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 ​$67​, ​(b) between ​$89 and ​$120​, and​ (c) more than ​$130.

Respuesta :

Answer

1. less than $67:

Z = (67 - 100) / 13

= -2.54 (based on Z-table value)

P(X < 67) = 0.0056

2. between $89 and $120:

Z(89) = (89 - 100) / 13

= -0.85

Z(120) = (120 - 100) / 13

= 1.54

P(89 < X < 120) = 0.9915 - 0.2033

= 0.7882

3. more than $130:

Z = (130 - 100) / 13

= 2.31

= 0.9896

P(X > 130) = 1 - 0.9896

= 0.0104

note: all the value above found by seek from Z-score table