Results 1 to 3 of 3

Math Help - Please Help, probability

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    6

    Please Help, probability

    The number of train passengers who fail to pay for their tickets in a certain region has a poison distribution with a mean of 4 per train

    i) all passengers pay their ticket?
    ii) More than 1 passenger fail to pay for their tickets

    If there are 4800 trains per month in the region,

    iii) Find the expected number of trains with at least 1 non paying passenger.
    iv) Calculate the average cost to the train company during a month if the average cost of a ticket is $12.54
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Mar 2008
    Posts
    39
    Quote Originally Posted by whleow View Post
    The number of train passengers who fail to pay for their tickets in a certain region has a poison distribution with a mean of 4 per train

    i) all passengers pay their ticket?
    ii) More than 1 passenger fail to pay for their tickets

    If there are 4800 trains per month in the region,

    iii) Find the expected number of trains with at least 1 non paying passenger.
    iv) Calculate the average cost to the train company during a month if the average cost of a ticket is $12.54
    i)= P(X=0) = e^-4 = 0.01832
    ii)= 1-P(X=1)-P(X=0) = e^-4(1+4) = 0.09157
    iii)+iv) Not sure, this could mean I've done the previous ones wrong...
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Nyoxis View Post
    i)= P(X=0) = e^-4 = 0.01832
    ii)= 1-P(X=1)-P(X=0) = e^-4(1+4) = 0.09157
    iii)+iv) Not sure, this could mean I've done the previous ones wrong...
    iii) X is the random variable number of non-paying passengers on a train.

    X ~ Poisson(mean = 4).

    Let Y be the random variable number of trains with at least one non-paying passenger.

    Then Y ~ Binomial(n = 4800, p = 0.98168)

    where p = \Pr(X > 0) = 1 - \Pr(X = 0) \approx 1 - 0.01832 = 0.98168.

    Then E(Y) = np ( ) = .....

    iv) (Answer to iii)($12.54) = .....
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: October 10th 2014, 10:02 PM
  2. Replies: 10
    Last Post: January 21st 2011, 12:47 PM
  3. Replies: 3
    Last Post: May 29th 2010, 08:29 AM
  4. Replies: 1
    Last Post: February 18th 2010, 02:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 07:30 AM

Search Tags


/mathhelpforum @mathhelpforum