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    There are two questions bothering me on this exercise:

    Find the value of Z, given each of the following probabilities;

    P(z<Z<0.27) = 0.54585

    I have no idea how to go about this :/ I tried drawing it but I can't see it.. Any help is greatly appreciated.

    P(z<Z<-1.25) = 0.09493

    I just don't see how to picture this :/
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    (Original post by MrToodles4)
    There are two questions bothering me on this exercise:

    Find the value of Z, given each of the following probabilities;

    P(z<Z<0.27) = 0.54585

    I have no idea how to go about this :/ I tried drawing it but I can't see it.. Any help is greatly appreciated.

    P(z<Z<-1.25) = 0.09493

    I just don't see how to picture this :/
    Well, P(z&lt; Z &lt; 0.27) = P(Z &lt; 0.27) - P(Z &lt; z) = 0.54585

    What is P(Z &lt; 0.27) ?? Hence what is P(Z&lt;z) ?? Hence what is z??
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    (Original post by RDKGames)
    Well, P(z&lt; Z &lt; 0.27) = P(Z &lt; 0.27) - P(Z &lt; z) = 0.54585

    What is P(Z &lt; 0.27) ?? Hence what is P(Z&lt;z) ?? Hence what is z??
    Yes I've gotten that far, I got P(z<-z) = 0.06057... and then??
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    so you have the standardised normal curve with Z along the horizontal axis with z = 0 in the middle.

    from the tables you find that when z = 0.27 the area to the left is 0.60642
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    (Original post by MrToodles4)
    Yes I've gotten that far, I got P(z<-z) = 0.06057... and then??
    Why is it -z?
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    (Original post by the bear)
    so you have the standardised normal curve with Z along the horizontal axis with z = 0 in the middle.

    from the tables you find that when z = 0.27 the area to the left is 0.60642
    Yes I have gathered that
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    (Original post by RDKGames)
    Why is it -z?
    My mistake sorry..
    So P(z<Z) = 0.05057,,, I don't understand the next step
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    (Original post by MrToodles4)
    My mistake sorry..
    So then use the inverse table in your formula booklet to go back from a probability to a z value.
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    (Original post by RDKGames)
    So then use the inverse table in your formula booklet to go back from a probability to a z value.
    Im assuming you're talking about table 4 in the AQA booklet - there is no value for 0.06057...
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    (Original post by MrToodles4)
    Im assuming you're talking about table 4 in the AQA booklet - there is no value for 0.06057...
    Indeed there isn't. But if you consider the z value for 1-(your probability) then take the -ve of that you'll be fine.
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    (Original post by RDKGames)
    Indeed there isn't. But if you consider the z value for 1-(your probability) then take the -ve of that you'll be fine.
    How would I draw this value on the curve? I'd like to see whats going on :/
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    (Original post by MrToodles4)
    How would I draw this value on the curve? I'd like to see whats going on :/


    Your z value is the red one and the red shaded area is what you ended up with (0.06057). Note that due to symmetry you get that the black area is the same as well. It can be concluded that P(Z&gt;-z) = 0.06057. This means that P(Z&lt;-z) = 1-0.06057. From this you can use the table to determine what -z is, hence you take the -ve then to get your z value.
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    (Original post by RDKGames)


    Your z value is the red one and the red shaded area is what you ended up with (0.06057). Note that due to symmetry you get that the black area is the same as well. It can be concluded that P(Z&gt;-z) = 0.06057. This means that P(Z&lt;-z) = 1-0.06057. From this you can use the table to determine what -z is, hence you take the -ve then to get your z value.
    Now that makes perfect sense! I got a bit confused and wrote that red Z on the left as a negative (as I mentioned above) - thanks a lot.

    I'll attempt the second question I had again now.
 
 
 
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