Statistics, Probability, Normal distribution

Question

Given a normal distribution where mu = 72 and sigma = 6, what is the probability that (please show all work and corresponding Z values):

a.) X>68

b.) X<65 or X>75

c.) 10% of the values are less than what X value?

d.) 60% of the values are between what 2 X values (symmetrically distributed around the mean)?

Answer 1

z = (x - mean)/SD = (x - 72)/6

P(x > 68) = P(z > (68-72)/6) = P(z > -0.67) = 1 - P(z < -0.67) = 1 - 0.2514 = 0.7486

P(x < 65) or P(x > 75) = P(x < 65) + P(x > 75) = P(z < (65-72)/6) + P(z > (75-72)/6) =

P(z <-1.167) + P(z > 0.5) = P(z < -1.167) + 1 - P(z < 0.5) = 0.122 + 1 - 0.6915 = 0.4305

z-score corresponding to 10th percentile = -1.28

-1.28 = (x - 72)/6

x = 64.32

central 60% includes data between 20th and 80th percentile

z-score corresponding to 80th percentile = 0.84

0.84 = (x - 72)/6

x = 77.04

z-score corresponding to 20th percentile = -0.84

-0.84 = (x - 72)/6

x = 66.96

66.96, 77.04