A population of values has a normal distribution with μ = 217.4 and σ = 40.9 .

Question

1. Find the probability that a single randomly selected value is greater than 221.2. Round your answer to four decimal places.  
2. Find the probability that a randomly selected sample of size  has a mean greater than 221.2. Round your answer to four decimal places.  

Answer 1

P(x > a) = P(z > (a - mean)/SD)

where x is N(217.4,40.9) and z is N(0,1)

so

P(x > 221.2) = P(z > (221.2 - 217.4)/40.9) = P(z > 0.09) = P(z < -0.09) = 0.4641

P(M > a) = P(z > (a - mean)/(SD)/sqrt(N))

so

P(M > 221.2) = P(z > (221.2 - 217.4)/(40.9/sqrt(75)) = P(z > 0.8) = P(z < -0.8) = 0.2119