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 the following pmf.

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

Compute E(X) and V(X).

E(X)	=  batches
V(X)	=  batches2

Compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of X.]

expected weight left	lb
variance of weight left	lb2

Answer 1

Concepts and reason

Mathematical expectation is the product of the probability of an event occurring and the value corresponding with the actual observed occurrence of the event.

Fundamentals

The expected value of the random variable can be defined as,

$E\left( X \right) = \sum {x{\rm{ }}p\left( x \right)}$

And

$E\left( {{X^2}} \right) = \sum {{x^2}{\rm{ }}p\left( x \right)}$

$E\left( {aX} \right) = aE\left( X \right)$

The variance of the random variable is,

$V\left( X \right) = E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2}$

$V\left( {a + bX} \right) = {b^2}V\left( X \right)$

Let X be the number of batches ordered by a randomly chosen customer. The pmf of X is,

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

The expected value of X is,

$\begin{array}{c}\\E\left( X \right) = \sum\limits_{i = 1}^n {{x_i} \cdot p\left( {{x_i}} \right)} \\\\ = 1\left( {0.3} \right) + 2\left( {0.5} \right) + 3\left( {0.1} \right) + 4\left( {0.1} \right)\\\\ = 0.3 + 1 + 0.3 + 0.4\\\\ = 2\\\end{array}$

The expected number of batches ordered by a randomly chosen customer is 2 batches.

The expected value of square of X is,

$\begin{array}{c}\\E\left( {{X^2}} \right) = \sum\limits_{i = 1}^n {x_i^2 \cdot p\left( {{x_i}} \right)} \\\\ = {1^2}\left( {0.3} \right) + {2^2}\left( {0.5} \right) + {3^2}\left( {0.1} \right) + {4^2}\left( {0.1} \right)\\\\ = 0.3 + 2 + 0.9 + 1.6\\\\ = 4.8\\\end{array}$

The variance of number of batches orders is,

$\begin{array}{c}\\V\left( X \right) = E\left( {{X^2}} \right) - {\left\{ {E\left( X \right)} \right\}^2}\\\\ = 4.8 - {2^2}\\\\ = 4.8 - 4\\\\ = 0.8\\\end{array}$

Let Y be the chemical supply company currently has in stock 100 lb of a certain chemical which it sells to customers in 5 lb batches. The relationship between X and Y is,

$Y = 100 - 5X$

The expected number of pounds left after the next customer order is shipped is,

$\begin{array}{c}\\E\left( Y \right) = E\left( {100 - 5X} \right)\\\\ = 100 - 5E\left( X \right)\\\\ = 100 - 5\left( 2 \right)\\\\ = 100 - 10\\\\ = 90\,lb\\\end{array}$

The variance of weight left is,

$\begin{array}{c}\\V\left( Y \right) = V\left( {100 - 5X} \right)\\\\ = {\left( { - 5} \right)^2}V\left( X \right) & & \left( {{\rm{Since }}V\left( {{\rm{constant}}} \right) = 0} \right)\\\\ = 25\left( {0.8} \right)\\\\ = 20\\\end{array}$

Ans:

The expected number of batches ordered by a randomly chosen customer is 2 batches.

The variance of number of batches orders is 0.8 batches².

The expected number of pounds left is 90 lb.

The variance of pounds left is 20 lb².