Easy standard deviation problem

[Femme_physics]

1. The problem statement, all variables and given/known data

The average monthly expense of a certain family during 11 months is 4000 shekels, with a standard deviation of 100 shekels.

Afterwards they added to the calculations the family's expense at the 12th month, and they found the average remained unchanged.

A) What was the family expense in the 12th month
B) Calculate the standard deviation


3. The attempt at a solution

Question A is easy, 4000.

Question B I thought was also easy, but I'm getting the wrong answer.

In my mind it should stay a 100!

The standard deviation so far was a 100, so I kept it like that, added the 12th month expense minus the average squared, all under square root, and I get a 100. Turns out it's not the answer. I'm not sure where my booboo is. Can anyone help me see it?

http://img200.imageshack.us/img200/7995/71798093.jpg (http://imageshack.us/photo/my-images/200/71798093.jpg/)

[Mark44]

Since the average remained the same, their expenditures in the 12th month were the same as the average, which you already know. However, the st. dev. should decrease.

In the calculation for the st. dev. for the first 11 months, there is a divisor of 10. For the calculation of the standard deviation for the whole year, the divisor will be 11, and the sum of the squared differences will be the same as before, since the squared difference between the expenditure for the month and the average expenditure are equal, x - x-bar is zero.

[Femme_physics]

Ah, thanks Mark! I again must be using the wrong formula that I miswrote in my notebook. But some sources make the number you divide by "n", some make it "n-1", I'm not clear on which is true!

[I like Serena]

Ah, thanks Mark! I again must be using the wrong formula that I miswrote in my notebook. But some sources make the number you divide by "n", some make it "n-1", I'm not clear on which is true!

They are both true, since standard deviations come in 2 flavors.
If this is the same notebook as the previous time, they'll be using "n".

Knowing this the calculation of the standard deviation in this case is still not trivial....

[Femme_physics]

Ah. I think I get it. Its depends whether it's an even or an odd number :)

[I like Serena]

Ah. I think I get it. Its depends whether it's an even or an odd number :)

No not quite. :wink:

I see you're eager to learn, so I'll try to explain.
Skip to the end if you think it's boring. ;)

It depends on whether you know the mean exactly beforehand, or whether you need the numbers to calculate the mean.

If you need the numbers to calculate the mean, the mean isn't quite accurate, and this shows in the fact that the standard deviation is slightly bigger.
This means you divide by "n-1".
This is called the "sample mean" respectively the "sample standard deviation".

If you know the mean beforehand exactly, it means there is no uncertainty left in the mean.
This shows in a standard deviation that is slightly smaller, and you divide by "n".
This is called the "population mean" respectively the "population standard deviation".

In your field of expertise, I don't think you'll ever know any mean beforehand exactly.
So you should always use the divider "n-1".

However, in your notebook they obviously use the other form, which is not wrong if you consider the numbers to represent the entire "population". So if you want the same answers as your solution manual, you'll probably have to use the divider "n".


So, that was quite a long story (and you may need a couple of examples to get it).
Now could you please do the problem? :wink:

[Femme_physics]

I'll use the divider "n" from now on then :)

Oh yea -- the problem! You did notice I was trying to smile and thank my way out of it, didn't you? ^^ Darn it, can't get anything around you! Okay, okay!

Since 1100/11 = 100

Then 1100/12 = 91.666667

[I like Serena]

I'll use the divider "n" from now on then :)

Oh yea -- the problem! You did notice I was trying to smile and thank my way out of it, didn't you? ^^ Darn it, can't get anything around you! Okay, okay!

Since 1100/11 = 100

Then 1100/12 = 91.666667

Not quite! :wink:

You're using a normal average here to find the standard deviation.
But standard deviations are made up from a square root of a sum of squares divided by a number.

So what you would have for the first 11 months is:

\sqrt{ \frac{(month1 - 4000)^2 + (month2 - 4000)^2 + ... + (month11 - 4000)^2}{11} } = 100

and what you need, is:

\sqrt{ \frac{(month1 - 4000)^2 + (month2 - 4000)^2 + ... + (month11 - 4000)^2 + (4000 - 4000)^2}{12} } = ?

Can you calculate this?

[Femme_physics]

Hmm. Am I suppose to guess what's in month 1, 2, 3... etc? It's as though I have 11 unknowns in one equation!

[I like Serena]

Hmm. Am I suppose to guess what's in month 1, 2, 3... etc? It's as though I have 11 unknowns in one equation!

Yes, you have 11 unknowns in one equation. :)

But suppose we define:
S =(month1 - 4000)^2 + (month2 - 4000)^2 + ... + (month11 - 4000)^2

and substitute that.
Suddenly those 11 unknowns have become 1 unknown! :smile:

[Femme_physics]

Do I plug it in like that? I get zero! (no scanner, but I got a lame webcam :D - hope img is clear)

[I like Serena]

Do I plug it in like that? I get zero! (no scanner, but I got a lame webcam :D - hope img is clear)

The image is clear :)

Seems you plugged the webcam in ok! :smile:

It seems right, although you left off the square root.
Furthermore you have another equation from which you can deduce the value of S.

[Femme_physics]

:)

Alright, I'll try the 1st equation first.
It seems right, although you left off the square root.

So I'm left with an unknown, under a square root, divided by 12 equals 0?

That's...still S = zero, no?

[I like Serena]

:)

Alright, I'll try the 1st equation first.

So I'm left with an unknown, under a square root, divided by 12 equals 0?

That's...still S = zero, no?

Uhh, no it does not equal 0, and neither does S.

What you wrote down, is what you need to calculate. The outcome is as yet unknown.

[Femme_physics]

According to my calculations, S must equal zero for the equation to make any sense. Either my algebra is messed, or I'm not sure what...

[I like Serena]

According to my calculations, S must equal zero for the equation to make any sense. Either my algebra is messed, or I'm not sure what...

You're setting the equation to zero, which it is not.

Try this equation:

\sqrt{ \frac{S}{11} } = 100


Can you solve it for S?

And then substitute that value of S in your expression. You'll find it is not zero, but the answer you're looking for.

Btw, be careful with the square root symbol. In your last scan you put down the square root sign wrong.

[Femme_physics]

Let's see. If I do the manipulation it's

s/11 = 1002
s/11 = 10000
s = 10000 x 11
s = 110000

Right?

[I like Serena]

Let's see. If I do the manipulation it's

s/11 = 1002
s/11 = 10000
s = 10000 x 11
s = 110000

Right?

Right! :smile:

[Femme_physics]

So according to my calculation, if we add something that adds up to 0, then all we need to do is


110000/12 = 9166.6666

Take the square root of it and...

S = 95.742

[I like Serena]

So according to my calculation, if we add something that adds up to 0, then all we need to do is


110000/12 = 9166.6666

Take the square root of it and...

S = 95.742

Yep! :smile:

Does it match your solution manual?

[Femme_physics]

I love hearing a string of "yep!" with that smiley ;)

And yes, it does match :)


You rock!

[I like Serena]

I love hearing a string of "yep!" with that smiley ;)

And yes, it does match :)

You rock!

You have just ensured yourself of my continued attention! :smile:
Let's rock together!!! :cool: