Respuesta :
Answer:
The probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.2241.
Step-by-step explanation:
Let X = number of students arriving at the 10:30 AM time slot.
The average number of students arriving at the 10:30 AM time slot is, λ = 3.
A random variable representing the occurrence of events in a fixed interval of time is known as Poisson random variables. For example, the number of customers visiting the bank in an hour or the number of typographical error is a book every 10 pages.
The random variable X is also a Poisson random variable because it represents the fixed number of students arriving at the 10:30 AM time slot.
The random variable X follows a Poisson distribution with parameter λ = 3.
The probability mass function of X is given by:
[tex]P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...,\ \lambda>0[/tex]
Compute the probability of X = 2 as follows:
[tex]P(X=2)=\frac{e^{-3}3^{2}}{2!}=\frac{0.0498\times 9}{2}=0.2241[/tex]
Thus, the probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.2241.
Answer:
Probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.224.
Step-by-step explanation:
We are given that a statistics professor finds that when he schedules an office hour at the 10:30 am, time slot, an average of three students arrive.
Let X = number of students arriving at the 10:30 AM time slot.
The above situation can be represented through Poisson distribution because a random variable representing the occurrence or arrival of events in a fixed period of time is known as Poisson random variables.
The probability distribution function of Poisson distribution is given by;
[tex]P(X =x) = \frac{e^{-\lambda }\times \lambda^{x} }{x!} ; x=0,1,2,3,.......[/tex]
where, [tex]\lambda[/tex] = arrival rate of students
As we know that the mean of Poisson distribution = [tex]\lambda[/tex]
So, X ~ Poisson([tex]\lambda=3[/tex])
Thus, the probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is given by = P(X = 2)
P(X = 2) = [tex]\frac{e^{-3 }\times 3^{2} }{2!}[/tex]
= [tex]\frac{e^{-3} \times 9 }{2}[/tex] = 0.224
Therefore, probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive is 0.224.
