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Advice on Exponential, Binomial, & Normal Distributions

  1. Sep 8, 2007 #1
    Hey,

    I'm new to all this so cut me some slack, but have been trying to work through some questions, and I cant seem to find answers to these questions... Or atleast find the confidence that my answers/working is correct...


    1. (Exponential Distribution) Telephone calls arrive at the information desk of a large
    computer software company at the rate of 15 per hour.
    What is the probability that the next call will arrive within 15 minutes?
    (What would eb the formula for this one???) I got 0.98347


    2. (Normal Distribution) For a group of trucks, it was found on an annual basis that the distance travelled per truck is normally distributed with a mean of 50.0 thousand km and a standard deviation of 12.0 thousand km.

    (i) What proportion of trucks can be expected to travel between 34.0 and 38.0
    thousand km in the year?


    I get around 6.64%, but im not sure that Im doing it right as I tried to do the same thing on the computer and I get something like 6.674438...

    (ii) How many km will be travelled by at least 80% of the trucks?

    (This one really has me stumpted as im not sure if it means I look at the normal distro curve, and sort of 40% on each side or, do I try and figure out the 20% value there, and say that at least 80% of the trucks travel >than that distance. The latter makes more sense to me...) I tried the latter and I got 39,920km


    3. (Binomial Distribution) A student is taking a multiple-choice exam in which each
    question has four choices.


    Assuming that he/she has no knowledge of the correct answers to any of the questions, he decided on a strategy in which he will place four balls (marked A, B, C and D) into a box.
    He randomly selects one ball (with replacement)for each question. The marking on the ball will determine hisanswer to the question.
    There are five multiple-choice questions on the exam. What is the probability the he will get:


    (i) five questions correct?
    (ii) at least four questions correct?


    I think i have this one sorted now, i)=1/1024 ii)=5/1024


    4. (Poisson Distribution) Assume that the number of network errors experienced in a day on
    a local area network (LAN) is distributed as a Poisson random variable. The average
    number of network errors experienced in a day is 2.4. What is the probability that in any
    given day
    (i) exactly one network error will occur?
    (ii) two or more network errors will occur?


    i) This one I get ~21.7%
    ii) I did 1-P(X=2) = ~78%


    Any help please?


    Thanks,

    NW
     
    Last edited: Sep 8, 2007
  2. jcsd
  3. Sep 8, 2007 #2

    EnumaElish

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    Homework Helper

    You are doing a 2-sided test, so you need 10% on each side.

    For 4.(ii), 1-P(X=2) is the probability that X is not 2. Is that what you wanted?

    It would help if you describe how you got each answer.
     
    Last edited: Sep 8, 2007
  4. Sep 8, 2007 #3
    Hey,

    Thanks for your reply!

    I seem to have it all sorted cept for question 2:

    2. (Normal Distribution) For a group of trucks, it was found on an annual basis that the distance travelled per truck is normally distributed with a mean of 50.0 thousand km and a standard deviation of 12.0 thousand km.

    (i) What proportion of trucks can be expected to travel between 34.0 and 38.0
    thousand km in the year?

    (ii) How many km will be travelled by at least 80% of the trucks?

    Could you explain a little more about this one...
     
  5. Sep 9, 2007 #4

    EnumaElish

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    Science Advisor
    Homework Helper

    (i) is the probability {34 < X < 38}. Do you know how to standardize the 34 and the 38 (turning them into a z score each)? Once you have that, you'd just need to look it up in a normal probability table.

    (ii) is the answer to Prob{Z < z} = 0.8 solved for the little z. If F is the cumulative normal distribution so that F(z) = Prob{Z < z}, F(z) = 0.8 solved for z is your answer. Again, you can look up which value of z satisfies this from a normal probability table. Most probability tables assume a two-sided test, but you are running a one-sided test. So you need to look for the z value that leaves 0.2 probability "to the right." In a two-sided test, that z value will be stated as the z value corresponding to a 0.4 tail probability (0.2 to the left of -z, 0.2 to the right of +z). You are concerned only about the +z.
     
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