i need help solving this

Question

A population of values has a normal distribution with  and . You intend to draw a random sample of size . Please show your answers as numbers accurate to 3 decimal places.

Find the probability that a single randomly selected value is greater than 50.3.

P(X > 50.3) =

Find the probability that a sample of size 

 is randomly selected with a mean greater than 50.3.

P( > 50.3) =

Answer 1

Mean = 51.5

SD = 89.2

n=215

x = value

For single value z = (x-mean)/SD

So P(X>50.3) = P(Z> (50.3-51.5)/89.2) = P( Z> -0.1345) = 1-P(Z<0.01345)=1 - 0.49463 = 0.50537

For sample size

P(X > 50.3) = P(z > (50.3 -51.5)/(89.2/sqrt(215)) = P(z > -0.1973) = 1 - P(z < 0.1973) = 1 - 0.4218 = 0.5782