Question

a statistics professor finds that when he schedules an office hour at the 10:30a, time slot, an average of three students arrive. Find the probability that in a randomly selected office hour in the 10:30 am time slot exactly two students will arrive

Answer 1

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:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...,\ \lambda>0$

Compute the probability of X = 2 as follows:

$P(X=2)=\frac{e^{-3}3^{2}}{2!}=\frac{0.0498\times 9}{2}=0.2241$

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 2

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;

$P(X =x) = \frac{e^{-\lambda }\times \lambda^{x} }{x!} ; x=0,1,2,3,.......$

where, $\lambda$ = arrival rate of students

As we know that the mean of Poisson distribution = $\lambda$

So, X ~ Poisson($\lambda=3$)

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) = $\frac{e^{-3 }\times 3^{2} }{2!}$

                     = $\frac{e^{-3} \times 9 }{2}$ = 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.