Question 856151
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Hi,
mean = 75, SD = 4
P(65 &#8804; x &#8804; 90) Using TI: The syntax is normalcdf(smaller, larger, µ, &#963;).
P(65 &#8804; x &#8804; 90)= normalcdf(65,90,75,4)

Or using z-score
P(65 &#8804; x &#8804; 90) = NORMSDIST(15/4) - NORMSDIST(-10/4)

What test score gives the 95th percentile? |NORMSINV(.05/2)| = 1.96
 1.96 = (X-75)/4
4*1.96 + 75 = 82.84, test score of 83 gives the 95th percentile