Normalizing the bounds,
y=186, z = (186-200)/12 = -1.17
y = 210, z = (210 - 200)/12 = 0.83
We want the area of the standard normal between those. Looking it up in the cumulative table,
Ф(1.17) = 0.87900
Ф(0.83) = 0.79673
We need Ф(-1.17) which is 1 - Ф(1.17)
Then the probability we seek is
p = Ф(0.83) - Ф(-1.17) = Ф(0.83) + Ф(1.17) - 1 = 0.87900 + 0.79673 - 1 = 0.67573
Answer: 67.6%
All are CORRECT except for D.57
The variable X is equal to 16
(9-(6-3))x2 hopefully this helps you out!!!
