Fresh produce is delivered to a restaurant every four days. Every 6 days, the restaurant receives a delivery of a dried and cann
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Answer:
Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>
Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>
Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>
Part d: <em>The probability of at least one arrival in a 15-second period is 0.9179.</em>
Step-by-step explanation:
Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,
The probability mass function of X can be written as,
Substitute the value of λ=10 in the formula as,
Part a:
The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,
<em>The probability of no arrivals in a one-minute period is 0.000045.</em>
Part b:
The probability mass function of X can be written as,
The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,
<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>
Part c:
Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as
That is,
So, the probability mass function of Y is,
The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:
<em>The probability of no arrivals in a 15-second is 0.0821.</em>
Part d:
The probability that there is at least one arrival in a 15-second period is calculated as,
<em>The probability of at least one arrival in a 15-second period is 0.9179.</em>