Question

A population is distributed normally with a mean of 110 and a standard deviation of 30.

a) P(X > 130)

b) P( 70 < X < 130)

Answer 1

This is a normal distribution question with

a) P(x > 130.0)=?
The z-score at x = 130.0 is,

z = 0.6667
This implies that
P(x > 130.0) = P(z > 0.6667) = 1 - 0.7475181106016127

b) P(70.0 < x < 130.0)=?

This implies that
P(70.0 < x < 130.0) = P(-1.3333 < z < 0.6667) = P(Z < 0.6667) - P(Z < -1.3333)
P(70.0 < x < 130.0) = 0.7475181106016127 - 0.09121668684984685

PS: you have to refer z score table to find the final probabilities.
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