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CaptainBlack
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Question by KS, reposted from Yahoo Questions
Since z is used for the variable we may assume that this is a normal distribution question, and that Z is a RV with a standard normal distribution.
In which case we have a problem asking us to do an inverse table look-up in a table of standard normal distribution. These come in two varieties one gives exactly the probability you require the area under the curve from 0 to a, the other gives the area from -infinity to a. In the latter case for you need:
##P(0<z<a)=P(-\infty<z<a) - P(-\infty< z<0) = P(-\infty< z< a) - 1/2##.
So for this type of table we look up:
##P(-\infty< z< a)=0.8554##
The way an inverse table look up is done is to look in the body of the table for the value of the probability and the value of a is then the corresponding value you would have looked up. This is shown in the attachment:
View attachment 85
CB
P(0<z<a) = 0.3554 solution
help!
i have to find a
Since z is used for the variable we may assume that this is a normal distribution question, and that Z is a RV with a standard normal distribution.
In which case we have a problem asking us to do an inverse table look-up in a table of standard normal distribution. These come in two varieties one gives exactly the probability you require the area under the curve from 0 to a, the other gives the area from -infinity to a. In the latter case for you need:
##P(0<z<a)=P(-\infty<z<a) - P(-\infty< z<0) = P(-\infty< z< a) - 1/2##.
So for this type of table we look up:
##P(-\infty< z< a)=0.8554##
The way an inverse table look up is done is to look in the body of the table for the value of the probability and the value of a is then the corresponding value you would have looked up. This is shown in the attachment:
View attachment 85
CB
Attachments
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