Answer
This equation is a good representation.
Work Step by Step
We know the following equation:
$P(x)=\frac{.93^x\cdot e^{-.93}}{x!}$
$P(x)=\frac{.395(.93^x)}{x!}$
Using this, we find:
$P(0)=\frac{.395(.93^0)}{0!}=.395$
$P(1)=\frac{.395(.93^1)}{1!}=.367$
$P(2)=\frac{.395(.93^2)}{2!}=.171$
$P(3)=\frac{.395(.93^3)}{3!}=.053$
$P(4)=\frac{.395(.93^4)}{4!}=.012$
$P(5)=\frac{.395(.93^5)}{5!}=.002$
$P(6)=\frac{.395(.93^6)}{6!}=.0004$
$P(7)=\frac{.395(.93^7)}{7!}=.00005$
Comparing these to the actual probabilities, we see that this equation is a fairly good representation of the actual numbers.
