Question

The monthly utility bills in a city are normally​ distributed, with a mean of ​$100 and a standard deviation of ​$13. Find the probability that a randomly selected utility bill is​ (a) less than ​$66​, ​(b) between ​$81 and ​$110​, and​ (c) more than ​$120.

Answer 1

Solution :

Given that ,

mean = = $100

standard deviation = = $13

(a)

P(x < $66) = P[(x - ) / < (66 - 100) / 13]

= P(z < -2.62)

= 0.0044

Probability = 0.0044

(b)

P($81 < x < $110) = P[(81 - 100)/ 13) < (x - ) /  < (110 - 100) / 13) ]

= P(-1.46 < z < 0.77)

= P(z < 0.77) - P(z < -1.46)

= 0.7793 - 0.0721

= 0.7072

Probability = 0.7072

(c)

P(x > $120) = 1 - P(x < 120)

= 1 - P[(x - ) / < (120 - 100) / 13)

= 1 - P(z < 1.54)

= 1 - 0.9382

= 0.0618

Probability = 0.0618