A z-score, also called a standard score, simply rescales a data point by subtracting the mean and dividing by the standard deviation, $z=(x-\mu)/\sigma$. The result tells you the distance from the centre measured in standard deviation units. Applying this transformation to a normal curve produces the standard normal distribution centred at 0 with a spread of 1, which is exactly why the z-score is tied to the normal distribution. The binomial distribution handles discrete success counts and does not rely on the z statistic. The chi square test works on frequency tables, and the t-test uses the t statistic when samples are small or the population variance is unknown. None of these define the z-score, so the normal distribution is the right link. $$\boxed{\text{Normal distribution}}$$