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Alice2005
December 28th 2005, 12:18 PM
Hello Everyone.

After consulting 5 different sources, (online sources as well as textbooks) I have come across an inconsistancy.

(This is a question regarding the standard normal distribution)

What is the percentage of the area covered by two standard deviations of the mean. Or What is P( mean + or - 2(standard deivation)). A few sources say that the answer is 95% while others say it is 96%.

When I calculate it, I get a .9544

However in doing a problem, I assumed P(-2<z<2)= 95% and got the problem completely wrong.

SirVLCIV
December 28th 2005, 12:26 PM
Use .9544.

urban
January 9th 2006, 09:48 PM
In my AP Stat class, we use the empirical rule (or 68, 95, 99.7 rule) as an approximation to the true values. This is actually cited in some textbooks as the "be all and end all" to solving the problems. Clearly, as n gets large, we can see the problems with this approximation.

This is a good approximation for MC problems in which there is no "close call". I tell them it's a "quick and dirty" for those contrived and obvious problems that some standardized tests are plagued with.

However, the true proportion between (according to my TI-84 :D ):

-1sd and +1sd is 0.6826894809
-2sd and +2sd is 0.954499876
-3sd and +3sd is 0.9973000656

mallkins
January 9th 2006, 10:05 PM
Alice

In my ASM manual Dr. O has put a sentence saying

"It is worth rememberin that the 95th percentile of the standard normal distribution is 1.645, while Pr(-1.96=<Z=<1.96)=0.95."

Try the 1.645, and maybe Krzys' will explain where it comes from.

Miles