Question

A statistics professor finds that when he schedules an office hour at the​ 10:30 a.m. time​ slot, an average of three students arrive. Use the Poisson distribution to find the probability that in a randomly selected office hour in the​ 10:30 a.m. time slot exactly five students will arrive.

Answer 1

Answer:

Using poisson distribution the probability that in a randomly selected office hour in the 10:30 a.m. time slot exactly five students will arrive is 0.10

Solution:  

The professor expected the students will arrive at 10:30 pm

The average of students arrive in time = 3

By using poisson equation,

A Poisson distribution is a statistical distribution showing the likely number of times that an event will occur within a specified period of time. It is used for independent events which occur at a constant rate within a given interval of time.

$P(x)=\frac{e^{-\mu} \times \mu^{x}}{x !}$

where $\mu$ is mean or average.

For exactly 5 students , x = 5 and $\mu$ = 3

$P(5)=\frac{e^{-3} \times 3^{5}}{5 !}$

We know that $5 !=120 ; 3^{5}=243 ; e^{-3}=0.0497$

Hence, $P(5)=0.0497 \times \frac{243}{120}$

$=\frac{12.09}{120} = 0.10$

Hence using poisson distribution the probability that in a randomly selected office hour in the 10:30 a.m. time slot exactly five students will arrive is 0.10