2
$\begingroup$

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.

enter image description here

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?

$\endgroup$

1 Answer 1

1
$\begingroup$
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

enter image description here

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.

$\endgroup$
1
  • $\begingroup$ 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... $\endgroup$ Jan 10, 2018 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.