Question 844302
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Hi,
using a TI Calculator
The syntax is normalcdf(smaller, larger, µ, &#963;)
mean age normalcdf(73, 9999, 75,6)= .6305   0r 63.05%
Note: The 9999 is used as the larger value to be at least 5 standard deviations from the mean.

Or  z = {{{(73-75)/6 = -1/3}}}
  P(z&#8804; -1/3) =  .3695   and 1 - .3695 = .6305