Results 1 to 4 of 4

Math Help - Help!!!! I need help!!!

  1. #1
    Newbie
    Joined
    Jun 2006
    Posts
    16

    Help!!!! I need help!!!

    Past records show that at a given college 20% of the students who began as psychology majors either changed their major or dropped out the school. An incoming class has 110 beginning psychology majors. What is the probability that as many as 30 of these students leave the psychology program?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,678
    Thanks
    611
    Hello, bombo31!

    The problem is not clearly worded . . .

    ast records show that at a given college 20% of the students who began as psychology majors
    either changed their major or dropped out the school.
    An incoming class has 110 beginning psychology majors.
    What is the probability that as many as 30 of these students leave the psychology program?
    The phrase "as many as 30" does not have a standard interpretation.

    Does it mean up to 30?
    . . If so, the probability is unreasonably huge.
    We must find the probabilities that: 0 leave, 1 leaves, 2 leave, 3 leave, 4 leave, 5 leave, 6 leave,
    . . 7 leave, 8 leave, 9 leave, 10 leave, 11 leave, 12 leave, . . . 28 leave, 29 leave, and 30 leave
    . . and add them up.

    If it means exactly 30 leave, it is still an unwieldly number: . C(110,30)(0.2)^{30}(0.8)^{80}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by bombo31
    Past records show that at a given college 20% of the students who began as psychology majors either changed their major or dropped out the school. An incoming class has 110 beginning psychology majors. What is the probability that as many as 30 of these students leave the psychology program?
    Note sure what you mean.

    If exactly 30
    Then, the probability is,
    {110 \choose 30}(.2)^{30}(.8)^{80}\approx .0159

    If at most 30
    Then, the probability is,
    \sum_{k=0}^{30} {110 \choose k}(.2)^k(.8)^{110-k}\approx .9753

    If at least 30
    Then, the probability is,
    \sum_{k=30}^{110} {110\choose k}(.2)^k(.8)^{110-k}\approx .0406


    (by trichtonomy there cannot be any other case)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by bombo31
    Past records show that at a given college 20% of the students who began as psychology majors either changed their major or dropped out the school. An incoming class has 110 beginning psychology majors. What is the probability that as many as 30 of these students leave the psychology program?
    Lets assume as Soroban suggests that the question means to ask what is
    the probability that 30 or more leave the psychology program.

    The number who leave the program is a Binomialy distributed random variable,
    and PerfectHacker has indicated how to calculate the required probability.
    However, with numbers like 30+ from 110 in the question we are usually
    expected to use the Normal approximation to the Binomial Distribution to do
    the calculations.

    This means we treat the number who leave the program as a normal random
    variable with the same mean and standard deviation as the binomial
    distribution.

    Here the mean is:

    \mu=N\times p=110\times 0.2=22,

    and:

    \sigma=\sqrt{N\times p \times (1-p)}=\sqrt{110\times 0.2 \times 0.8}\approx 4.1952.

    Now with the normal approximation when we ask for the probability of M
    success we evaluate the area under the normal distribution between M-0.5
    and M+0.5. So to evaluate the probability of 30 or more drop-outs we
    want to know the probability in the normal approximation of 29.5 or more
    drop-outs.

    The critical z-score corresponding to 29.5 is:

    z_c=\frac{29.5-22}{4.1952}=1.7877

    Looking this up on a standard normal table or calculator tells us that the
    required probability is \approx 0.037 or about 3.7 \%

    Now this is different from PH's result which is correct, but it should
    be slightly different. They do agree to the nearest percentage point.

    RonL
    Last edited by CaptainBlack; June 9th 2006 at 01:04 AM.
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum