Mathematics: 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.

where is the link or photoStep-by-step explanation:...

A grocery store counts the number of customers who

harleycochran2ovyt3n

17.04.2021

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:

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

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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, $\mu = \frac{10}{60} = \frac{1}{6}$

This is:

$P(X \geq 1) = 1 - P(X = 0)$

In which

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

$P(X = 0) = \frac{e^{-\frac{1}{6}}*(\frac{1}{6})^{0}}{(0)!} = 0.8465$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.8465 = 0.1535$

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, $\mu = \frac{10*4}{60} = \frac{4}{6} = \frac{2}{3}$

This probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

P(X < 2) = P(X = 0) + P(X = 1)

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

$P(X = 0) = \frac{e^{-\frac{2}{3}}*(\frac{2}{3})^{0}}{(0)!} = 0.5134$

$P(X = 1) = \frac{e^{-\frac{2}{3}}*(\frac{2}{3})^{1}}{(1)!} = 0.3423$

P(X < 2) = P(X = 0) + P(X = 1) = 0.5134 + 0.3423 = 0.8557

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8557 = 0.1443$

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

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