Problem in statistics...pls help

A population comprises 5 members. The number of all possible samples of size 2 that can be drawn from it with replacement is

a) 100

b) 15

c) 125

d) 25

pls give the reason too as i already know the answer...thnx in advance :)

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.