Respuesta :
Answer:
The probability that 5 messages are received in 1 hour is [tex]P(5)\approx 0.1755[/tex]
The probability that 10 messages are received in 1.5 hours is [tex]P(10)\approx 0.0858[/tex]
The probability that less than 2 messages are received in 1/2 hour is [tex]P(X<2) \approx 0.2873[/tex]
Step-by-step explanation:
The probability distribution of a Poisson random variable representing the number of successes occurring in a given time interval or a specified region of space is given by the formula:
[tex]P(X)=\frac{e^{-\mu}\mu^{x} }{x!}[/tex] where
[tex]\mu =[/tex] mean number of successes in the given time interval or region of space.
[tex]e=2.71828[/tex]
a) What is the probability that 5 messages are received in 1 hour?
From the information given we know that [tex]\mu=5[/tex]
Applying the Poisson random variable formula we get:
[tex]P(5)=\frac{e^{-5}\cdot 5^{5}}{5!}\\P(5)\approx 0.1755[/tex]
b) What is the probability that 10 messages are received in 1.5 hours?
We know from the information given that in 1 hour, 5 messages are received. In 0.5 hours we received 2.5 messages therefore in 1.5 hours we received 7.5 messages [tex]\mu=7.5[/tex]
Applying the Poisson random variable formula we get:
[tex]P(10)=\frac{e^{-7.5}\cdot 7.5^{10}}{10!}\\P(10)\approx 0.0858[/tex]
c) What is the probability that less than 2 messages are received in 1/2 hour?
We know from the information given that in 1 hour, 5 messages are received therefore in 0.5 hours we received 2.5 messages [tex]\mu=2.5[/tex]
To find the probability that less than 2 messages are received in 1/2 hour we find the values when [tex]P(0)[/tex] and [tex]P(1)[/tex] because [tex]P(X<2)=P(0)+P(1)[/tex]
Applying the Poisson random variable formula we get:
[tex]P(0)=\frac{e^{-2.5}\cdot 2.5^{0}}{0!}\\P(0)\approx 0.0821[/tex]
[tex]P(1)=\frac{e^{-2.5}\cdot 2.5^{1}}{1!}\\P(1)\approx 0.2052[/tex]
[tex]P(X<2)=P(0)+P(1)\\P(X<2)\approx 0.0821+0.2052\\P(X<2) \approx 0.2873[/tex]
