z = (x - μ) / σ
Substituting these values into the formula, we can calculate the z-score for x = 0:
z = (0 - 50) / 10
z = -50 / 10
z = -5
The z-score of -5 indicates that the value x = 0 is 5 standard deviations below the mean. In a standard normal distribution (mean = 0, standard deviation = 1), such an extreme z-score would be considered highly unusual