Probability question

Discussion in 'Probability' started by R H, Nov 30, 2011.

  1. R H

    R H Guest

    I don't know how to calculate the following:

    Out of 67 students in an exam, coming from various schools, 10
    students didn't pass. Out of the 10 that didn't pass, 7 came from a
    particular school that sent 9 students in total. What is the
    probability that the 7 students that failed from the same school were
    just bad students versus the probability that they were not as
    adequetly prepared as the rest who passed the exam?

    Thanks a lot for the help. This is not homework. I am teachning hair
    styling.
     
    R H, Nov 30, 2011
    #1
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  2. R H

    Ray Koopman Guest

    In situations like this it helps to put the numbers in a table:

    particular other
    school schools total

    pass 2 55 57

    fail 7 3 10

    total 9 58 67

    The pass rate for students from the particular school was 2/9 = 22%,
    compared to 55/58 = 95% for students from other schools.

    We can not say what the probability was that the students from the
    particular school were bad or were not as adequately prepared as
    those from other schools. What we can say is how likely it is that
    you would get a difference as big as or bigger than the one you got
    if the students were from a single population, with only random
    differences among them. The easiest way to get that probability
    is with an online calculator such as the one at
    http://www.iancampbell.co.uk/twobytwo/calculator.htm

    For your data it says P < .0001, which means that the chances of
    seeing a difference as big as the one you got if both sets of
    students were from the same population are less than 1 in 10,000. So
    either a very rare random event has occurred, or the assumption that
    the two groups of students were from the same population is wrong
    and there are systematic differences between the two groups.
     
    Ray Koopman, Dec 1, 2011
    #2
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  3. R H

    R H Guest

    Thanks Ray, it makes sense. I admire people like you who can solve
    this type of problems.
     
    R H, Dec 1, 2011
    #3
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