Respuesta :
Answer:
Part a: The probability of no arrivals in a one-minute period is 0.000045.
Part b: The probability of three or fewer passengers arrive in a one-minute period is 0.0103.
Part c: The probability of no arrivals in a 15-second is 0.0821.
Part d: The probability of at least one arrival in a 15-second period is 0.9179.
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,
[tex]X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)[/tex]
The probability mass function of X can be written as,
[tex]P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots[/tex]
Substitute the value of λ=10 in the formula as,
[tex]P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}[/tex]
Part a:
The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,
[tex]\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}[/tex]
The probability of no arrivals in a one-minute period is 0.000045.
Part b:
The probability mass function of X can be written as,
[tex]P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots[/tex]
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,
[tex]\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}[/tex]
The probability of three or fewer passengers arrive in a one-minute period is 0.0103.
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 [tex]\frac{X}{4}[/tex]
[tex]\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}[/tex]
That is,
[tex]Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)[/tex]
So, the probability mass function of Y is,
[tex]P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots[/tex]
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:
[tex]\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}[/tex]
The probability of no arrivals in a 15-second is 0.0821.
Part d:
The probability that there is at least one arrival in a 15-second period is calculated as,
[tex]\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}[/tex]
[tex]\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}[/tex]
The probability of at least one arrival in a 15-second period is 0.9179.
