Customers arrive at a drive-up window of a fast-food restaurant at a rate of 3 per minute during lunch hour. a.What is the probability that the next customer will arrive within 2 minutes? b.What is the probability that the next customer will arrive within 4 minutes? c.During the dinner time period,the arrival rate is 1 per minute.What are your answers to a and b for this period? Round to four decimal places as needed.
Question
Customers arrive at a drive-up window of a fast-food restaurant at a rate of 3 per minute during lunch hour. a.What is the probability that the next customer will arrive within 2 minutes? b.What is the probability that the next customer will arrive within 4 minutes? c.During the dinner time period,the arrival rate is 1 per minute.What are your answers to a and b for this period? Round to four decimal places as needed.
Solution
The exponential distribution is defined as:
f(x|λ) = λ * exp(-λx) for x >= 0, 0 otherwise
Where λ is the rate parameter. The cumulative distribution function, which gives the probability that a random variable X following an exponential distribution is less than or equal to x, is given by:
F(x|λ) = 1 - exp(-λx)
Given λ = 3 (for lunch hour) and λ = 1 (for dinner time), we can calculate the probabilities as follows:
a. P(X < 2) during lunch hour = F(2|3) = 1 - exp(-3*2) = 1 - exp(-6) = 1 - 0.0025 = 0.9975
b. P(X < 4) during lunch hour = F(4|3) = 1 - exp(-3*4) = 1 - exp(-12) = 1 - 0.000006 = 0.999994
c. P(X < 2) during dinner time = F(2|1) = 1 - exp(-12) = 1 - exp(-2) = 1 - 0.1353 = 0.8647 P(X < 4) during dinner time = F(4|1) = 1 - exp(-14) = 1 - exp(-4) = 1 - 0.0183 = 0.9817
So, the answers are:
a. 0.9975 b. 0.9999 c. 0.8647 (for 2 minutes) and 0.9817 (for 4 minutes)
Please note that these are rounded to four decimal places.
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