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Mathematics 28 Online
OpenStudy (anonymous):

standard normal variable, find the probability P(z> 0.59)

OpenStudy (amistre64):

1 - z(0.59)

OpenStudy (amistre64):

just gotta table up the z(0.59)

OpenStudy (amistre64):

or negate the z, and find the table value of z(-0.59)

OpenStudy (anonymous):

How do i what z is? thanks

OpenStudy (anonymous):

I will put in the entire problem. the probability that z lies between -2.41 and z and 0 P(z>0.59) thanks

OpenStudy (amistre64):

do you have a calculator, perhaps a ti83? if not, there are ztables written up that can be utilized

OpenStudy (amistre64):

the probability that z lies between -2.41 and 0 is the cumulative probability from -2.41 to 0 of a mean of 0 and sd of 1

OpenStudy (anonymous):

i have a t-83

OpenStudy (amistre64):

under that stats button, you can hit the normalCDF function

OpenStudy (anonymous):

did that

OpenStudy (amistre64):

that function takes 4 arguments: lower, upper, mean, sd

OpenStudy (amistre64):

to find the cumulative probability of z, from -2.41 to 0 normalCDF(-2.41,0,0,1) is the inputs

OpenStudy (amistre64):

with any luck, you should get about .492

OpenStudy (amistre64):

to find the cumulative probability of z > 0.59, its the same function lower is .59 and upper is say 9999 for some arbitrarily large upper value

OpenStudy (anonymous):

ok, got the .492 and used it as the z score but is coming upw thi different answer what is p(z > .59)

OpenStudy (amistre64):

z>.59 is about .2776 from the tables

OpenStudy (anonymous):

ok, thanks you are best

OpenStudy (amistre64):

youre welcome :)

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