document.write( "Question 364036: If the random z variable is the standard normal score, it is that P(-3 < z < 3) could easily be approximated without referring to a table? Why or Why not?\r
\n" ); document.write( "\n" ); document.write( " I'm not really sure to tell the truth. For pretty every question, I have had to look up the number in the table to get my answer. \r
\n" ); document.write( "\n" ); document.write( "I would appreciate any help given to me. \r
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\n" ); document.write( "\n" ); document.write( "Sam\r
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Algebra.Com's Answer #259661 by robertb(5830) $\"\"$ $\"About$

You can put this solution on YOUR website!
The standard normal curve has a few neat properties that make it convenient to use. (Aside from its being a benchmark for all computations involving any normal distribution.) Among them are its zero mean and standard deviation of 1. There is the 68-95-99.7 rule, which says that 68% of the observations fall between -1 and 1 (within 1 standard deviation of the mean of 0), 95% fall between -2 and 2 (within 2 standard deviations of the mean) and 99.7% fall between -3 and 3 (within 3 standard deviations of the mean). Thus, P(-1 < z <1) = 0.68, P(-2 < z< 2) = 0.95, and P(-3 < z <3) = 0.997. You don't need a standard normal table for these. =) \n" ); document.write( "

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