Question

The monthly utility bills in a city are noramlly distributed, with a mean of $100 and a standard deviation of $12. A utility bill is randomly selected,
a. Find the probability that the utility bill is less than $70.
b. Find the probability that the utility bill is between $90 and $120.
c. Find the probability that the utility bill is more than $140.
d. What percent of the utility bills are more than $125?
e. If 300 utility bills are randomly selected, about how many would one expect to be less than $90.

Answer 1

Solution:- Given that mean = 100, standard deviation = 12

a. P(X < 70) = P((X-μ)/σ < (70-100)/12) = P(Z < -2.5) = 0.0062

b. P(90 < X < 120) = P((90-100)/12 < Z < (120-100)/12) = P(-0.8333 < Z < 1.6667) = 0.7492

c. P(X > 140) = P(Z > (140-100)/12) = P(Z > 3.3333) = 0.0004

d. P(X > 125) = P(Z > (125-100)/12) = P(Z > 2.0833) = 0.0188

e. For n = 300

= P((X-μ)/(σ/sqrt(n)) < (90-100)/(12/sqrt(300)))
= P(Z < -14.4337)
= 0