Answer :
Answer: a) 30, b) 0.2138, c) 0.0820
Step-by-step explanation:
Since we have given that
X be the number of calls in 1 hour = 60 minutes
Rate of one call every two minutes.
[tex]\lambda=\dfrac{1}{2}[/tex]
(a) What is the expected number of 911 calls in one hour?
[tex]\lambda=\dfrac{1}{2}\times 60=30[/tex]
(b) What is the probability of three 911 calls in five minutes? If required, round your answer to four decimal places.
[tex]\lambda=\dfrac{1}{2}\times 5=2.5[/tex]
Number of calls = 3
So, P(X=3) is given by
[tex]P(X=3)=\dfrac{e^{-\lambda}\lambda^x}{x!}\\\\P(X=3)=\dfrac{e^{-2.5}\times (2.5)^3}{3!}\\\\P(X=3)=0.2138[/tex]
(c) What is the probability of no 911 calls during a five-minute period? If required, round your answer to four decimal places.
[tex]P(X=0)=\dfrac{e^{-2.5}(2.5)^0}{0!}=0.0820[/tex]
Hence, a) 30, b) 0.2138, c) 0.0820
Using the Poisson distribution, it is found that:
a) The expected number of calls is of 30 per hour.
b) There is a 0.2138 = 21.38% probability of three 911 calls in five minutes.
c) There is a 0.0821 = 8.21% probability of no 911 calls during a five-minute period.
Poisson distribution:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
In which
x is the number of successes
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean.
Item a:
- One call every 2 minutes, thus, by proportions, in one hour(60 minutes), 30 calls.
Item b:
- Period of 5 minutes, one call each 2 minutes, so in 5 minutes, 2.5 calls, which means that the mean is [tex]\mu = 2.5[/tex].
- The probability is P(X = 3), so:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 3) = \frac{e^{-2.5}(2.5)^{3}}{(3)!} = 0.2138[/tex]
0.2138 = 21.38% probability of three 911 calls in five minutes.
Item c:
- The probability is P(X = 0), so:
[tex]P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-2.5}(2.5)^{0}}{(0)!} = 0.0821[/tex]
0.0821 = 8.21% probability of no 911 calls during a five-minute period.
A similar problem is given at https://brainly.com/question/24225387