Question

A sample of 7 drivers in Ohio reported the number of trips they took outside the county from where they lived. The data is reproduced below. Compute the variance of the distribution. 2.14 3.62 1.9 2.8

Answer 1

`E(X) = \sum_(i=1)^n X_i P(X_i)`

And replacing we got:

`E(X) = 1 *(2)/(7) + 2*(0)/(7)+ 3*(3)/(7) + 4*(1)/(7)+ 5*(1)/(7) =(20)/(7)`

Now we can find the econd moment given by this formula:

`E(X^2) = \sum_(i=1)^n X^2_i P(X_i)`

And replacing we got:

`E(X^2) = 1^2 *(2)/(7) + 2^2*(0)/(7)+ 3^2*(3)/(7) + 4^2*(1)/(7)+ 5^2*(1)/(7) =(70)/(7)`

And we can find the variance with this formula:

`Var(X) = E(X^2) -[E(X)]^2 = (70)/(7) -((20)/(7))^2 = 1.9`

And the best answer would be:

Var (X) =1.9

Explanation:

For this case we assume the following distribution obtained from internet:

X F p

1 2 2/7

2 0 0/7

3 3 3/7

4 1 1/7

5 1 1/7

_________

Tot 7 1

We can begin finding the expected value with this formula:

`E(X) = \sum_(i=1)^n X_i P(X_i)`

And replacing we got:

`E(X) = 1 *(2)/(7) + 2*(0)/(7)+ 3*(3)/(7) + 4*(1)/(7)+ 5*(1)/(7) =(20)/(7)`

Now we can find the econd moment given by this formula:

`E(X^2) = \sum_(i=1)^n X^2_i P(X_i)`

And replacing we got:

`E(X^2) = 1^2 *(2)/(7) + 2^2*(0)/(7)+ 3^2*(3)/(7) + 4^2*(1)/(7)+ 5^2*(1)/(7) =(70)/(7)`

And we can find the variance with this formula:

`Var(X) = E(X^2) -[E(X)]^2 = (70)/(7) -((20)/(7))^2 = 1.9`

And the best answer would be:

Var (X) =1.9