Joint Distribution (Means,Medians....) PLEASE HELP!

Economics2012

Hi, I'm wondering if somebody could help me understand this.....
If you have a joint distribution with

Hourly Wage (Y)
Years of Education (X) 9 15 30
10 0.07 0.02 0.01
14 0.10 0.30 0.10
16 0.02 0.10 0.28


Sorry the 0.07,0.10 and 0.02 are under the 9 and the 0.02,,0.30 and 0.10 are under 15 etc.

How would you find the mean,median or st dev?

I'm so very confused with this stuff?

Any help would be greatly appreciated?

mathman

It is not clear what you want. If you want the mean of everything, just add up all the numbers and divide by the number of items (9). Ti get the median, arrange them in order and take the middle number. To get the variance, first get the average of the squares and then subtract the square of the mean.

Economics2012

Well see the table looks a bit odd there Y is along the top and X is along the left hand side.
9,15 and 30 are belong to Y and 10,14 and 16 are belong to X. then the numbers are in the table.
It says find the means, medians and st devs of X and Y.
I did what you said before I posted here, but that would be the whole table wouldn't it?
I'm confused as in how you get one for each, is that what this means?

mathman

You need to clarify what you want the mean, etc. of. Do you want it over X for fixed Y (3 answers) or over Y for fixed X (also 3 answers) or something else?

Economics2012

I found the x and y mean.....
u=Ep(x)x

ux=(10)(0.1)+(14)(0.5)+)(16)(0.4) = 14.4 -> mean for X by using the marginals
uy= same procedure I got 21.61.

Do you know how you get the median for this, if you understand what I mean?

mathman

Where did 0.1, 0.5, and 0.4 come from?

Economics2012

Adding the x's across and then you get the y's by adding down.

mathman

It looks like you are asking for the statistics of X independent of Y and the statistics of Y independent of X. ux is then the average wage and uy is the average years. What is the underlying question?