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  1. #1
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    backwards Normal distribution problem

    I feel like I should be able to do this but I feel absolutely lost.


    Assume that the weight, X, of male college students is normally distributed. Assume further that P(X <= 160) = 1/2 and P(X <= 140) = 1/4.

    a.) What find the value of 'mu' and 'sigma'
    b.) Calculate P(X >= 190)
    c.) Calculate what percentage of students who weigh over 190 pounds will weigh more than 200 pounds.



    I think b and c should be simple enough once I have mu and sigma but how do I get those? Any help would be appreciated.
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    Quote Originally Posted by BklynKid View Post
    I feel like I should be able to do this but I feel absolutely lost.


    Assume that the weight, X, of male college students is normally distributed. Assume further that P(X <= 160) = 1/2 and P(X <= 140) = 1/4.

    a.) What find the value of 'mu' and 'sigma'
    b.) Calculate P(X >= 190)
    c.) Calculate what percentage of students who weigh over 190 pounds will weigh more than 200 pounds.



    I think b and c should be simple enough once I have mu and sigma but how do I get those? Any help would be appreciated.
    P(X <= 160) = 1/2 tells you straight away that \mu = 160.

    To get \sigma:

    Z = \frac{X-\mu}{\sigma} \Rightarrow z_{140} = \frac{140 - 160}{\sigma} = \frac{-20}{\sigma}

    where \Pr(Z < z_{140}) = \frac{1}{4} \Rightarrow z_{140} = -0.6745 (from inverse tables, correct to four decimal places).

    Therefore -0.6745 = \frac{-20}{\sigma} \Rightarrow \sigma = .......


    The slightly more difficult question would have been if they'd given you P(X <= 200) = 0.9113 and P(X <= 140) = 1/4.

    Then you'd need to solve the following simultaneous equations:

    z_{200} = \frac{200 - \mu}{\sigma} .... (1)

    z_{140} = \frac{140 - \mu}{\sigma} .... (2)

    for \mu and \sigma ......
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    Wow that's amazing, thanks! Yea that makes sense! I was able to do part (b) just fine using that.


    A few questions though, what is the reason behind \mu = 160 ? Just because the probability is 1/2 that makes it an automatic?
    And how would I then approach part (c) of this problem when P(X >= 190) = 0.154


    Again, thank you.
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    Quote Originally Posted by BklynKid View Post
    Wow that's amazing, thanks! Yea that makes sense! I was able to do part (b) just fine using that.


    A few questions though, what is the reason behind \mu = 160 ? Just because the probability is 1/2 that makes it an automatic?
    And how would I then approach part (c) of this problem when P(X >= 190) = 0.154


    Again, thank you.
    The normal distribution is a symmetric around its mean. It's always true for a normal distribution that \Pr(X \geq \mu) = \Pr(X \leq \mu) = \frac{1}{2}.

    c.) Conditional probability: You want \Pr(X > 200 | X > 190) = \frac{\Pr(X > 200 \, \text{and} \, X > 190)}{\Pr(X > 190)} = \frac{\Pr(X > 200)}{\Pr(X > 190)} = .....
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    Thumbs up

    Yea that makes sense also. Thanks a ton!




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