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).

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

And replacing we got:

And then the variance would be given by:

Step-by-step 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:

And replacing we got:

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

And replacing we got:

And then the variance would be given by:

.24

step-by-step explanation:

the probability is   .24

the first player had the highest probability of getting a silver marble

the game could be made fair by adding back the marble that was drawn after each draw

12.5 feet

step-by-step explanation: