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Given the length percentiles data the WHO has published for girls. That's length in cm at for certain months. e.g. at birth the 50% percentile is 49.1 cm.

Month   L   M   S   SD  P01 P1  P3  P5  P10 P15 P25 P50 P75 P85 P90 P95 P97 P99 P999
0   1   49.1477 0.0379  1.8627  43.4    44.8    45.6    46.1    46.8    47.2    47.9    49.1    50.4    51.1    51.5    52.2    52.7    53.5    54.9
1   1   53.6872 0.0364  1.9542  47.6    49.1    50  50.5    51.2    51.7    52.4    53.7    55  55.7    56.2    56.9    57.4    58.2    59.7
2   1   57.0673 0.03568 2.0362  50.8    52.3    53.2    53.7    54.5    55  55.7    57.1    58.4    59.2    59.7    60.4    60.9    61.8    63.4
3   1   59.8029 0.0352  2.1051  53.3    54.9    55.8    56.3    57.1    57.6    58.4    59.8    61.2    62  62.5    63.3    63.8    64.7    66.3

P01 is the 0.1% percentile, P1 the 1% percentile and P50 is the 50% percentile.

Say, I have a certain (potentially fractional) month, say 2.3 months. (a height measurement would be done at a certain number of days after birth and you can divide that by 30.4375 to get a fractional month)

How would I go about approximating the percentile for a specific height at a fraction month? i.e. instead of just seeing it "next to P50", to say, well that's about "P62"

One approach I thought of would be to do a linear interpolation, first between month 2 and month 3 between all fixed percentile values. And then do a linear interpolation between P50 and P75 (or those two percentiles for which there is data) values of those time-interpolated values.

What I fear is that because this is a bell curve the linear values near the middle might be too far off to be useful.

So I am thinking, is there some formula, e.g. a quad curve that you could use with the fixed percentile values and then get an exact value on this curve for a given measurement?

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  • $\begingroup$ This bell curve is a normal distribution, and I suppose there is a formula by which you can get values on the curve. The temporal interpolation can probably still be done linear without causing much distortion. $\endgroup$ Oct 19 '20 at 5:22
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I would calculate the mean and standard deviation at each time period. Then perform a linear (or quadratic) fit to each separately, so you know what the mean is at an interpolated time, and a standard deviation at that time. The rest is straightforward.

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  • $\begingroup$ Sorry, I am a programmer and not a mathematician. Please give an example calculation. $\endgroup$ Oct 23 '20 at 3:57
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The row for Month 0 represents $0$ days to $30.4375$ days (per your standard) and Month 1 represents $30.4375$ to $2 \times 30.4375$ days and so on. If an age value falls exactly on the end limit for a range (such as 30.4375), we consider it to be part of that range and not the next one (even though the start value matches exactly).

Now, you should be able to lookup the row and standard percentile figures for a child of certain age. Then use the actual length measurement for that child to determine which percentile she falls under. Both can be linear interpolation. Remember that the percentiles are the expected distribution curve for that age range based on average statistics.

If you need the exact percentile where this data point falls, use the formula for the normal distribution curve to calculate $x$ by inverting the formula (Hint: take logarithms and solve for $x$):

$$y = {1 \over \sigma \sqrt {2 \pi} }e^{-{(x - \mu)^2 \over 2 \sigma^2}}$$

where, $\mu$ is the mean and $\sigma$ is the standard deviation.

You have $(x_0, y_0)$ for the known percentile start and $(x_1, y_1)$ for the known percentile end. Then use linear interpolation to find the exact percentile as you know the bounds of the percentile in which you have determined it to fall using the two lookups.

$$y - y_0 = {(x_1 - x_0) \over (y_1 - y_0)}{(x - x_0)}$$

WHO standard graph

The procedure is graphically displayed above. For a given age, we take the closest percentile curves and then take the two points on the respective curves. X-axis is age and Y-axis is the secondary Y-axis on the graph (ie., percentile) These are the $(x_0, y_0), (x_1, y_1)$. We then linearly interpolate between these points to obtain the linear equation between age and percentile. Then we substitute the actual length measurement to obtain the actual percentile.

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  • $\begingroup$ I agree with linear interpolation for the age, but not that linear interpolation would work reasonably for the percentile on a normal distribution curve. $\endgroup$ Oct 24 '20 at 9:06
  • $\begingroup$ If you look at the percentile values for Month 0: $43.4, 44.8, 45.6, 46.1, 46.8, 47.2 , 47.9,, \dots$, where would you place a length measurement of $43.5$? It requires only a simple range check. $P_0 \lt 43.4, 43.4 \lt P_1 \le 44.8, \dots$. So, you assign $P_1$. for the measurement 43.5. Why is that not reasonable? $\endgroup$
    – vvg
    Oct 24 '20 at 9:49
  • $\begingroup$ It is. It sufficient because the values are distinct percentile values. I.e 50% but I want to have an approximate percentile value in between those Percentile values. Also percentiles values are not equally spaced and this a linear interpolation would be too inaccurate for my purpose $\endgroup$ Oct 25 '20 at 10:09
  • $\begingroup$ I see. I have edited my answer to handle this. $\endgroup$
    – vvg
    Oct 25 '20 at 19:15
  • $\begingroup$ Sorry, I am am so hopeless mathematically challenged... I am going to accept your answer and give you the bounty... but could you spell out the "hint: take logarithms and solve for x" - also I don't understand the y, y0, x0, y1, x1 in the second formula.... $\endgroup$ Oct 26 '20 at 20:08
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I worked through the question based on two examples. The first was my older daughter which was initially quite long/tall.

