Question

One airline averages about 2.1 fatalities per month. Assume that the probability distribution for​ x, the number of fatalities per​ month, can be approximated by a Poisson probability distribution. Complete parts​ (a) through​ (c).
(a) What is the probability that no fatalities will occur during any given month? (b) What is the probability that one fatality will occur during any given month?
(c) Find the standard deviation of x.

Answer 1

Answer:

a) 12.25% probability that no fatalities will occur during any given month.

b) 25.72% probability that one fatality will occur during any given month.

c) The standard deviation of x is 1.45.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval. The variance is the same as the mean

One airline averages about 2.1 fatalities per month.

This means that $\mu = 2.1$

(a) What is the probability that no fatalities will occur during any given month?

This is $P(X = 0)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2.1}*(2.1)^{0}}{(0)!} = 0.1225$

12.25% probability that no fatalities will occur during any given month.

(b) What is the probability that one fatality will occur during any given month?

This is $P(X = 1)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-2.1}*(2.1)^{1}}{(1)!} = 0.2572$

25.72% probability that one fatality will occur during any given month.

(c) Find the standard deviation of x.

The standard deviation is the square root of the variance. So

$\sqrt{2.1} = 1.45$

So the standard deviation of x is 1.45.