Question

The monthly utility bills in a city are normally​ distributed, with a mean of ​$100 and a standard deviation of ​$12. Find the probability that a randomly selected utility bill is​ (a) less than ​$65​, ​(b) between ​$84 and ​$100​, and​ (c) more than ​$130.

Answer 1

solution:-

given that mean = 100 , standard deviation = 12

(a) less than ​$65
P(X < 65) = P(Z < (65-100)/12)
= P(Z < -2.92)
= 1−P(Z < 2.92)
= 1-0.9982
= 0.0018

(b) between ​$84 and ​$100
P(84 < X < 100) = P((84-100)/12 < Z < (100-100)/12)
= P(-1.33 < Z < 0)
= P(Z < 0) - P(Z < -1.33)
= 0.5 - 0.0918
= 0.4082

(c) more than ​$130
P(X > 130) = P(Z > (130-100)/12)
= P(Z > 2.5)
= 1-P(Z < 2.5)
= 1-0.9938
= 0.0062

all values from Z standard normal table