Question

Norbert Radyk asked Jul 27, 2022

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1 Answer

Corentin Branquet · Answer 1 · 2022-07-30T22:45:13+0000

Answer:

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.

Explanation:

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

$P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)$

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,
$\mu = (10)/(60) = (1)/(6)$

This is:

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

In which

$P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)$

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

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

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,
$\mu = (10*4)/(60) = (4)/(6) = (2)/(3)$

This probability is:

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

In which

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

So

$P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)$

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

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

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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