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Homework Help: Binomial distribution of coin tosses

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    1. A fair coin is tossed 100 times.
    (a) Find an approximate probability of getting at least 60 heads.

    (b) Find an approximate probability of getting exactly 60 heads.

    3. The attempt at a solution
    part b) would be b(60;100,.5)

    part a) we would need the table for the cumulative distribution, but this is a practice question for my midterm where I won't have access to the tables. Is there a way I can find this with the formula on the test?
  2. jcsd
  3. Oct 19, 2015 #2
    why don't you try approximating with a normal distribution
  4. Oct 19, 2015 #3


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    Will you have access to the normal (z) table? You could approximate this as a normal distribution.
    Otherwise, I would look at ways to simplify. Since p=1-p=.5, your terms if you were going to write out the sum of p(60)+p(61) + ... +p(100) would all be 100 C k (.5)^100.
    You will see that those terms shrink pretty quickly, so depending on how accurate you need your answer, you might be able to only use some of the largest terms--maybe the first 10 or so.
    Last option, you could assume that cumulative prob of 50 is .5 +p(50)/2. Then find p(51 - 59). The remaining part is p(at least 60).
  5. Oct 19, 2015 #4

    Ray Vickson

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    Perhaps the tester is looking for a convenient approximation in symbolic form, without necessarily needing a numerical answer. A normal approximation would be the way to go, and specifying correct "z-values" (maybe without being able to actually compute them) would get nearly full marks. Only you and your instructor can know for sure.
  6. Oct 19, 2015 #5
    I will have access to the normal distribution table. For the normal approximation to the binomial distribution

    Z = (X - np)/sqrt(np(1-p)) = (60 - 50)/sqrt(25) = 10/5 = 2

    we want
    1 - F(2)
    looking at the standard normal table for
    z = 2.00 we have F(z) = .9772
    so the answer is
    1 - .9772 = .0228
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