1. ## [SOLVED] Mean question

Hello, i am stuck on a question and i can't work it out.

if you have the number 1234, and you take all 24 permutations of it, you let X = the number of numbers that stay the same i.e 1324 = X=2 as 1 and 4 have stayed teh same, 4321 = X=0 as none have stayed the same.

Now the question says that E(X) = 1, yet when i work it out i get 2

X = {0,1,2,3,4} so E(X) = ((0+1+2+3+4)/5) = 2

where am i going wrong?

2. Originally Posted by Rapid_W
Hello, i am stuck on a question and i can't work it out.

if you have the number 1234, and you take all 24 permutations of it, you let X = the number of numbers that stay the same i.e 1324 = X=2 as 1 and 4 have stayed teh same, 4321 = X=0 as none have stayed the same.

Now the question says that E(X) = 1, yet when i work it out i get 2

X = {0,1,2,3,4} so E(X) = ((0+1+2+3+4)/5) = 2

where am i going wrong?
Hi Rapid W,

It is true that the possible values of X are 0, 1, 2, 3, and 4, but it is not true that they are all equally likely (which is what you seem to assume).

Suggestion: Write down all 24 permutations of 1234, then write down the value of X for each permutation. Then compute the average value of X.

3. crikey i completely forgot they they had different probablitys X=3 doesn't even occur. Thank you for pointing ou the obvious

4. mean = ((1/24)*4)+((6/24)*2)+(8/24) = 1

am i correct in therefore thinking that the variance is -4/5 ?

5. Originally Posted by Rapid_W
mean = ((1/24)*4)+((6/24)*2)+(8/24) = 1

am i correct in therefore thinking that the variance is -4/5 ?
? No, the variance can never be negative.