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Mathematics 17 Online
OpenStudy (anonymous):

Use a table to find a z-score that fits the given conditions. Interpolate if necessary. 10% of the area under the standard normal curve is above the score. Can anyone explain how to do this? I'm lost.

jimthompson5910 (jim_thompson5910):

sounds like you want to find the value of k such that P(Z > k) = 0.10

jimthompson5910 (jim_thompson5910):

are you allowed to use a calculator?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

yes, I am.

jimthompson5910 (jim_thompson5910):

ok one calculator you can use is wolfram alpha you would use the normalcdf function to get this http://www.wolframalpha.com/input/?i=normalcdf&a=*C.normalcdf-_*Formula.dflt-&f2=0.10&x=0&y=0&f=NormalProbabilities.pr_0.10&a=*FVarOpt.1-_***NormalProbabilities.pr--.***NormalProbabilities.z--.**NormalProbabilities.l-.*NormalProbabilities.r---.*--&a=*FVarOpt.2-_**-.***NormalProbabilities.mu--.**NormalProbabilities.sigma---.**NormalProbabilities.pr--- notice how there is a line on that page that says P(Z > 1.282) = 0.1 so k = 1.282 is the value we're looking for

OpenStudy (anonymous):

Okay that makes sense. I am still a little confused how you figured out how 0.1 = 1.282

jimthompson5910 (jim_thompson5910):

they aren't equal like that

jimthompson5910 (jim_thompson5910):

its more like P(Z > 1.282) = 0.10

jimthompson5910 (jim_thompson5910):

in other words the probability of picking a z-score larger than 1.282 is 0.10 (or 10%)

OpenStudy (anonymous):

okay so lets say another question asks 33% of the area under the standard normal curve is above the score. Then would the answer be -.44 ?

jimthompson5910 (jim_thompson5910):

on that calculator page, you now type in 0.33

jimthompson5910 (jim_thompson5910):

then you look for the line that has P(Z > ???) = 0.33

OpenStudy (anonymous):

okay so it is 0.44 not negative

jimthompson5910 (jim_thompson5910):

yep P(Z > 0.44) = 0.33

OpenStudy (anonymous):

Thank you for your help ! :)

jimthompson5910 (jim_thompson5910):

you're welcome

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