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Math Help - Finding standard deviation and mean

  1. #1
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    Finding standard deviation and mean

    I draw a standard normal curve and filled out everything but I couldn't figure it out. Can someone give me some hints?

    ok. so mean and standard deviation are unknown. This guy spends less than one minute doing something about 40% of time and spends more than two minutes doing something 2% of time.

    Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by loadup2 View Post
    I draw a standard normal curve and filled out everything but I couldn't figure it out. Can someone give me some hints?

    ok. so mean and standard deviation are unknown. This guy spends less than one minute doing something about 40% of time and spends more than two minutes doing something 2% of time.

    Any help would be appreciated.
    \Pr(X < 1) = 0.4:

    \Pr(Z < z_1) = 0.4 \Rightarrow z_1 = -0.2533. Therefore:

    Z = \frac{X - \mu}{\sigma} \Rightarrow -0.2533 = \frac{1 - \mu}{\sigma} .... (1)


    \Pr(X > 2) = 0.02:

    \Pr(Z > z_2) = 0.02 \Rightarrow z_2 = 2.054. Therefore:

    Z = \frac{X - \mu}{\sigma} \Rightarrow 2.054 = \frac{2 - \mu}{\sigma} .... (2)


    Solve equations (1) and (2) simultaneously.
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  3. #3
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    Question

    Why is \Pr(Z < z_1) = 0.4 \Rightarrow z_1 = -0.2533?
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  4. #4
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    Quote Originally Posted by kburk46 View Post
    Why is \Pr(Z < z_1) = 0.4 \Rightarrow z_1 = -0.2533?
    Because Pr(Z < -0.2533) = 0.4.
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  5. #5
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    but how did you come up with -.2533?
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  6. #6
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    Quote Originally Posted by kburk46 View Post
    but how did you come up with -.2533?
    Pr(Z < z1) = 0.4 => Pr(Z > -z1) = 0.4 => Pr(Z < -z1) = 0.6 at which point you go to your tables and get the value of -z1 and hence z1.

    Since z1 is known to be negative before any calculation is done (why?) it might be easier to see it as:

    Pr(Z < -a) = 0.4 => Pr(Z > a) = 0.4 => Pr(Z < a) = 0.6 and now go to tables. Then a = 0.2533 => -a = -0.2533.
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