Cambridge Senior Spec book Probabilities cooked and dogy questions

[Jemin99]

Page 666, question 3
The weight of a large loaf of a bread is norm with mean 420 and sd 30.( say X)
The weight of a small loaf of a bread is norm with mean 220 and sd 10 (Say Y)
Sd of total of 5 large loaves and 10 small loaves are?

My Answer: 10(55)^1/2. Since each of a big loaf is INDEPENDENT HENCE 5Var(X)+10Var(Y)=Overall var
But answer says its 25Var(X)+100Var(Y) square root.....

[jamonwindeyer]

Page 666, question 3
The weight of a large loaf of a bread is norm with mean 420 and sd 30.( say X)
The weight of a small loaf of a bread is norm with mean 220 and sd 10 (Say Y)
Sd of total of 5 large loaves and 10 small loaves are?

My Answer: 10(55)^1/2. Since each of a big loaf is INDEPENDENT HENCE 5Var(X)+10Var(Y)=Overall var
But answer says its 25Var(X)+100Var(Y) square root.....

Hey Jemin99! The rule for taking the variance of a linear function of a variable is:



I think I've got a proof somewhere, but it's pretty tedious from memory. In any case, you need to take the square of the coefficient and put it out the front of your variance! That's where that 25 and 100 come from

[Jemin99]

Hey Jemin99! The rule for taking the variance of a linear function of a variable is:



I think I've got a proof somewhere, but it's pretty tedious from memory. In any case, you need to take the square of the coefficient and put it out the front of your variance! That's where that 25 and 100 come from

Hey thanks man but thing is
Wouldnt it be Var(X1+X2+X3+X4+X5) since every big loaf is independent to each other rather than Var(5X) which is 25 Var(X)?

[keltingmeith]

Hey Jemin - I'm inclined to agree with you here. The problem with the books is I think they're mostly relying on probability, and not thinking statistically enough about it all!

Hey Jemin99! The rule for taking the variance of a linear function of a variable is:



I think I've got a proof somewhere, but it's pretty tedious from memory. In any case, you need to take the square of the coefficient and put it out the front of your variance! That's where that 25 and 100 come from

It's actually quite nice.

[Jemin99]

Hey Jemin - I'm inclined to agree with you here. The problem with the books is I think they're mostly relying on probability, and not thinking statistically enough about it all!

It's actually quite nice.

Hey man! So u r saying i can be right?

[keltingmeith]

I mean, assuming your arithmetic was right, yup. I definitely think it should 5var(X) + 10 var(Y) over 25 and 100.

[1234helloooo]

So when do we actually use Var(5X)=5^2Var(X) and when do we use Var(5X)=5xVar(X)

Still so confused :/://

[keltingmeith]

So when do we actually use Var(5X)=5^2Var(X) and when do we use Var(5X)=5xVar(X)

Still so confused :/://

You NEVER have var(5X)=5var(X). That therein lies your confusion.

Let's consider the case of the 5 large loaves alone for now. Then, the variance of the weight of all the loaves is given by var(L1+L2+L3+L4+L5), where L is the weight of the large loaf. Now, let's try and calculate the variance using var(5L). Since they're equal, we get:

var(5L)=var(L1+L2+L3+L4+L5)

which gives us:

5L=L1+L2+L3+L4+L5

There's one massive problem with this, though. This assumes that each large loaf weighs exactly the same - but that can't be true all the time! Statistically speaking, yes, you COULD get each of the loaves to be the same weight, but it's unlikely. These things are normally distributed, it's much more likely that they're each different. That's because the weight of each load is INDEPENDENT of the other loaves.

Because of this, the problem isn't choosing the "correct formula". The problem is you're tackling the question completely wrong. For these cases where each thing you're measuring is INDEPENDENT of the other, you shouldn't say X1+X2=2X, because that completely destroys independence.

So instead, the way you'd tackle it is like this:

Total variance = var(L1+L2+L3+L4+L5)
since each L are independent, we then get:
Total = var(L1)+var(L2)+var(L3)+var(L4)+var(L5)
=5var(L)


So you see, we do get this a*var formula highlighted above, but it's because we DON'T have 5L. What you need to be able to do in your exam is distinguish when each draw is independent, and when they're not. If they're independent, then you use the sum formula. If they're not independent (and are exactly equal to the first event), then you use the square formula.

[1234helloooo]

Thanks form your answer!!!
But I'm still confused, my textbook says Var(aX+bY) =a^2Var(X) + b^2Var(Y) for independent events, is this a different thing??

[jamonwindeyer]

Thanks form your answer!!!
But I'm still confused, my textbook says Var(aX+bY) =a^2Var(X) + b^2Var(Y) for independent events, is this a different thing??

We would use that formula (and it's easy to confuse with the situation above, as I did) if we applied a function to some variable!! For example, say we had a random variable X that was the weight of a parcel to be sent in the post. It has its own mean and variance, but say we care only about the COST of sending the parcel, where the rate is $2.50 per kilogram.

Now, the random variable we care about is 2.5X, and the variance would be \(2.5^2Var(X)\).

So if we apply a function directly to our random variable, that's when the formula above comes in handy. Not when adding the variances of different choices