Question

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

Answer 1

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