Question

The number of requests for assistance received by a towing service is a Poisson process with rate θ = 4 per hour.a. Compute the probability that exactly ten requests are received during a particular 2-hour period.b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?c. How many calls would you expect during their break?

Answer 1

Answer:

a) 9.93% probability that exactly ten requests are received during a particular 2-hour period

b) 13.53% probability that they do not miss any calls for assistance

c) 2

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.

Poisson process with rate θ = 4 per hour.

This means that $\mu = 4n$, in which n is the number of hours.

a. Compute the probability that exactly ten requests are received during a particular 2-hour period.

n = 2, so $\mu = 4*2 = 8$

This is P(X = 10). So

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

$P(X = 10) = \frac{e^{-8}*8^{10}}{(10)!} = 0.0993$

9.93% probability that exactly ten requests are received during a particular 2-hour period

b. If the operators of the towing service take a 30-min break for lunch, what is the probability that they do not miss any calls for assistance?

n = 0.5, so $\mu = 4*0.5 = 2$

This is P(X = 0). So

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

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

13.53% probability that they do not miss any calls for assistance

c. How many calls would you expect during their break?

n = 0.5, so $\mu = 4*0.5 = 2$