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Mathematics 18 Online
OpenStudy (anonymous):

Find the area under the standard normal curve for the following: (A) P(z > 2.06) (B) P(0 < z < 0.18) (C) P(-1.31 < z < 1.09) @ap statistics

OpenStudy (anonymous):

Yeah, you need a calculator for this. You'll want to use the normalcdf on a ti calculator

OpenStudy (anonymous):

I have been using the one I have TI-30XA. I am looking it up in my book to see if I can figure it out. But I do not get what to start with. I guess I will figure it out. If you can help I would appreciate it. I am having a rough time with these.

OpenStudy (anonymous):

So the probability like this is the area under the normal graph. So probability z > 2.06 means the area under the graph from z=2.06 to z=infinity. normalcdf(a,b) gives the area under the curve form z=a to z=b. So for problem A we want to do normalcdf(2.06,999999).

OpenStudy (anonymous):

This is what I came up with: (A) P(z > 2.06) 2.06=0.4803 P(z> 2.06)=0.5000-0.4803=0.0197 (B) P(0<z<0.18) 0.18=0.0714 P(0<z<0.18)=0.0714 (C) P(-1.31<z<1.09) -1.31=0.4049 1.09=0.3621 P(-1.31<z<1.09)=0.4049 - 0.3621=0.0428

OpenStudy (anonymous):

For the last one you'll want to add them instead of subtracting. Draw a diagram to go with situation in order to help the operations

OpenStudy (anonymous):

because the first part of the question is a negative, right?

OpenStudy (anonymous):

so the last ones answer is 0.7670

OpenStudy (anonymous):

Because the 0.4049 is area from -1.31 to 0 and 0,3621 is area from 0 to 1.09. You want area from -1.31 to 1.09 so you add

OpenStudy (anonymous):

and yes

OpenStudy (anonymous):

Thank you.

OpenStudy (anonymous):

|dw:1320291343807:dw| I usually draw a diagram like this to help me

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