Question

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let X = the number of batches ordered by a randomly chosen customer, and suppose that X has pmf below:

x 1 2 3 4
p(x) 0.3 0.5 0.1 0.1

Compute E(X) and V(X).

Answer 1

`E(X) = 1*0.3 +2*0.5 +3*0.1 +4*0.1 = 2`

In order to find the variance first we need to find the second moment given by:

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

And replacing we got:

`E(X^2)=1^2*0.3 +2^2*0.5 +3^2*0.1 +4^2*0.1 = 4.8`

And then the variance would be given by:

`Var(X) = E(X^2) -[E(X)]^2 = 4.8 -(2)^2 = 0.8`

Explanation:

Previous concepts

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).

Solution to the problem

For this case we have the following probability distribution:

X 1 2 3 4

P(X) 0.3 0.5 0.1 0.1

And we can calculate the expected value with this formula:

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

And replacing we got:

`E(X) = 1*0.3 +2*0.5 +3*0.1 +4*0.1 = 2`

In order to find the variance first we need to find the second moment given by:

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

And replacing we got:

`E(X^2)=1^2*0.3 +2^2*0.5 +3^2*0.1 +4^2*0.1 = 4.8`

And then the variance would be given by:

`Var(X) = E(X^2) -[E(X)]^2 = 4.8 -(2)^2 = 0.8`