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
