Suppose that during the hours of service the number of passe

Suppose that during the hours of service, the number of passengers coming to a bus stop follows a purely random arrival process. Assume that on average, 12 passengers arrive per hour.

(a) What is the probability that there are exactly 10 passengers arriving in one hour?

(b) What is the probability that there are two or fewer passengers arriving in 15 minutes?

(c) What is the probability that there are exactly 20 passengers arriving in two hours?

(d) What is the probability that there are exactly six passengers arriving in 30 minutes?

(e) Find the probability that no more than 5 minutes elapses between two successive passengers arriving.

(f) Find the mean and standard deviation of the time between two successive passengers arriving.

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Solution

a)

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    12      
          
x = the number of successes =    10      
          
Thus, the probability is          
          
P (    10   ) =    0.104837256 [ANSWER]

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b)

There is an average of 12*(15/60) = 3 passengers in 15 minutes.

Using a cumulative poisson distribution table or technology, matching          
          
u = the mean number of successes =    3      
          
x = the maximum number of successes =    2      
          
Then the cumulative probability is          
          
P(at most   2   ) =    0.423190081 [ANSWER]

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c)

There is an average of 12*2 = 24 passengers in 2 hours.

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    24      
          
x = the number of successes =    20      
          
Thus, the probability is          
          
P (    20   ) =    0.062378173 [ANSWER]

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d)

There is an average of 12*(30/60) = 6 passengers in 30 minutes.

Note that the probability of x successes out of n trials is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes =    6      
          
x = the number of successes =    6      
          
Thus, the probability is          
          
P (    6   ) =    0.160623141 [ANSWER]

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