Suppose that the body mass index (BMI) of a person, defined as the person\'s bod

Question

Suppose that the body mass index (BMI) of a person, defined as the person's body mass (in kilograms) divided by the square of their height (in meters), follows a lognormal distribution. For a specific population of men between the ages of 25 and 34, BMI is lognormally distributed with parameters 3.215 and = 0.157. a) What is the median BMI for this population? b) Find the 80h percentile of BMI for this population. c) What proportion of men has a BMI in the healthy range, defined as 18.5 to 24.9 kg/m2?

Explanation / Answer

(a) Here BMI is lognormally distributed with parameters = 3.215 and = 0.157

So median BMI = e3.215 = 24.9034

(b) Here 80th percentile means the cumulative probability would be 0.80

[(ln (X) - )/ ] = 0.80

so as per Z - table

[(ln (X) - )/ ] = -1 (0.80 ) = 0.8416

(ln (X) - 3.215 )/0.157 = 0.8416

ln (X) = 3.215 + 0.157 * 0.8416 = 3.3471

X = 28.42

(c) Here,

proportion of men in the health range , 18.5 to 24.9

Pr(18.5 kg/m2 < X < 24.9 kg/m2) = NORMCDF(ln (24.9) ; 3.215; 0.157) - NORMCDF(ln (18.5) ; 3.215; 0.157)

= (Z2) - (Z1)

where is the normal standard cumulative distribution function

Z2 = [ln (24.9) - 3.215] / 0.157 = 0

Z1 = [ln (18.5) - 3.215) / 0.157] = -1.8932

Pr(18.5 kg/m2 < X < 24.9 kg/m2) = (0) - (-1.8932) = 0.5 - 0.0292 = 0.4708

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