Respuesta :
Answer:
a) 38.4% probability that on a given day this item is requested more than 5 times.
b) 0.67% probability that on a given day this item is not requested at all.
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 interval.
An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day.
This means that [tex]\mu = 5[/tex]
What is the probability that on a given day this item is requested
(a) more than 5 times?
Either it is requested 5 times or less, or it is requested more than 5 times. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 5) + P(X > 5) = 1[/tex]
We want P(X > 5). So
[tex]P(X > 5) = 1 - P(X \leq 5)[/tex]
In which
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)[/tex]
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067[/tex]
[tex]P(X = 1) = \frac{e^{-5}*(5)^{1}}{(1)!} = 0.0337[/tex]
[tex]P(X = 2) = \frac{e^{-5}*(5)^{2}}{(2)!} = 0.0842[/tex]
[tex]P(X = 3) = \frac{e^{-5}*(5)^{3}}{(3)!} = 0.1404[/tex]
[tex]P(X = 4) = \frac{e^{-5}*(5)^{4}}{(4)!} = 0.1755[/tex]
[tex]P(X = 5) = \frac{e^{-5}*(5)^{5}}{(5)!} = 0.1755[/tex]
[tex]P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 + 0.1755 = 0.616[/tex]
[tex]P(X > 5) = 1 - P(X \leq 5) = 1 - 0.616 = 0.384[/tex]
38.4% probability that on a given day this item is requested more than 5 times.
(b) not at all?
[tex]P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067[/tex]
0.67% probability that on a given day this item is not requested at all.
