This is a question of queuing theory and the Poisson process would be applicable in this question. Sally typically serves 10 customers per hour.
The number of customers served in a given 15 minute window
= 10 customer/hour x 15 minutes x 1/60 hour/minute = 2.5 customers
This can also be termed as mean serving rate (number of customers served per unit time) or `lambda`. Thus,
`lambda` = 2.5 customers
The probability of serving less...
This is a question of queuing theory and the Poisson process would be applicable in this question. Sally typically serves 10 customers per hour.
The number of customers served in a given 15 minute window
= 10 customer/hour x 15 minutes x 1/60 hour/minute = 2.5 customers
This can also be termed as mean serving rate (number of customers served per unit time) or `lambda`. Thus,
`lambda` = 2.5 customers
The probability of serving less than 3 customer per block of 15 minutes can be determined as
P (x < 3) = P (x = 0) + P (x = 1) + P (x = 2)
where, x is the number of customers served in the 15 minute time period.
Thus, P (x < 3) = `(2.5^0 xx e^(-2.5))/(0!) + (2.5^1 xx e^(-2.5))/(1!) + (2.5^2 xx e^(-2.5))/(2!) `
= 0.544.
Thus, the probability that less than 3 customers are served in a given 15 minute time period is 0.544.
Hope this helps.
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