If n = 100 and p = 0.02 in a binomial experiment, does this satisfy the rule for a normal approximation? Why or why not?
No. The expectation value is n*p = 2. The only lower possibile results are 1 and 0. There are many possible larger number of "successes" after 100 trials. The distribution will be skewed to larger numbers than 2.
One rule for validity of a normal distributuion is that both x=np and n(1 − p) must be greater than 5. That condition is not met here.