Statistics II Question

[vince_24]

To all the smart students on this forum, I have a question that needs answering & you guys are my final hope

My question is with regards to variance calculations.

I am really confused as to when to square the co-efficient and when we do not in terms of variance calculations.

Some examples are set out below

• If we have X~Po(4) and Y~Po(6) - How do we calculate the variance of 2x+4y? Would it be 2^2Var(x) + 4^2Var(y) or 2Var(x) + 4Var(y)
• If we have a fair coin being tossed 5 times, If we want to calculate the variance after being doubled, would it be 2^2Var(x) or 2Var(X)?

Any help on this matter is greatly appreciated. I wish you all the best in your examinations.

[astarmathsandphysics]

For a poisson the variance is the mean in this case 2 and 4 resp so var(2x+4y) =2*4+4*6

[astarmathsandphysics]

I dont understand the second question. What is doubled? The variance for a binomial distribution is np(1-p) If the number of throws is doubled then the variance is doubled

[vince_24]

It is from the May June 03 Paper.

A fair coin is tossed 5 times and the number of heads is recorded.

(i) The random variable X is the number of heads. State the mean and variance of X [Understand]
(ii) The number of heads is doubled and denoted by the random variable Y. State the mean and variance of Y. [Do Not Understand]

[astarmathsandphysics]

The vaiance of y is orriginally 5*0.5(1-0.5) anf when y is doubled the new variance is 2^2*5*0.5(1-0.5)

[astarmathsandphysics]

I just want to remind you that for a poisson distribution the mean is equal to the variance so to find the variance, find the mean first

[vince_24]

Okay, yes. The original Variance is just npq.

When Y is doubled, why is the new variance 2^2 * 5 * 0.5(1-0.5) ? I don't understand the 2^2 bit. Why is it not just 2? Why are some questions 2^2 and some just 2?

[astarmathsandphysics]

V=sum(x-m)^2/(n-1) so if all the x's are doubled
V=sum(2x-2m)^2/(n-1)=V=2^2sum(x-m)^2/(n-1)

[astarmathsandphysics]

mean =sumx/n if the x's are doubled then mean=sum(2x)/n=2sumx/n so the mean is doubled

[vince_24]

mean =sumx/n if the x's are doubled then mean=sum(2x)/n=2sumx/n so the mean is doubled

Yeah, but I am confused with the variance part.

[astarmathsandphysics]

It is because in the variance (x-mean) is squared so if you double x, you double the mean and get(2x-2*mean)^2 which is 2^2(x-mean)^2

[Superstrings]

When a co-efficient is given for any variables, then it's always squared.


The problem is that the examiner may not be nice, and you have to form the equation yourself, in which case, things can go something like

X + X + X + X + X + Y + Y + Y

or

5X + 3Y.

This generally causes a lot of confusion.

In this case, coin toss =/= poisson so mean =/= variance.
Therefore you can't just find the mean and assume it's the same as the variance.

I think your question is, "To square or not to square, the variance co-efficient?"

Well, there is no straightforward answer.
Let's say, there is a long jumper, whose distances have a mean of 5m and s.d. of 0.5m.
No two jumps will ever be the same... (although they can be the same distance)
If you have currency, $50 will always = $50, so it is squared.

I'm sorry I haven't been more clear, but in time, you will be used to it.
I'll stick around this topic ^^

[vince_24]

When a co-efficient is given for any variables, then it's always squared.


The problem is that the examiner may not be nice, and you have to form the equation yourself, in which case, things can go something like

X + X + X + X + X + Y + Y + Y

or

5X + 3Y.

This generally causes a lot of confusion.

In this case, coin toss =/= poisson so mean =/= variance.
Therefore you can't just find the mean and assume it's the same as the variance.

I think your question is, "To square or not to square, the variance co-efficient?"

Well, there is no straightforward answer.
Let's say, there is a long jumper, whose distances have a mean of 5m and s.d. of 0.5m.
No two jumps will ever be the same... (although they can be the same distance)
If you have currency, $50 will always = $50, so it is squared.

I'm sorry I haven't been more clear, but in time, you will be used to it.
I'll stick around this topic ^^

Thanks for the help. For the coin question, the mark scheme states you have to square the co-efficient of the variance, thus giving the mean = 5 and variance = 5.

You are getting to my question, if it isnt too much trouble, could you give a further example? Thanks for your time.

[astarmathsandphysics]

If the average price of petrol triples the variance will be multiplied by 3^2. Noticed the standard deviation is only multiplied by 3 cos it is sqrt(variance)

[Superstrings]

For the coin question the variance is squared. This is because you are repeating the same test.

If you found this information useful, please mail max g's to
Iksa
Thaurissan
Alliance


Thanks :p