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Normally distributed probability problem

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data
    an exam is normally distributed with a mean of 75 and a variance of 64. students with a mark of 90 or above recieve an A+
    a) what percentage of students recieve an A+?
    b) if 12 students write the exam what is the probability 3 get A+?

    2. Relevant equations
    z-score- z=mark-mean/standard deviation
    my teacher gave us a chart that gives probabilities as area under a curve for normal distributions

    3. The attempt at a solution

    a) since the standard deviation is the square root of the variance:
    SD= 8
    z-score= 90-75/8= 1.88
    now using the area under the curve chart, for a z-score of 1.88 it reads 0.9699. from this i subtract 1 because i want the amount above 90 and this value is below 90, so i get:
    = 1-0.9699
    = 0.0301 or 3.01%

    b) if 12 students write and 3 get an A+ then 9 students must get something else:
    p(x=3)= 3(0.0301)/9(0.9699)= 0.0103
    since the probability of getting over A+ is 0.0301 and anything else is one subtract 0.0301.

    do my answers seen correct? if not can someone guide me in the right direction?
  2. jcsd
  3. Feb 24, 2009 #2
    The answer to part a) is correct, however, I don't really understand what calculation you've done for part b).

    Personally I would just use the Choose function i.e. 12C3 * 0.0301^3 * 0.9699^9
  4. Feb 24, 2009 #3
    Ok, would I also put in 12C9 when multlying0.9699?. Also why areyou putting 0.0301 with an exponent also for the other0.9699?
  5. Feb 24, 2009 #4
    nope you just have one 12C3 - it's how you use the binomial theorem:

    P(X=r) = nCr * p^r * q^(n-r) where q = 1-p :)
  6. Feb 24, 2009 #5
    oh and n = total so in this case 12
  7. Feb 24, 2009 #6
    Alright! Thank you
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