Assuming that your data are more or less normally distributed, you can convert to z scores as follows:

For each case, subtract the mean score for the sample from the score obtained by that individual. Then divide by the standard deviation for the sample. You can probably use the calculator at http://vassarstats.net/standard.html to make this happen quickly & easily.

The z score, of course, has a mean of 0 and a standard deviation of 1. To transform these into Wechsler-type scaled scores (mean of 10, SD of 3), I think you'd simply multiple by 3 and then add 10.

If the distribution is badly skewed or bimodal, you can still use this approach but you will not have the "expected" number of cases with each possible score. I haven't seen this issue discussed with regard to the Wechsler scaled scores, but it crops up in the literature on T Scores (more common in personality tests). The standard scores you get using the basic formula above preserve the original distribution of the data. If you want to force your scores to fall along a normal curve, there are extra steps to take. (The first approach yields a linear T Score; the second, a uniform or normalized T Score.) Erik Mortensen of the U. of Copenhagen (who is on ResearchGate) wrote a paper on normalized T scores, and Auke Tellegen (U. of Minnesota, now retired) wrote one on uniform T Scores, both in the 1990s.

Usually a normal rank transformation is done to transform the data to normality and retrieve the according norm values. A simple z transformation often deviates to much, esp. in the extreme sections of the distribution (the place we are most interested in psychological research). I would rather suggest to be very very cautious with this and usually the normal rank transformation is the better option. However, if the deviation from normality is too pronounced, none of these approaches really work. It can be a good option just to use percentile ranks as a form of distribution free measure for the probability of the outcome.

P.S.: An online calculator for the rank transformation: https://www.psychometrica.de/norming.html

Thank you very much for your valuable answers!