document.write( "Question 856151: a statistics exam has test scored with a normal distribution. The mean score was a 75 with a standard deviation of 4.\r
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Algebra.Com's Answer #515783 by ewatrrr(24785)\"\" \"About 
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document.write( "Hi,
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document.write( "mean = 75, SD = 4
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document.write( "P(65 ≤ x ≤ 90) Using TI: The syntax is normalcdf(smaller, larger, µ, σ).
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document.write( "P(65 ≤ x ≤ 90)= normalcdf(65,90,75,4)\r
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document.write( "Or using z-score
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document.write( "P(65 ≤ x ≤ 90) = NORMSDIST(15/4) - NORMSDIST(-10/4)\r
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document.write( "What test score gives the 95th percentile? |NORMSINV(.05/2)| = 1.96
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document.write( " 1.96 = (X-75)/4
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document.write( "4*1.96 + 75 = 82.84, test score of 83 gives the 95th percentile \n" );
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