Calculate percentile of a raw score from normative data of a psychological test

I have the manual of a psychological test. The raw score (ranging from 0 to 3) has to be converted in the percentile, using the normative data in the table below.

Is there a way to compute the exact percentile of the raw scores, using the table? For example, if I have a row score, on the first dimension, of 1.30, how can I calculate the percentile?

0.01    .4329   .1214   .0429   .4465   .0000
0.025   .4801   .1671   .0660   .5491   .0000
0.05    .5434   .2551   .1029   .6267   .0000
0.10    .6274   .3557   .1400   .7127   .0556
0.25    .8272   .5152   .2757   .8736   .2230
0.33    .9181   .5895   .3657   .9363   .3154
0.50    1.0961  .7562   .5526   1.0668  .5417
0.66    1.2508  .9358   .7592   1.2100  .7959
0.75    1.3387  1.0398  .9007   1.2910  .9663
0.90    1.5761  1.3068  1.2412  1.4927  1.4142
0.95    1.7094  1.4822  1.4918  1.6180  1.6466
0.975   1.8418  1.5962  1.6626  1.7373  1.8685
0.99    1.9556  1.8100  1.8980  1.8714  2.2157


Plotting the first two columns

Assuming normal distribution the quantile q is linked to the value x by

q = 1/2 Erfc[(μ - x)/(Sqrt[2] σ)]

or in Mathematica syntax: q = CDF[NormalDistribution[μ, σ], x]

Taking the mean (media) as 1.0997 and solving for σ for each point

Quantile    Value   Std. Dev.
0.01    0.4329  0.28663
0.025   0.4801  0.316128
0.05    0.5434  0.338206
0.10    0.6274  0.368538
0.25    0.8272  0.404009
0.33    0.9181  0.412809
0.50    1.0961  ComplexInfinity
0.66    1.2508  0.366336
0.75    1.3387  0.354342
0.90    1.5761  0.371737
0.95    1.7094  0.370671
0.975   1.8418  0.378629
0.99    1.9556  0.367916


This shows a large range of standard deviations suggesting the results are not normally distributed.

It may be expedient simply to interpolate from the curve.

• Ok, thanks. I notice that it is a bit intricate to calculate exact percentile. I wonder why in the manual there are not all percentile points... Jan 10, 2018 at 7:57