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

x | 1 | 2 | 3 | 4 |
---|---|---|---|---|

$p(x)$ | 0.2 | 0.4 | 0.3 | 0.1 |

Compute $E(X)$ and $V(X)$. Then compute the expected number of pounds left after the next customer's order is shipped and the variance of the number of pounds left.

#### Solution

The expected value of $X$ is

` $$ \begin{aligned} E(X)&= \sum x*p(x)\\ &= 1*p(1)+2*p(2)+3*p(3)+4*p(4)\\ &= 1*0.2 + 2*0.4 + 3*0.3 + 4*0.1\\ &= 2.3 \text{ lots} \end{aligned} $$ `

To find $V(x)$ let us find $E(X^2)$.

` $$ \begin{aligned} E(X^2)&= \sum x^2*p(x)\\ &= 1^2*p(1)+2^2*p(2)+3^2*p(3)+4^2*p(4)\\ &= 1*0.2 + 4*0.4 + 9*0.3 + 16*0.1\\ &= 6.1 \end{aligned} $$ `

Variance of $X$ is

` $$ \begin{aligned} V(X) &= E(X^2) - [E(X)]^2 \\ &= 6.1 - 2.3^2 \\ &= 6.1 - 5.29 \\ &= 0.81 \end{aligned} $$ `

Let $Y$ = the number of pounds left after first customer.

`$Y=100-5X$`

.

The expected number of pounds left after the next customer's order is

` $$ \begin{aligned} E(Y) & =E(100 - 5X)\\ & = 100 - 5*E(X)\\ & = 100 - 5*2.3\\ & = 88.5 \end{aligned} $$ `

The variance of number of pounds left after the next customer's order is

` $$ \begin{aligned} V(Y) & =V(100 - 5X)\\ & = 25*V(X)\\ & = 25*0.81\\ & = 20.25 \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators