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  1. #1
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    Need help with bernoulli trails

    A recent national study showed that approximately 45% of college students binge drink. Let X equal the number of students
    in a random sample of size n = 12 who binge drink. Find the probability that
    (a) X is at most 5
    (b) X is at least 6
    (c) X is equal to 7
    (d) Give the mean, variance, and standard deviation of X

    I tried solving the answers and received the following
    (a) 0.5269
    (b) 0.96615
    (c) 0.148945
    (d) mean = 0.45, variance = 0.2475, standard deviation = 0.4975
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  2. #2
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    Quote Originally Posted by MathRules! View Post
    A recent national study showed that approximately 45% of college students binge drink. Let X equal the number of students
    in a random sample of size n = 12 who binge drink. Find the probability that
    (a) X is at most 5
    (b) X is at least 6
    (c) X is equal to 7
    (d) Give the mean, variance, and standard deviation of X

    I tried solving the answers and received the following
    (a) 0.5269
    (b) 0.96615
    (c) 0.148945
    (d) mean = 0.45, variance = 0.2475, standard deviation = 0.4975
    The following answers are wrong:

    (b) Try again.
    (d) All of it. eg. Mean = np .... how can you get 0.45 from this?
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  3. #3
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    Originally Posted by MathRules!
    A recent national study showed that approximately 45% of college students binge drink. Let X equal the number of students
    in a random sample of size n = 12 who binge drink. Find the probability that
    (a) X is at most 5
    (b) X is at least 6
    (c) X is equal to 7
    (d) Give the mean, variance, and standard deviation of X

    I tried solving the answers and received the following
    (a) 0.5269
    (b) 0.96615
    (c) 0.148945
    (d) mean = 0.45, variance = 0.2475, standard deviation = 0.4975


    Quote Originally Posted by mr fantastic View Post
    The following answers are wrong:

    (b) Try again.
    (d) All of it. eg. Mean = np .... how can you get 0.45 from this?

    Ok, I see my error in d
    (d) So for the mean = np, therefore 12(.45) = 5.4
    variance = np(1-p) = 5.4 (1-.45) = 2.97
    standard deviation = sqrt(2.97) = 1.72

    for b I am still working on it. I still don't see my error.
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  4. #4
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    I think I found the answer to part b.

    b) so when X is at least 6: 1 - P(X >=5) = 1- 0.5269 = 0.4731
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  5. #5
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    Quote Originally Posted by MathRules! View Post
    I think I found the answer to part b.

    b) so when X is at least 6: 1 - P(X >=5) = 1- 0.5269 = 0.4731
    I think you meant to post 1 - \Pr(X {\color{red}\leq} 5) since your answer is correct.
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