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: $\lambda=2\cdot 2.1=4.2$

$P(X\geq 5)=1-P(X<5)\\\\P(X\geq 5)=1-P(X\leq 4)\\\\P(X\geq 5)=1-\sum_{x=0}^4 \frac{4.2^x\cdot e^{-4.2}}{x!}\\\\P(X\geq 5)=1-0.590\\\\P(X\geq 5)=0.410$

Therefore, the probability is P=0.410.

Answer 2

Suppose X indicates the number of customers that enter a grocery store within one minute.

$\to X \sim \text{Poisson}(2.1)$

X's probability mass function:

$\to P(X=x)=\frac{e^{-2.1} \times 2.1^{x}}{x!} ,x =0,1,2,..$

For question 1:

The likelihood of exactly two consumers arriving in a minute:

$P(X=2)=\frac{e^{-2.1} \times 2.1^{2}}{2!}= 0.270016 \approx 0.270$

For question 2:

Suppose Y represents the average group of consumers who enter a grocery shop in a two-minute period.

$Y \sim \text{Poisson}(2.1\times 2) \ or\ Y \sim \text{Poisson}(4.2)$

Y's probability mass function:

$P(Y = y) = \frac{e^{-4.2} \times 4.2^{y}}{y!}, y =0,1,2..$

This is possible that more than three clients will arrive within a two-minute timeframe.

$= P(Y > 3) \\\\= 1 - P(Y \leq 3)\\\\=1- \Sigma^{3}_{y=0} \frac{e^{-4.2} \times 4.2^y}{y!}\\\\=1-0.395403\\\\= 0.604597\approx \ 0.605$

For question 3:

Given that exactly two customers arrive in the first minute, the likelihood of at least seven clients arriving in three minutes is  

$= \text{P( at least 5 customers arrive in two minutes)} \\\\= P(Y \geq 5)\\\\= 1 - P(Y<5)\\\\=1-P(Y \leq 4)\\\\= 1 - 0.589827\\\\= 0.410173 \approx \ 0.410$

Learn more:

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