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  1. #1
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    Probability question help

    Q#1. The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

    (a) at least 10 survive
    (b) from 3 to 8 survive
    (c) exactly 5 survive

    Q#2. Given a normal distribution with population Mean u=40 and , Standard Deviation= 6 find
    (a) the area below 32
    (b) the area above 27
    (c) the area between 42 and 51
    (d) the x value that has 45% of the area below it
    (e) the x value that has 13% of the area above it
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nisoo-1 View Post
    Q#1. The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:

    (a) at least 10 survive
    (b) from 3 to 8 survive
    (c) exactly 5 survive

    Q#2. Given a normal distribution with population Mean u=40 and , Standard Deviation= 6 find
    (a) the area below 32
    (b) the area above 27
    (c) the area between 42 and 51
    (d) the x value that has 45% of the area below it
    (e) the x value that has 13% of the area above it
    your first question was asked by another user. it was answered here

    (a) and (b) of your second question was answered here

    This is my 22th post!!!!!
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  3. #3
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    Quote Originally Posted by nisoo-1 View Post
    Q#1. The probability that a patient recovers from a rare blood disease is 0.4. If 15 people are known to have contracted this disease, what is the probability that:
    Using the Hackerian Gamma Approximation techiqnue
    That \sum_{k=a}^b {n\choose k} p^k (1-p)^{n-k} \approx \int_a^b \frac{n!}{\Gamma (n-t) \Gamma (t-1)} p^t (1-p)^{n-t} dt
    (a) at least 10 survive
    \sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0195
    (b) from 3 to 8 survive
    \sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .797
    (c) exactly 5 survive
    This is easy,
    {15\choose 5}(.4)^ 5 (.6)^{10}

    Note: These problems were done by numerical integration.
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  4. #4
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    Quote Originally Posted by ThePerfectHacker View Post
    Using the Hackerian Gamma Approximation techiqnue
    That \sum_{k=a}^b {n\choose k} p^k (1-p)^{n-k} \approx \int_a^b \frac{n!}{\Gamma (n-t) \Gamma (t-1)} p^t (1-p)^{n-t} dt

    \sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0195

    \sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .797

    This is easy,
    {15\choose 5}(.4)^ 5 (.6)^{10}

    Note: These problems were done by numerical integration.
    Which in this case is not terribly accurate:

    \sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0175

    \sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .878

    RonL
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  5. #5
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    Quote Originally Posted by CaptainBlank View Post
    Which in this case is not terribly accurate:

    \sum_{k=10}^{15} {15\choose k}(.4)^k (.6)^{15-k} = .0175

    \sum_{3}^8 {15\choose k} (.4)^k (.6)^{15-k} = .878

    RonL
    I think I maybe made a mistake in the algorithm programmed in the computer. The other times I used it, it worked well.
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