Variance

[wtrow]

Twelve passengers enter an elevator at the basement and independently choose to exit
randomly at one of the ten above-ground floors. What is the variance of the number of
stops that the elevator will make? (Assume no one else calls the elevator).

So I have already solved for the expected value:
E(X)=10(1-(9/10)^12) = 7.176
the teacher confirmed that this is correct.

Now to find the variance, I tried using Var(x)=E(X-E(X))^2, however I get an answer much too large (490). What am I doing wrong?

 

[JeffM]

Twelve passengers enter an elevator at the basement and independently choose to exit
randomly at one of the ten above-ground floors. What is the variance of the number of
stops that the elevator will make? (Assume no one else calls the elevator).

So I have already solved for the expected value:
E(X)=10(1-(9/10)^12) = 7.176
the teacher confirmed that this is correct.

Now to find the variance, I tried using Var(x)=E(X-E(X))^2, however I get an answer much too large (490). What am I doing wrong?

Not quite sure what you are doing wrong. E{[X - E(X)][sup:3k9yd7tj]2[/sup:3k9yd7tj]} =
E{X[sup:3k9yd7tj]2[/sup:3k9yd7tj] - [2E(X) * X] + [E(X)][sup:3k9yd7tj]2[/sup:3k9yd7tj]} =
E(X[sup:3k9yd7tj]2[/sup:3k9yd7tj]) - [2E(X) * E(X)] + [E(X)][sup:3k9yd7tj]2[/sup:3k9yd7tj] =
E(X[sup:3k9yd7tj]2[/sup:3k9yd7tj]) - 2[E(X)][sup:3k9yd7tj]2[/sup:3k9yd7tj] + [E(X)][sup:3k9yd7tj]2[/sup:3k9yd7tj] =
E(X[sup:3k9yd7tj]2[/sup:3k9yd7tj]) - [E(X)][sup:3k9yd7tj]2[/sup:3k9yd7tj]

Is that the formula you used?

 

[wtrow]

I get different answers strangely enough. Using E(X^2)-E(X)^2 I get:

10*(1-.9^12)^2 - (10*(1-.9^12))^2
=5.149 - 51.495
=-46.436

I dont think a negative answer can be right.

 

[JeffM]

I get different answers strangely enough. Using E(X^2)-E(X)^2 I get:

10*(1-.9^12)^2 - (10*(1-.9^12))^2
=5.149 - 51.495
=-46.436

I dont think a negative answer can be right. That is for sure!

I think your problem is that, in general, E(X[sup:3g36qw2r]2[/sup:3g36qw2r]) DOES NOT EQUAL [E(X)][sup:3g36qw2r]2[/sup:3g36qw2r].

Try this approach. Write down in one column the possible values of X. Write down in an adjacent column the probability of that value of X. Write down in a third column the products of the probability and the value. The sum of the third column is the expected value of X. In a fourth column, write down the squares of the values. In a fifth column, write down the products of the probability and the square. The sum of the fifth column is the expected value of X[sup:3g36qw2r]2[/sup:3g36qw2r]. So subtracting the sum of column three from the sum of column five gives you your answer. This approach is very easy to set up on a spread sheet.