Outcome x –1 0 1 2 3 4
f(x) 0.05 0.20 0.25 0.35 0.10 0.05

calculate the variance of x.

the formula that my teacher gave me is E(x^2) - U^2= V

so i applied it and i got -1^2*0.05+0^2*0.2+1^2*0.25+2^2*0.35+3^2*0.1+4^2*0.05 = 3.4
i still have to subtract the u^2 which is 1.4^2 and that should give me 3.4-1.4^2=1.44

the problem is i checked the answer and the final answer is 3.4. i seem to be getting the final answer half way so i'm obviously doing something really wrong or there is something i'm not seeing. PLEASE HELP

Question

Outcome x	–1	0	1	2	3	4
f(x)	              0.05	0.20	0.25	0.35	0.10	0.05

calculate the variance of x.

the formula that my teacher gave me is E(x^2) - U^2= V

so i applied it and  i got -1^2*0.05+0^2*0.2+1^2*0.25+2^2*0.35+3^2*0.1+4^2*0.05 = 3.4
i still have to subtract the u^2 which is 1.4^2 and that should give me 3.4-1.4^2=1.44

the problem is i checked the answer and the final answer is 3.4. i seem to be getting the final answer half way so i'm obviously doing something really wrong or there is something i'm not seeing. PLEASE HELP

tkhunny · Answer

You are doing it exactly correctly.  It is NOT 3.4.  It is 1.44.

Does that instill confidence in you or does it call into question whoever or whatever came up with the incorrect answer.

You can test it out using another method.  I used a spreadsheet and created the distribution:

-1 five times
0 twenty times
1 twenty-five times
2 thirty-five times
3 ten times
4 five times

This recreates the distribution exactly.

Without any fancy formulas, I get $\mu = 1.44\;and\;\sigma^{2} = 1.44$
With the fancy formula, which is correct, I also get  $\mu = 1.44\;and\;\sigma^{2} = 1.44$

Good work.  Have no fear, except for the grading mechanism breaking down.  :-(

anonymous · Answer

i will upload the correction sheet from the exam. it is the question 5.

tkhunny · Answer

I do have a typo, which I repeated.  I have $\mu = 1.44$.  This should have been $\mu = 1.4$.  Having said that, the sheet is wrong.