#### koraal

##### New Member
This really drives me crazy, because i know its simple statistics.

According to the World Health Organization (WHO), a child growth of age by weight should be like this :
http://www.who.int/childgrowth/standards/tab_wfa_boys_p_0_13.txt

you dont have to go into the link, here is a screenshot of that link: if a child age is 0 weeks (just got birth!), and if his weight is for example 2.6 KG we know that he is in the percentile 5% , if his weight is 3 KG, he is in the percentile 25%
and so on

my question is, how can i know the percentilein between?
for example, for a given weight of 3.1 KG, what is the percentile? should be between 25%-50%
How can i know? do we need extra info for that? im not sure what L M S is but i assume that L is lengh , M is median and S --> i dont know

This is not linear equation, this is how the Graph looks like: Thank you
Koral

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#### obh

##### Well-Known Member
"for a given weight of 3.1 KG, what is the percentile?
must be for a specific week, for example for week 0, you can see 3.1KG is between 15% orange line and 50% green line.
The exact value is not linear calculation as the distribution for each week is a normal distribution (bell curve)

#### GretaGarbo

##### Human
The simplest thing would be to do linear interpolation.

LMS stand for Lambda, mu and sigma and was sugested by Cole(1988) and Cole & Green (1992).
Now gamlss is the WHO standard ( look at their home page).

The distribution is (in general not) normally distributed. Suggested distributions are Box-Cox-Cole-Green, Box-Cox-power exponential and Box-Cox-T. Look at their book.

I guess that it would not be impossible to get the original data from WHO and you could rerun the estimation with the gamlss-package in R and calculate the centile you want. But linear interpolation is easier.

#### obh

##### Well-Known Member
Hi Greta Where do you see the distribution is not normally distributed?
Thanks
O

#### koraal

##### New Member
Hi Greta Where do you see the distribution is not normally distributed?
Thanks
O
The simplest thing would be to do linear interpolation.

LMS stand for Lambda, mu and sigma and was sugested by Cole(1988) and Cole & Green (1992).
Now gamlss is the WHO standard ( look at their home page).

The distribution is (in general not) normally distributed. Suggested distributions are Box-Cox-Cole-Green, Box-Cox-power exponential and Box-Cox-T. Look at their book.

I guess that it would not be impossible to get the original data from WHO and you could rerun the estimation with the gamlss-package in R and calculate the centile you want. But linear interpolation is easier.
Ok ... so... for age 0, with 3.1 KG,
how do i calculate the percentile? assuming normal distribution

#### obh

##### Well-Known Member
Per Greta, this is not a normal distribution.(and you can count she is correct )
But I assume using a normal will give a better result than linear, Greta ?

you need to calculate for a specific time.

Example For t=0, x=3.1kg You have two points (x,p): (2.9kg, 0.15), (3.3kg, 0.5)
z=(x-μ)/σ ~ normal
μ=3.3kg
p(2.9-3.3/σ)=0.15

(2.9-3.3)/σ =inv z (0.15) = -1.036433
σ = -0.3859
Z=(3.1-3.3)/-0.3859=0.5182
P( z ≤ -0.5182 ) = 0.302159

just for comparison, if you use linear extrapulation:
0.15+(0.5-0.15)/(3.3-2.9)*(3.1-2.9)=0.15+0.35/0.4*0.2=0.325

#### hlsmith

##### Less is more. Stay pure. Stay poor.
Hey, once upon a time years ago, I tried to recreate the weight percentile or BMI calculator for pediatric patients and finally said it isn't worth the time. There are non-linear relationships and feature engineering if I remember, then add on top of that gender differences. Depending on how much you are planning on using this, I would recommend just using an online calculator where you input your values. Though, if you think you will use this regularly, then go forth and figure out the intricacies.

I haven't looked into the above, but perhaps via the R package you can just score your own values are get at the transformation functions.

#### koraal

##### New Member
Per Greta, this is not a normal distribution.(and you can count she is correct )
But I assume using a normal will give a better result than linear, Greta ?

you need to calculate for a specific time.

Example For t=0, x=3.1kg You have two points (x,p): (2.9kg, 0.15), (3.3kg, 0.5)
z=(x-μ)/σ ~ normal
μ=3.3kg
p(2.9-3.3/σ)=0.15

(2.9-3.3)/σ =inv z (0.15) = -1.036433
σ = -0.3859
Z=(3.1-3.3)/-0.3859=0.5182
P( z ≤ -0.5182 ) = 0.302159

just for comparison, if you use linear extrapulation:
0.15+(0.5-0.15)/(3.3-2.9)*(3.1-2.9)=0.15+0.35/0.4*0.2=0.325
Thanks, this helped alot 