Answer :
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:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given time interval.
Poisson process with rate θ = 4 per hour.
This means that [tex]\mu = 4n[/tex], 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 [tex]\mu = 4*2 = 8[/tex]
This is P(X = 10). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 10) = \frac{e^{-8}*8^{10}}{(10)!} = 0.0993[/tex]
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 [tex]\mu = 4*0.5 = 2[/tex]
This is P(X = 0). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353[/tex]
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 [tex]\mu = 4*0.5 = 2[/tex]