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Urgent help required!

  1. Sep 11, 2005 #1
    Hi guys.

    There was a question in my SAC which I still have an hour to finish, that I had no Idea about.

    The question was like following.

    5 students graduated from a school which has a apprenticeship probability of .15.

    I answered a) b) and c) which was to find the probability of none getting an apprenticeship and one getting it and at least one gettingit.


    In part D it say using Binomial method find the probability that 15 to 20 students (Inclusive) out of 110 will get apprenticeship.


    Can someone help me plz.

    I have made no progress from getting any info from the question :(

    Cheers!
     
  2. jcsd
  3. Sep 12, 2005 #2

    HallsofIvy

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    Science Advisor

    I suspect your time is up by now but with numbers as large as 110 students, I would be inclined to use the normal approximation. For large N, a binomial distribution with probability of "success" p can be approximated by a normal distribution with mean pN and standard deviation [tex]\sqrt{p(1-p)N}[/tex]. Here p= 0.15 so the normal approximation has mean 16.5 and standard deviation 3.75 (approx). Since the normal distribution is continuous we interpret "between 15 to 20 students" as the interval [14.5, 20.5].
     
  4. Sep 12, 2005 #3

    Can you please clarify how I can get the answer? I took a look at my text bookbut there are many diffrent methods of doing it :(

    I absloutly hate probability so I am abit lost :(

    Thanx for your reply
     
  5. Sep 13, 2005 #4
    anyone knows how I could get an answer?
     
  6. Sep 13, 2005 #5
    can anyone confirm if this is right?
     
  7. Sep 13, 2005 #6
    damnz,

    I accidently deleted the other post
    here is it again :(

    [tex]X ~ N (16.5,14.06) Pr(15 < Z < 20)[/tex]
    [tex]Pr((\frac{15-16.5}{3.75})(\frac{20-16.5}{3.75}))[/tex]
    [tex]Pr(-.4 < Z < .933)[/tex]
    [tex]Pr(Z < .933)- Pr(Z > .4)[/tex]
    [tex]Pr(Z <.933) - (1-Pr(Z < .4))[/tex]
    [tex]1.5-1.25=.25[/tex]

    How does it look? any obvious mistakes?

    i only have another 5 hours left before I have that class again.

    Cheers
     
    Last edited: Sep 13, 2005
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