Problem in statistics...pls help

Hey your problem is tooo complex in front of mine...i hope you can help in this..... here's my question...


how to compute pooled variance??


If we are given N (Population Size) Mean (X) and Variance for say 2 groups how can we compute the combined/pooled variance???????

sample size with replacement = N^n = 5^2 = 25

Lets say we have 5 items - Sample Space = [B1, B2, B3, B4, B5]

From this we are selecting 2 with replacement. Lets say we are selecting B1 first. Now the sample space would be [B2,B3,B4,B5]. B1 in our hand. We are selecting with replacement. Hence we keep the item B1 back into the sample space. Now the sample space is [B1, B2, B3, B4, B5]. Now select the 2nd item which could be either B1, or B2 or B3 or B4 or B5. ie the pair of items selected would be

(B1,B1) , (B1,B2) , (B1,B3) , (B1,B4) , (B1,B5) - 5 ways of selecting.

Now the same could be applied with selecting B2 as the first item.

 (B2,B1) , (B2,B2) , (B2,B3) , (B2,B4) , (B2,B5) - 5 ways of selecting.

Likewise for B3,B4 and B5.

(B3,B1) , (B3,B2) , (B3,B3) , (B3,B4) , (B3,B5) - 5 ways of selecting.

(B4,B1) , (B4,B2) , (B4,B3) , (B4,B4) , (B4,B5) - 5 ways of selecting.

(B5,B1) , (B5,B2) , (B5,B3) , (B5,B4) , (B5,B5) - 5 ways of selecting.

Thus the total ways of selecting 2 items from 5 items with replacement is 25. It is generally given by the formula N^n.

N-  Total size  and n - sample size.

Hope the explanation is clear. Thanks.