Consider a normal distribution with mean 30 and standard deviation 2. What is the probability that a value selected at random from this distribution is greater than 30

The probability that the value selected at random is greater than 30 will be found as follows:

z-score is given by:

z = (x-mu) / sig

z = (30-30) / 2=0

thus

P (x>30) = 1-P (X<30) = 1-P (z<0) = 1-0.5=0.5

Answer P (x>30) = 0.5