Girl Age 49 days, 60 cm divide by 30.4375 = 1.61 months

So that's between month 1 and month 2:

Month  P01 P1  P3  P5  P10 P15 P25 P50 P75 P85 P90 P95 P97 P99 P999
1   47.6    49.1    50  50.5    51.2    51.7    52.4    53.7    55  55.7    56.2    56.9    57.4    58.2    59.7
2   50.8    52.3    53.2    53.7    54.5    55  55.7    57.1    58.4    59.2    59.7    60.4    60.9    61.8    63.4

subtract the lower month: 1.61 - 1 = 0.61

So the value is 61% the way to month 2.

I would get a percentile line for this by linear interpolation

For each percentile I can interpolate values from the month row before and after it.

e.g. for P01

p1 = 47.6, p2 = 50.8

P01 = p1 * (1.0 - 0.61) + p2 * (0.61) P01 = 18.564 + 30.988 = 49,552

Month   P01 P1  P3  P5  P10 P15 P25 P50 P75 P85 P90 P95 P97 P99 P999
1   47.6    49.1    50.0    50.5    51.2    51.7    52.4    53.7    55.0    55.7    56.2    56.9    57.4    58.2    59.7
2   50.8    52.3    53.2    53.7    54.5    55.0    55.7    57.1    58.4    59.2    59.7    60.4    60.9    61.8    63.4
1.6 49.552  51.052  51.952  52.452  53.213  53.713  54.413  55.774  57.074  57.835  58.335  59.035  59.535  60.396  61.957

60 cm is between 59,535 (P97) and 60,396 (P99).
0.465 away from the lower, 0.396 away from the higher value. 0.465 is 54% of the distance between them (0,861)

P = (1-0.54) * 97 + 0.54 * 99 = 44.62 + 53.46 = 98,08 Rounded P98

Turns out that this is a bad example. At the extremes the percentiles are very closely spaced so that linear interpolation would give similar results. But my problem is that linear interpolation in the middle is inaccurate. Let’s do a better example. This time with my second daughter who was more in the „middle of the road“ after birth.

Girl Age 119 days, 60.5 cm
divide by 30.4375 = 3.91 months - so we interpolate between month 3 and month 4:

Month   P01 P1  P3  P5  P10 P15 P25 P50 P75 P85 P90 P95 P97 P99 P999
3   53.3    54.9    55.8    56.3    57.1    57.6    58.4    59.8    61.2    62.0    62.5    63.3    63.8    64.7    66.3
4   55.4    57.1    58.0    58.5    59.3    59.8    60.6    62.1    63.5    64.3    64.9    65.7    66.2    67.1    68.8
3.91    55.211  56.902  57.802  58.302  59.102  59.602  60.402  61.893  63.293  64.093  64.684  65.484  65.984  66.884  68.575
        

60.5 cm is between 60.402 (P25) and 61.893 (P50) 0.098 of the distance 1.491 = 6.6%

P = 25 * (1-0.066) + 50 * 0.066 = 23.35 + 3.3 = 26.65
round up to P27

To compare that to approximating it on a bell curve, I used an online calculator/plotter:

This needed a mean and a standard deviation, which I think I found on the percentile table left-most columns. But I also need to interpolate these for month 3.91:

Month   L   M   S   SD
    3    1.0 59.8029 0.0352  2.1051
4   1.0 62.0899 0.03486 2.1645
3.91    1.0 61.88407    0.0348906   2.159154
    


I have no idea what L and S mean, but M probably means MEAN and SD probably means Standard Deviation.

Plugging those into the online plotter…

μ = 61.88407
σ = 2.159154
x = 60.5

The online plotter gives me a result of P(X < x) = P26

This is far enough from the P27 I arrived at by linear interpolation, to warrant a more accurate approach.

I searched around a bit and stumbled across a great explanation of z-Scores.

Z-Score is the number of standard deviation from the mean that a certain data point is.

(x - M) / SD = -0.651

Then I was able to convert that into a percentile by consulting a z-score table.

Looking up -0.6 on the left side vertically and then 0.05 horizontally I get to 0.25785

So that rounds to be also P26.

So that’s one workable approach, although it would require me to implement these z-Tables so that I can implement it in an app. I found a Swift package that offers multiple statistics functions.

The function for „Normal distribution“ is described as

Returns the normal distribution for the given values of x, μ and σ. The returned value is the area under the normal curve to the left of the value x.

I tried it out for the second example, to see what result I would get for this value between P25 and P50:

let y = Sigma.normalDistribution(x: 60, μ: 55.749061, σ: 2.00422)
// result 0.2607534748851712

That seems very close enough to P26. It is different than the value from the z-tables, 0.25785 but it rounds to the same integer percentile.

For the first example, between P97 and P99, we also get within rounding distance of P98.

let y = Sigma.normalDistribution(x: 60, μ: 55.749061, σ: 2.00422)
// result 0.9830388548349042
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