Ask Me Help Desk - Finding standard deviation

in sample of 36 drivers it was found that the average driving seniority of driver is 14 year,with a standard deviation 5 year. The investigator decide to increase to the sample more 4 drivers so in the sum of this sample are include 40 drivers.
the seniority of the new 4 driver is (in years) 16,5,10,25.
the standard deviation of all the driver together is:
A)5.245 years
B)5.296 years
C)6.225 years
D)the standard deviation will bigger but we can't know the exact value of it.
E)the standard deviation will get smaller and we can't know the exact value of it.
thanks.

This will take some time to calculate... you only need to know two formulae:

$Mean = \frac{\Sigma x}{n}$

$Standard\ Deviation = \sqrt{\frac{\Sigma x^2}{n} - $\frac{\Sigma x}{n}$^2}$

Then use what you know.

$14 = \frac{\Sigma x}{36}$

$\Sigma x= 504$

$5 = \sqrt{\frac{\Sigma x^2}{36} - (14)^2}$

$\Sigma x^2 = 7956$

The new summation x is:

$Sigma x= 504 + 16 + 5 + 10 + 25$

The new summation x^2 is:

$\Sigma x^2 = 7956 + 16^2 + 5^2 + 10^2 + 25^2$

Then, find the new standard deviation;

$\sigma = \sqrt{\frac{\Sigma x^2}{n} - $\frac{\Sigma x}{n}$^2}$

hi.
in the picture I mark in blue "V" the section that I understand and in red circle the section that I am not understand.
in the first circle how did you get this value?
and in the 2nd circle how can I know the value there?
http://p1cture.me/images/68891610722702199694.jpg
thank you.

Quote:

Originally Posted by Unknown008

$Standard\ Deviation = \sqrt{\frac{\Sigma x^2}{n} - $\frac{\Sigma x}{n}$^2}$

Hey, wait a second...

Are you sure about this, Unknown008?

I thought the formula for standard deviation was

$\sigma = \sqrt{\frac 1n \sum_{i=1}^n(x_i-\bar x)^2}$

Note that that's NOT equivalent to your formula.

I think in general you can't compute the new standard deviation without knowing the old data points because you don't know how far they are from the new mean. However, this case is an exception because the mean didn't change:

$\bar x_{old} = 14$

$\bar x_{new} = \frac{\sum{x_{old}}+\sum{x_{new}}}{n_{new}}=\frac{ 36 \cdot 14+16+5+10+25}{40}=14$

The new standard deviation is

$\sigma_{new} = \sqrt{\frac 1{n_{new}} \left ( \sum_{i=1}^{n_{old}}(x_i-\bar x_{new})^2+\sum_{i=n_{old}+1}^{n_{new}}(x_i-\bar x_{new})^2 \right )}$

Again, please note that usually we wouldn't be able to know the first term in the brackets, but in this case since the old mean and new mean are the same, we can get that term directly from the old standard deviation.

I'll show you how in the next post.

Pop, I had to leave rather abruptly before finishing my last post. I'll elaborate a little more.

Starting with the equation I left off with:

$\sigma_{new} = \sqrt{\frac 1{n_{new}} \left ( \underbrace{\sum_{i=1}^{n_{old}}(x_i-\bar x_{new})^2}+\sum_{\small{i=n_{old}+1}}^{n_{new}}(x _i-\bar x_{new})^2 \right )}$

If you examine the term I highlighted with a brace underneath it and recall that $\bar x_{new}=\bar x_{old}=14$ , you'll see that you already have that value from the information you already knew.

$\sigma_{old} = \sqrt{\frac 1{n_{old}} {\sum_{i=1}^{n_{old}}(x_i-\bar x_{old})^2}} =5$ (This you know from the information you were given, along with the definition of standard deviation).

${\frac 1{n_{old}} {\sum_{i=1}^{n_{old}}(x_i-\bar x_{new})^2}} =25$ (Here we've just gotten rid of the square root and replaced $\bar x_{old}$ with $\bar x_{new}$ ).

$\underbrace{{\sum_{i=1}^{n_{old}}(x_i-\bar x_{new})^2}} =25 \cdot n_{old}=25 \cdot 36=900$

Now notice that the left side of the equation (with a brace under it) is the same as the highlighted term from the equation at the top. Thus you can replace that whole term in the top equation with 900.

$\sigma_{new} = \sqrt{\frac 1{n_{new}} \left ( 900+\sum_{\small{i=n_{old}+1}}^{n_{new}}(x_i-\bar x_{new})^2 \right )}=\sqrt{\frac 1{40} \left ( 900+\sum_{i=37}^{40}(x_i-14)^2 \right )}$

Now you should be able to just plug in the four new data points for $x_{37}$ through $x_{40}$ .

hi.
well I feel I almost Understand it, I know I need to use with the x=16,5,10,25
"Now you should be able to just plug in the four new data points for X37 through X40 ."
this I also not Understand where to plug it?
in the red circle what value need to be there?
http://p1cture.me/images/04533570649516267854.jpg

thanks for your help.

May the answer is C?

You just plug in each of your four new values for those terms.

$\sigma =\sqrt{\frac 1{40} \left ( 900+(16-14)^2 +(5-14)^2+(10-14)^2+(25-14)^2 \right )}=\sqrt{\frac 1{40} \left ( 900+4 +81+16+121 \right )}$

What do you get for an answer?

OH OK yep now I can see.
So sure the answer is B)5.296 years :)
Thanks

Yes, I'm sure it's the correct formula for ungrouped data.

Anyway, my book has those two versions for ungrouped data:

$\sigma = \sqrt{\frac{\Sigma (x - \bar{x})^2}{n}}$

$\sigma = \sqrt{\frac{\Sigma x^2}{n} - \bar{x}^2}$

I saw the formula you gave on wikipedia, and mine is another version, displayed as:

$\sigma = \sqrt{E[X^2] - (E[X])^2}$

And expectation is the same as mean, and mean is the sum divided by the total number of terms, n.

$\sigma = \sqrt{E[X^2] - (E[X])^2} = \sqrt{\frac{\Sigma x^2}{n} - $\frac{\Sigma x}{n}$^2}$

pop000: second circle: this is the formula for standard deviation, the last term is the mean, to the square.

first circle: solving for sigma x^2

$5 = \sqrt{\frac{\Sigma x^2}{36} - (14)^2}$

Square both sides:

$25 = \frac{\Sigma x^2}{36} - (14)^2$

add 14^2

$221 = \frac{\Sigma x^2}{36}$

Multiply by 36:

$7956 = \Sigma x^2$

And with that I get 5.296225071 as the standard deviation.

Ah, yes. I assumed you must be correct (you almost always are :) ). I just hadn't seen that formula before. I guess I should have just finished doing the problem your way. Then I would have seen that it gives the same answer. :)

And as usual, your way turns out to be easier.

I learned it that way, and to be honest, I don't understand the summation symbol, you know the one you used in your formula. I've seen it quite often on here but was never really sure how it worked. Must be simple, but hey, they never taught me this at school.

$\sum ^n_{i = k}$

where n and k are constants.

Do you maybe have another awesome site which would help me with that? ;)