<PRE><B>Question 780945</B>
The number of words in the active vocabulary of children of a particular age is normally distributed with a mean of 3000 and a standard deviation of 500. 
Using the normal curve approximation rules, 
how many words would a child of this age have to know to be 
in the top 2%:::
invNorm(0.98) = 2.0537
x = 2.0537*500 + 3000 = 4026.87
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the top 16%
invNorm(0.84) = 0.9945
x = 0.9945*500+3000 = 3497.23
-----------------------
the top 98%?
invNorm(0.02) = -2.0537
x = -2.0537*500+3000 = 1973.13
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Cheers,
Stan H.
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