Question

For the given values of the random variable X, fill in the corresponding probabilities

Short answer

$X=x$	role="math" localid="1648582898375" $P(X=x)$
$0$	$P(X=0)$$=\frac{e^{-8}{8}^0}{0!}=0.00033$
$1$	role="math" localid="1648582946177" $P(X=1)=\frac{e^{-8}{8}^1}{1!}=0.00268$
$2$	$P(X=2)$$=\frac{e^{-8}{8}^2}{2!}=0.01073$
$3$	role="math" localid="1648583006514" $P(X=3)=\frac{e^{-8}{8}^3}{3!}=0.02862$
$4$	$P(X=4)$$=\frac{e^{-8}{8}^4}{4!}=0.05725$

Step-by-step solution

:Given

Eighteen teens on average dies due to motor vehicle accident every day.
Formula used:
$P(X=x)=\frac{e^{-{\lambda }^x}}{x!}$

:Explanation of Solution

Interpretation:
$X$is a arbitrary variable who represents the no. of teens who die because of motor vehicle injuries.$X$can take any value $0,1,2,3,4...$According to the given situation, we can clearly say that $X$follows the Poisson Distribution with parameter mean $\left(\lambda \right)$is 8 per day and variance$\left(\lambda \right)$ is also 8 per day. In Poisson distribution Variance is equals to mean.
Poisson distribution occurs when there is not definite no. of trials or$n\to \infty$ and occurrence of no. of events in a fixed time interval is fixed.
Probability mass function of Poisson distribution is$P(X=x)=\frac{e^{-x^x{x}^x}}{x!}$ x=0,1,2,3,4,
We have used above mentioned probability mass function to get the probabilities