help with my statistics homework? (basic stuff)

[iamme]

i've done the work, but i think i might have done something wrong...

three amounts:

1,000 (0.25 probability)
2,000 (0.60 probabilty)
5,000 (0.15 probability)

i've found the mean = 1000(0.25) + 2000(0.60) + 5000(0.15) = 2200

now i need to find the variance, but i get some huge number (over a million).

here's what i did:

(1000-2200)^2 (0.25) + (2000-2200)^2 (0.60) + (5000-2200)^2 (0.15) = 1,560,000

is this correct?

 

[alexeikgb]

Originally posted by: iamme
i've done the work, but i think i might have done something wrong...

three amounts:

1,000 (0.25 probability)
2,000 (0.60 probabilty)
5,000 (0.15 probability)

i've found the mean = 1000(0.25) + 2000(0.60) + 5000(0.15) = 2200

now i need to find the variance, but i get some huge number (over a million).

here's what i did:

(1000-2200)^2 (0.25) + (2000-2200)^2 (0.60) + (5000-2200)^2 (0.15) = 1,560,000

is this correct?

Correction: divide by 3 to get the answer

but I haven't taken stat for 4 years now......

 

[pancho619]

The variance would be ((1000-2200)^2 (0.25) + (2000-2200)^2 (0.60) + (5000-2200)^2 (0.15) )/3

= 520000

Variance = [Sum(X-M)^2] / N

where X= your scores (1000(0.25 probability)
M = mean
N = number of scores

Hope this helps.

 

[iamme]

Originally posted by: pancho619
The variance would be ((1000-2200)^2 (0.25) + (2000-2200)^2 (0.60) + (5000-2200)^2 (0.15) )/3

= 520000

Variance = [Sum(X-M)^2] / N

where X= your scores (1000(0.25 probability)
M = mean
N = number of scores

Hope this helps.

strange, in my book, there's no mention of dividing the entire sum by N (number of scores). 😕

 

[her209]

Drink more :beer:

 

[ingenuiti]

For a population variance divide by N. However, for a probability variance, you are not suppose to divide by N.

 

[iamme]

ok, i'm looking for the variance of a discrete probability distribution.

dividing by N is for just a population (or sample, N-1).

 

[pancho619]

For a population variance divide by N. However, for a probability variance, you are not suppose to divide by N.

My apologies. Sorry for the confusion.

-- pancho

 

[ingenuiti]

Originally posted by: iamme
ok, i'm looking for the variance of a discrete probability distribution.

dividing by N is for just a population (or sample, N-1).

Then your answer is correct.

 

[iamme]

Originally posted by: pancho619

For a population variance divide by N. However, for a probability variance, you are not suppose to divide by N.

My apologies. Sorry for the confusion.

-- pancho

no prob. i should have included more information in the beginning.

my answer just seemed awfullly high 🙂

thanks guys!

 

[ingenuiti]

Remember it is very hard to interpret the variance as a number. Generally you should calculate the standard deviation from the variance to see if the answer makes sense.

 

[welst10]

your number IS correct. You should look at standard deviation, which is SQRT(1560000)=1249, which doesn't look abnormmaly high.