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Moivre-Laplace theorem (homework)

  1. Apr 24, 2015 #1
    < Mentor Note -- thread moved to HH from the technical math forums, so no HH Template is shown >

    I have no idea what I am doing wrong. Here is my problem:

    S(n) - the number of successes in 10 Bernoulli trials, each with probability 0.7

    I am supposed to use normal approximation to compute :

    P{|S(n)/10-0.7|>0.2}

    We are allowed to use a calculator.

    I got:
    |S(n)-7|>2|
    5<S(n)>9

    I have computed S*:

    S(n)-7/1.44914
    7=mean
    1.44914- standard deviation

    5*1.44914-7<Z<9*1.44914-7

    Area (probability): 0.2451

    But there is no such answer in the options.
     
    Last edited by a moderator: Apr 24, 2015
  2. jcsd
  3. Apr 24, 2015 #2

    SammyS

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    What do you mean by 5<S(n)>9 ?

    That's an invalid compound inequality.
     
  4. Apr 24, 2015 #3
    I see. :( My line of thought was as follows:

    S(n)-7>2 or S(n)-7<-2 (this is an absolute value equation: |S(n)/10-0.7|>0.2)

    How to correct it?
     
  5. Apr 24, 2015 #4

    SammyS

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    What you have here is correct.

    S(n)-7>2 or S(n)-7<-2

    That says S(n) > 9 or S(n) < 5 . (You had S(n) > 5 )

    You can't make these into a compound inequality.
     
  6. Apr 24, 2015 #5

    Ray Vickson

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    Last edited: Apr 24, 2015
  7. Apr 25, 2015 #6
    Thank you so much. I have tried to work it out. I think I got it. My problem is that I still can't get an exact answer. The closest (with the correction you have suggested):
    if x = 5.5 I get 0.15031188 (I can choose 0.1675)
    http://www.danielsoper.com/statcalc3/calc.aspx?id=53

    The result: 0.08377348 is very imprecise.
    Actually some calculators demand x, some Z. It is a mess.
     
  8. Apr 25, 2015 #7
    Sorry, I get 0.71860226. Even worse.
     
  9. Apr 25, 2015 #8
    I got it at last. :) Many thanks. :) You have helped me a lot. :) :) :)
     
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