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    #1
    Let $ \displaystyle X$ be normally distributed with $ \displaystyle \mu =100cm$ and $ \displaystyle \sigma =5 cm$

    $ \displaystyle (a$) shade region $ \displaystyle P(X>105)$



    (b) Given that $ \displaystyle P(X<d)=P(X>105)$, find the value of $ \displaystyle d$.

    wasn't sure if this meant that $ \displaystyle d$ is the left of 105 which would be larger in volume than $ \displaystyle X>105$

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    #2
    What volume? The probabilities are areas...

    Anyway, it's asking you to set the two probabilities equal to each other. So if your area was measured from the left instead of from the right, what value on the x axis would you get to? Hint: The areas are equal and symmetrical about the mean.

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    #3 Thread Author
    Quote Originally Posted by Prove It View Post
    What volume? The probabilities are areas...

    Anyway, it's asking you to set the two probabilities equal to each other. So if your area was measured from the left instead of from the right, what value on the x axis would you get to? Hint: The areas are equal and symmetrical about the mean.
    yes area not volume

    so then $ \displaystyle d=95$ if $ \displaystyle p(X<d)$ for the same area as $ \displaystyle P(X>105)$

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    #4
    Quote Originally Posted by karush View Post
    yes area not volume

    so then $ \displaystyle d=95$ if $ \displaystyle p(X<d)$ for the same area as $ \displaystyle P(X>105)$
    Correct

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    #5 Thread Author
    (c) Given that $ \displaystyle P(X>105)=0.16$ (correct to $ \displaystyle 2$ significant figures), find $ \displaystyle P(d<X<105)$

    so that is within $ \displaystyle 68\%$ within
    $ \displaystyle 1$ standard deviation of the mean

    or do just $ \displaystyle (2)0.16 = 0.32$

    not sure??

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    #6
    Quote Originally Posted by karush View Post
    (c) Given that $ \displaystyle P(X>105)=0.16$ (correct to $ \displaystyle 2$ significant figures), find $ \displaystyle P(d<X<105)$

    so that is within $ \displaystyle 68\%$ within
    $ \displaystyle 1$ standard deviation of the mean

    or do just $ \displaystyle (2)0.16 = 0.32$

    not sure??
    Yeah. It's 68% within 1 standard deviation.
    But that means that P(d<X<105)=P(95<X<105)=0.68.

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    #7
    Quote Originally Posted by karush View Post
    (c) Given that $ \displaystyle P(X>105)=0.16$ (correct to $ \displaystyle 2$ significant figures), find $ \displaystyle P(d<X<105)$

    so that is within $ \displaystyle 68\%$ within
    $ \displaystyle 1$ standard deviation of the mean

    or do just $ \displaystyle (2)0.16 = 0.32$

    not sure??
    I would write (for clarity):

    $ \displaystyle P(d<X<105)=P(d<X<100)+P(100<X<105)$

    $ \displaystyle 0.68=P(d<X<100)+\left(0.5-P(X>105) \right)$

    $ \displaystyle 0.68=P(d<X<100)+0.34$

    $ \displaystyle P(d<X<100)=0.34$

    What do you find?

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    #8 Thread Author
    Quote Originally Posted by MarkFL View Post
    I would write (for clarity):

    $ \displaystyle P(d<X<105)=P(d<X<100)+P(100<X<105)$

    $ \displaystyle 0.68=P(d<X<100)+\left(0.5-P(X>105) \right)$

    $ \displaystyle 0.68=P(d<X<100)+0.34$

    $ \displaystyle P(d<X<100)=0.34$

    What do you find?
    i understand what you have here... but don't know how the 0.16 comes into this.

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    #9
    I used:

    $ \displaystyle P(X>105)=0.16$

    in my statement, as :

    $ \displaystyle 0.5-P(X>105)=0.5-0.16=0.34$

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