Ask your own question, for FREE!
Mathematics 22 Online
OpenStudy (anonymous):

The standard normal curve shown below models the population distribution of a random variable. What proportion of the values in the population does not lie between the two z-scores indicated on the diagram? http://media.apexlearning.com/Images/200712/20/394c2d3e-12bc-4812-be69-bb30bb5205f9.gif

OpenStudy (anonymous):

@amistre64

OpenStudy (amistre64):

access denied ....

OpenStudy (anonymous):

sorry! It is a standard z curve, left z score= -1.2, and the right z score= .85. @amistre64

OpenStudy (kropot72):

The proportion lying to the left of z = -1.2 can be found from a standard normal distribution table.. Can you do this step?

OpenStudy (kropot72):

@topoftheworld24 Are you there?

OpenStudy (dumbcow):

this is table http://en.wikipedia.org/wiki/Standard_normal_table#Cumulative_table

OpenStudy (dumbcow):

Also P(Z>z) = 1- P(Z<z) assuming z is positive P(Z<-z) = 1-P(Z<z)

OpenStudy (kropot72):

The table here has negative values of z: http://lilt.ilstu.edu/dasacke/eco148/ztable.htm

OpenStudy (anonymous):

i got the table but i didnt know what to do after

OpenStudy (kropot72):

If you are using this table, http://lilt.ilstu.edu/dasacke/eco148/ztable.htm , what value do you get when you look in the column headed '.00' to the right of the z value -1.2 ?

OpenStudy (kropot72):

@topoftheworld24 How are you going with the table?

OpenStudy (anonymous):

.1151

OpenStudy (kropot72):

Good work! Now find the probability value for z = 0.85 (lets call it p). Then the proportion of values to the right of z = 0.85 (call it p1) is given by: p1 = 1 - p. Finally the required proportion of values is: 0.1151 + p1

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!