Question

1. A restaurant serves an average of 180 customers per hour during the lunch time. (a)....

1. A restaurant serves an average of 180 customers per hour during the lunch time.

(a). What probability distribution is most appropriate for calculating the probability of a given number of customers arriving within one hour during lunch time?

(b). What are the mean and the standard deviation of the number of customers this restaurant serves in one hour during lunch time?

(c). Would it be considered unusually low if only 150 customers showed up to this restaurant in one hour during lunch time?

(d). Calculate the probability that this restaurant serves 170 customers in one hour during lunch time?

Homework Answers

Answer #1

(a) The Poisson distribution is most appropriate for calculating the required probability.

(b) Let X be the random variable denoting the number of customers the restaurant serves in one hour during lunch time. Thus, X ~ Poi(180).

We have, the mean, E(X) = 180 [From the properties of Poisson distribution] and variance, Var(X) = 180.

(c) The probability that 150 customers showed up to the restaurant during the lunch time = P(X = 150) = = 0.0023.

Hence, the probability is very low that only 150 customers showed up to the restaurant during lunch time.

(d) The probability that 170 customers showed up in one hour during lunch time = P(X = 170) = = 0.0230.

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