A grocery store counts the number of customers who arrive during an hour. The average over a year is 10 customers per hour. Assume the arrival of customers follows a Poisson distribution. (It usually does.) Find the probability that at least one customer arrives in a particular one minute period. Round your answer to 3 decimals. Find the probability that at least two customers arrive in a particular 4 minute period. Round your answer to four decimals.

0.1535 = 15.35% probability that at least one customer arrives in a particular one minute period.

0.1443 = 14.43% probability that at least two customers arrive in a particular 4 minute period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

In which

x is the number of sucesses

e = 2.71828 is the Euler number

is the mean in the given interval.

Find the probability that at least one customer arrives in a particular one minute period.

The average is of 10 customers per hour(60 minutes), so for this case,

This is:

In which

0.1535 = 15.35% probability that at least one customer arrives in a particular one minute period.

Probability that at least two customers arrive in a particular 4 minute period.

The average is of 10 customers per hour(60 minutes), so for this case,

This probability is:

In which

So

0.1443 = 14.43% probability that at least two customers arrive in a particular 4 minute period.

the answer is c. hope this !

step-by-step explanation:

answer: i wpould go with c

step-by-step explanation: