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Mathematics 20 Online
OpenStudy (anonymous):

Find the probabilities associated with the standard normal random variable z P(z>5)

OpenStudy (kirbykirby):

The probability is going to be very close to 0 (essentially could be considered 0 in practice) I don't think most standard normal tables show values beyond z=4, and so for z>4, you have less than a 0.0001 probabiliy, so expect even a lower probability for z>5

OpenStudy (anonymous):

thats what i assumed , but what would i write for my answer?

OpenStudy (kirbykirby):

I think writing \(\approx 0\) should be sufficient. My standard normal table shows values for \(P(Z \le z)\) for z at 3.99 as the highest value. and there it shows 1.0000, meaning that \(P(Z > 3.99)\), to 4 digits, would be 0.0000 ... so ya .. 0 should be good enough

OpenStudy (kirbykirby):

unless you have a calculator that can spit out the exact value.. which would be super small anyway

OpenStudy (anonymous):

i do, but how would i enter the data in the calculator?

OpenStudy (kirbykirby):

hmm I am not sure .. I never had a fancy calculator.. just a basic scientific one. Um, you can enter this on WolframAlpha though: http://www.wolframalpha.com/input/?i=P%28Z%3E5%29%2C+Z~N%280%2C1%29

OpenStudy (anonymous):

i dont understand how to enter it though ):

OpenStudy (kirbykirby):

on Wolfram? or on the calculator

OpenStudy (anonymous):

either /= we havent used calculators for this

OpenStudy (kirbykirby):

Wolfram shows you the value on the site. I already put the input. The value is the number that popped up.. \(2.88 \times 10^{-7}\) For your calculator, I am not sure... But if you only used tables to do find the probabilities, then ya just approximately 0 should be more than sufficient!

OpenStudy (anonymous):

oh! alright! thank you :)

OpenStudy (kirbykirby):

=]

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