Question

Suppose that the number of units produced daily at factory A is a random variable with mean $20$and standard deviation $3$and the number produced at factory B is a random variable with mean $18$and standard deviation of $6$. Assuming independence, derive an upper bound for the probability that more units are produced today at factory B than at factory A.

Short answer

An upper bound for the probability that more units are produced today at factory $B$ than at factory $A$ is $45/49=.9184$.

Step-by-step solution

Given Information

Let $X_A$represents the number of units produced daily at factory $A$and $X_B$represents the number of units produced daily at factory $B$. For these random variables is given that:

${\mu }_A=E\left(X_A\right)=20,{\sigma }_A=3,{\mu }_B=E\left(X_B\right)=18,{\sigma }_A=6$

Also, assume that random variables $X_A$and $X_B$are independent.

Explanation

The probability that more units are produced today at factory $B$than at factory $A$is

$P\left\{X_B>X_A\right\}=P\left\{X_B-X_A>0\right\}.$

In that case, let's consider the random variable $X=X_B-X_A$. The random variable $X$has mean

$\mu =E\left[X\right]={\mu }_B-{\mu }_A=-2$

and, because of independence, variance

${\sigma }^2=Var\left(X\right)={\sigma }_B^2+{\sigma }_A^2=45$

Now, using Corollary for $\underset{¯}{a=2}$, we get:

$P\{X>\underbrace{-2+2}\}\le \frac{{\sigma }^2}{{\sigma }^2+2^2}=\frac{45}{45+4}=\mathbf{.}\mathbf{9184}$