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NORMAL DISTRIBUTION - HELP s1 PLEASE

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 :/
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<Z<0.27)=P(Z<0.27)−P(Z<z)=0.54585P(z< Z < 0.27) = P(Z < 0.27) - P(Z < z) = 0.54585

What is P(Z<0.27)P(Z < 0.27) ?? Hence what is P(Z<z)P(Z<z) ?? Hence what is zz??
Reply 2
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MrToodles4
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Original post by RDKGames
Well, P(z<Z<0.27)=P(Z<0.27)−P(Z<z)=0.54585P(z< Z < 0.27) = P(Z < 0.27) - P(Z < z) = 0.54585

What is P(Z<0.27)P(Z < 0.27) ?? Hence what is P(Z<z)P(Z<z) ?? Hence what is zz??


Yes I've gotten that far, I got P(z<-z) = 0.06057... and then??
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
(edited 5 years ago)
Original post by MrToodles4
Yes I've gotten that far, I got P(z<-z) = 0.06057... and then??


Why is it -z?
Reply 5
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MrToodles4
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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
Reply 6
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MrToodles4
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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
(edited 5 years ago)
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 zz value.
Reply 8
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MrToodles4
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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 zz value.


Im assuming you're talking about table 4 in the AQA booklet - there is no value for 0.06057...
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 zz value for 1-(your probability) then take the -ve of that you'll be fine.
Original post by RDKGames
Indeed there isn't. But if you consider the zz 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 :/
Original post by MrToodles4
How would I draw this value on the curve? I'd like to see whats going on :/




Your zz 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>−z)=0.06057P(Z>-z) = 0.06057. This means that P(Z<−z)=1−0.06057P(Z<-z) = 1-0.06057. From this you can use the table to determine what −z-z is, hence you take the -ve then to get your zz value.
Original post by RDKGames


Your zz 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>−z)=0.06057P(Z>-z) = 0.06057. This means that P(Z<−z)=1−0.06057P(Z<-z) = 1-0.06057. From this you can use the table to determine what −z-z is, hence you take the -ve then to get your zz value.


Now that makes perfect sense! :smile: 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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