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On the average, 1.8 customers per minute arrive at any one of the checkout counters of a grocery store. What probability distribution can be used to find the probability that there will be no customers arriving at a checkout counter in the next minute

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anuksha0456

The probability that there will be no customers arriving at a checkout counter in the next minute is 0.16530.

The number of customers arriving per minute at the checkout counter of a grocery store is following Poisson Distribution.

A Poisson Distribution over a variable X, having a mean λ, has a probability for a random variable x as [tex]P(X = x) = e^{-\lambda} \frac{\lambda^{x}}{x!}[/tex] .

In the question, we are asked to find the probability that no customer arrives at the checkout, hence we put x = 0, and the mean number of customers per minute (λ) = 1.8.

Substituting the values of x and λ in the equation, we get:

[tex]P(X = 0) = e^{-1.8} \frac{1.8^{0}}{0!}[/tex] ,

or, [tex]P(0) = e^{-1.8} = 0.16530[/tex] .

Therefore, the probability that there will be no customers arriving at a checkout counter in the next minute is 0.16530.

Learn more about Poisson Distributions at

https://brainly.com/question/7879375

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