Hence you would do:

Unparseable latex formula:

\displaystyle[br]\begin{equation*}\text{Var} \left(\frac{B_1 + B_2 + \cdots + B_{12}}{12}\right) = \frac{1}{12^2}\text{Var}(B_1 + \cdots + B_{12}) = \frac{12}{12^2} \text{Var}(B) = \frac{\text{Var} B}{12}\end{equation*}



How did you know to divide by 12 here?
Reply 15
Original post by L'Evil Wolf
How did you know to divide by 12 here?


What's the mean? "sum all the terms and divide by the number of terms".
Original post by Zacken
What's the mean? "sum all the terms and divide by the number of terms".


The worked solution on page 4 of the edexcel book says to do 12Var(X) are they wrong?
Reply 17
Original post by L'Evil Wolf
The worked solution on page 4 of the edexcel book says to do 12Var(X) are they wrong?


No, I was answering a different question. My post was about the variance of the mean of the weights.

The variance of the weights in total, is 12Var(X).
Original post by L'Evil Wolf
How did you know to divide by 12 here?


I see what you are saying here.

If you are given two distributions for men and women, eg M~N(a,b) and W~N(c,d) and you are asked to find the probability of the men's weight being greather than 1.5 times the woman's weight, then we can call that a new distribution (say U), and M-1.5W = U ~ N (a - 1.5c, b + 1.5^2d) (and we are looking for P(U>0) but that is not the point I am trying to illustrate :tongue:)

When you're just adding people's weights up rather than multiplying the distribution by a number, you don't need to use Var(ax) = a^2Var(x).
Original post by SeanFM
I see what you are saying here.

If you are given two distributions for men and women, eg M~N(a,b) and W~N(c,d) and you are asked to find the probability of the men's weight being greather than 1.5 times the woman's weight, then we can call that a new distribution (say U), and M-1.5W = U ~ N (a - 1.5c, b + 1.5^2d) (and we are looking for P(U>0) but that is not the point I am trying to illustrate :tongue:)

When you're just adding people's weights up rather than multiplying the distribution by a number, you don't need to use Var(ax) = a^2Var(x).


Got it, thanks for the post :smile:

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