The number of female customers a coffee shop follow a Poisson process with a mea

Question

The number of female customers a coffee shop follow a Poisson process with a mean rate of 3 per hour. The number of male customers to the same coffee shop also follow a Poisson process with a mean rate of 6 per hour and is independent of the arrivals of female customers. a) What is the probability that the next customer will arrive within 5 minutes? b) What is the probability that exactly thee customers will arrive in the next 5 minutes? The number of female customers a coffee shop follow a Poisson process with a mean rate of 3 per hour. The number of male customers to the same coffee shop also follow a Poisson process with a mean rate of 6 per hour and is independent of the arrivals of female customers. a) What is the probability that the next customer will arrive within 5 minutes? b) What is the probability that exactly thee customers will arrive in the next 5 minutes? The number of female customers a coffee shop follow a Poisson process with a mean rate of 3 per hour. The number of male customers to the same coffee shop also follow a Poisson process with a mean rate of 6 per hour and is independent of the arrivals of female customers. a) What is the probability that the next customer will arrive within 5 minutes? b) What is the probability that exactly thee customers will arrive in the next 5 minutes?

Explanation / Answer

a)mean rate for female = 3per hour = 3/12 per 5 minutes = 1/4 per 5 minutes

mean rate for male =6 per hour = 6/12 per 5 minutes = 1/2 per 5 minutes

a)Probablity tha female customer arrive in 5 minutes = 1-P(0) = 1- e-1/4 =1- 0.7788 = 0.2212

Probablity tha male customer arrive in 5 minutes = 1-P(0) = 1- e-1/2 =1- 0.60653 = 0.39347

Probabilty that next customer arrive within 5 minutes = 0.2212+0.39347 -0.2212*0.39347 = 0.5276

b)Probabilty that exactly three customers will in next 5 minutes

= P(F = 0)*P(M=3)+ P(F = 1)*P(M=2)+ P(F = 2)*P(M=1)+ P(F = 3)*P(M=0)

P(F=0) = e-1/4 .= 0.7788 ,P(M=3) = e-1/2(1/2)3/3! = 0.0126

P(F=1) = e-1/4(1/4). = 0.1947 ,P(M=2) = e-1/2(1/2)2/2! = 0.0758

P(F=2) = e-1/4(1/4)2/2!. = 0.0243 ,P(M=1) = e-1/2(1/2)1/1! =0.3032

P(F=3) = e-1/4(1/4)3/3! = 0.0020 ,P(M=0) = e-1/2 = 0.60653

Probabilty that exactly three customers will in next 5 minutes

= 0.7788*0.0126+0.1947*0.0758+0.0243*0.3032+0.0020*0.60653 = 0.03315196

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