Respuesta :
Answer:
1. c. Poisson
2. 0.9592 = 95.92% probability that in any one minute at least one purchase is made.
3. 0.0017 = 0.17% probability that no one makes a purchase in the next 2 minutes.
Step-by-step explanation:
We have only the mean, which means that the Poisson distribution is used to solve this question, and thus the answer to question 1 is given by option c.
Poisson distribution:
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.
Mean of 3.2 minutes:
This means that [tex]\mu = 3.2n[/tex], in which n is the number of minutes.
2. What is the probability that in any one minute at least one purchase is made?
[tex]n = 1[/tex], so [tex]\mu = 3.2[/tex].
This probability is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-3.2}*3.2^{0}}{(0)!} = 0.0408[/tex]
So
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.0408 = 0.9592[/tex]
0.9592 = 95.92% probability that in any one minute at least one purchase is made.
3. What is the probability that no one makes a purchase in the next 2 minutes?
2 minutes, so [tex]n = 2, \mu = 3.2(2) = 6.4[/tex]
This probability is P(X = 0). So
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
[tex]P(X = 0) = \frac{e^{-6.4}*6.4^{0}}{(0)!} = 0.0017[/tex]
0.0017 = 0.17% probability that no one makes a purchase in the next 2 minutes.
