<span>The number of cell phone minutes used by high school seniors follows a normal distribution with a mean of 500 and a standard deviation of 50. what is the probability that a student uses more than 580 minutes?
Given μ=500 σ=50 X=580 P(x<X)=Z((580-500)/50)=Z(1.6)=0.9452 => P(x>X)=1-P(x<X)=1-0.9452=0.0548=5.48%
