SOLUTION: If I have a 5 yr. old boy who has a BMI of 18 and the BMI is considered a normally distributed variable, whose mean for this age boy is 15.41914 and whose standard deviation is 0.0

Question 440446: If I have a 5 yr. old boy who has a BMI of 18 and the BMI is considered a normally distributed variable, whose mean for this age boy is 15.41914 and whose standard deviation is 0.075992. How can I calculate the PERCENTILE ranking for this boy. Please show me HOW you achieve the numbers--I need to see HOW to calculate the result--PLEASE DO NOT USE A TABLE. I want to write an excel program for our school nurse and I need the EQUATION to plug in and test for in excel. Thank you.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
If I have a 5 yr. old boy who has a BMI of 18 and the BMI is considered a normally distributed variable, whose mean for this age boy is 15.41914 and whose standard deviation is 0.075992. How can I calculate the PERCENTILE ranking for this boy. Please show me HOW you achieve the numbers--I need to see HOW to calculate the result.
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1st: Find the z-value of 18
using the equation: z(x) = (x-u)/sigma
z(18) = (18-15.41914)/0.075992 = 2.580860/0.075992 = 33.962259
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Note: z is a measure of the number of standard deviations the
score is above the mean of the distribution.
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By the way, a z-score of 33.96.. is extremely unusual.
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2nd: Find the percent of BMI scores that are less than 18.
That percent is the percent of a normally distributed population
that is less than z = 33.962259
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3rd: Use a Statistical z-chart or an EXCEL function to find
P(x < 18) = P(z < 33.962259) is approximately 100%ile.
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Cheers,
Stan H.
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