This is a worked example of calculating (estimating) the lean body mass for a hypothetical patient:
- ht: 125 cm
- wt: 23kg
- gender: 'f'
- age: 7yrs
The formula for lean body mass(girls) is:
Equation 1
\[
ln(LBM) = -3.8345 + 0.954 \times \ln(height) + 0.6515 \times \ln(weight) - 0.0102 \times (BMI z-score)^2
\]
Looks pretty straight forward— except for getting a z-score for the body mass index (BMI). Lets do that first.
\[
BMI = \frac{wt}{ht^2}
\]
\[
BMI = \frac{23}{1.25^2}
\]
\[
BMI = \frac{23}{1.5625}
\]
\[
BMI = 14.72
\]
Next, we need the equation for the BMI z-score and the appropriate
data from the CDC's published tables:
Equation 2
\[
Z = \frac{((\frac{BMI}{M})^{L}) - 1}{L\cdot S}
\]
and the age-and-gender-specific L, M, S data:
\[
age (months) = 12 \times 7
\]
\[
age (months) = 84
\]
\[
L = -2.93, M= 15.45, S = 0.11
\]
So the full BMI z-score equation is now:
\[
Z = \frac{((\frac{14.72}{15.45})^{-2.93}) - 1}{-2.93\cdot 0.11}
\]
\[
Z = \frac{(0.9528^{-2.93}) - 1}{-0.3223}
\]
\[
Z = \frac{(1.1522) - 1}{-0.3223}
\]
\[
Z = \frac{0.1522}{-0.3223}
\]
\[
Z = \frac{0.1522}{-0.3223}
\]
\[
Z = -0.4722
\]
Back to equation 1:
\[
ln(LBM) = -3.8345 + 0.954 \times \ln(height) + 0.6515 \times \ln(weight) - 0.0102 \times (BMI z-score)^2
\]
and now substituting in our values:
\[
ln(LBM) = -3.8345 + 0.954 \times \ln(125) + 0.6515 \times \ln(23) - 0.0102 \times (-0.4722)^2
\]
\[
ln(LBM) = -3.8345 + 0.954 \times 4.8283 + 0.6515 \times 3.1355 - 0.0102 \times (-0.4722)^2
\]
\[
ln(LBM) = -3.8345 + 4.6062 + 2.0428 - 0.0102 \times 0.2230
\]
\[
ln(LBM) = -3.8345 + 4.6062 + 2.0428 - 0.0023
\]
\[
ln(LBM) = 0.7717 + 2.0428 - 0.0023
\]
\[
ln(LBM) = 2.8122
\]
\[
LBM = exp(2.8122)
\]
\[
LBM = 16.6465
\]