Question

Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the Poisson distribution. Provide answers to the following to 3 decimal places. What is the probability that exactly two customers arrive in a minute? Find the probability that more than three customers arrive in a two-minute period. What is the probability that at least seven customers arrive in three minutes, given that exactly two arrive in the first minute?

Answer 1

Answer:

a)  P(2)=0.270

b) P(X>3)=0.605

c)  P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the  Poisson distribution:

$\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}$

a)  In this case: $\lambda=2.1$

$P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270$

Therefore, the probability is P(2)=0.270.

b)  In this case: $\lambda=2\cdot 2.1=4.2$

$P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605$

Therefore, the probability is P(X>3)=0.605.

c)  We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: $\lambda=2\cdot 2.1=4.2$

$P(X\geq 5)=1-P(X$

Therefore, the probability is P=0.410.