z = (90-100)/20 = -.5
z table = 0.3085
z = (110-100)/20 = .5
z table = 0.6915
0.6915-0.3085 = 0.3830
D
A normal distribution has a mean of ƒÝ = 100 with ƒã = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 90 and X = 110?
a. 0.6915
b. 0.3085
c. 0.1915
d. 0.3830
1 answer