SOLUTION: Customers arrive randomly at a checkout counter at an average rate of 2 every five minutes. PLEASE USE 4 DECIMAL PLACES IN THE ANSWERS. Question content area bottom Part 1

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Question 1205908: Customers arrive randomly at a checkout counter at an average rate of 2 every five minutes.

PLEASE USE 4 DECIMAL PLACES IN THE ANSWERS.
Question content area bottom
Part 1
​a) What is the probability that none arrive in a five minute​ period?
The probability that none arrive in a five minute period is  
  
enter your response here.
​b) What is the probability that more than four arrive in a ten minute​ period?
The probability that more than four arrive in a ten minute period is
  
enter your response here.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Let X denotes the number of persons arriving at the counter.
a Poisson distribution with parameter: lambda=2 per 5 minutes
P%28x%2C+lambda%29=%28e%5E%28-lambda%29%2Alambda%5Ex%29%2Fx%21
P%28x%2C2%29=%28e%5E-2%2A2%5Ex%29%2Fx%21

the probability function is
f%28x%29=+%28e%5E-2%2A2%5Ex%29%2Fx%21
The probability that none arrive in a five-minute period is
f%280%29=+%28e%5E-2%2A2%5E0%29%2F0%21=e%5E%28-2%29
P%28X=0%29=e%5E%28-2%29
P%28X=0%29=1%2Fe%5E2
P%28X=0%29=0.1353


The probability that more+than+four customers arrive in a 5 minute period is
P%28X%3E4%29=1-P%28X%3C=4%29
P%28X%3E4%29=1-%28f%280%29%2Bf%281%29%2Bf%282%29%2Bf%283%29%2Bf%284%29%29
f%280%29=+%28e%5E-2%2A2%5E0%29%2F0%21=e%5E-2
f%281%29=+%28e%5E-2%2A2%5E1%29%2F2%21=2e%5E-2%2F2=e%5E-2
f%282%29=+%28e%5E-2%2A2%5E2%29%2F2%21=4e%5E-2%2F2=2e%5E-2
f%283%29=+%28e%5E-2%2A2%5E3%29%2F3%21=8e%5E-2%2F6=4e%5E-2%2F3
f%284%29=+%28e%5E-2%2A2%5E4%29%2F4%21=16e%5E-2%2F24=2e%5E-2%2F3

P%28X%3E4%29=1-%28e%5E-2%2Be%5E-2%2B2e%5E-2%2B4e%5E-2%2F3%2B2e%5E-2%2F3%29
P%28X%3E4%29=1-6e%5E-2
P%28X%3E4%29=0.1879883
P%28X%3E4%29=0.1880