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Math Help - pdf and mean problems

  1. #1
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    pdf and mean problems

    1) The length of time, Y, that a customer spends in line at a bank teller's window before being served is described by the exponential pdf:

    f(y) = 0.2e-0.2y, y 0.

    a) What is the probability that a customer will wait for more than 10 minutes?

    I used 1 - P( Y 10 ) = 1 - 0.2e-0.2y = 0.9729

    b) Suppose the customer will leave if the wait is more than 10 minutes. Assume that the customer goes to the bank 3 times next month. Let the random variable X be the number of times the customer leaves without being served. Find P(X=1).

    This I'm not sure how to do. Any hints?


    2) A warehouse contains 10 printing machines, 3 of which are defective. A company selects 5 of the machines at random. What is the mean number of defective printers selected?

    Since there are 3 defective machines, I found the probability of picking each one.
    1: 3/10
    2: 2/9
    3: 1/8
    Then I multiplied the number of the defective machine by its probability of being selected and summed them all up: (3/10) + [2*(2/9)] + [3*(1/8)] = 1.1194

    Does this seem right?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by engineer22 View Post
    1) The length of time, Y, that a customer spends in line at a bank teller's window before being served is described by the exponential pdf:

    f(y) = 0.2e-0.2y, y 0.

    a) What is the probability that a customer will wait for more than 10 minutes?

    I used 1 - P( Y 10 ) = 1 - 0.2e-0.2y = 0.9729

    b) Suppose the customer will leave if the wait is more than 10 minutes. Assume that the customer goes to the bank 3 times next month. Let the random variable X be the number of times the customer leaves without being served. Find P(X=1).

    This I'm not sure how to do. Any hints?
    You are confusing the pdf with the cumulative distribution function.

    <br />
f(y)=0.2 e^{-0.2y}, \ \ \ 0<y<\infty<br />

    is a pdf, it integrates up to 1, but not a cdf as it goes to zeros as y \to \infty.

    The cdf corresponding to this pdf is:

    <br />
F(y) = \int_0^y f(\zeta)\ d\zeta = 1 - e^{- 0.2\ x}<br />

    The second part of the problem is a binomial distribution problem, you will have found the probability that
    he leaves without being served on a single visit call this p, then the number of times he leaves without
    being served in three visits X has a binomial distribution B(3, p).

    RonL
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  3. #3
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    for number 2 you have a binomial distribution where the probability of picking a defective part is \frac{3}{10}. so

    E[X]=np=5\left(\frac{3}{10}\right)=\frac{3}{2}
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