SOLUTION: 9. A normal distribution of utility bills shows the mean to be $100 and the standard deviation of $12.
a) What percent of the utility bills are more than $125.
b) If 300 utilit
Algebra ->
Probability-and-statistics
-> SOLUTION: 9. A normal distribution of utility bills shows the mean to be $100 and the standard deviation of $12.
a) What percent of the utility bills are more than $125.
b) If 300 utilit
Log On
Question 1120406: 9. A normal distribution of utility bills shows the mean to be $100 and the standard deviation of $12.
a) What percent of the utility bills are more than $125.
b) If 300 utility bills are randomly selected, about how many of them would you expect to be less than $90?
You can put this solution on YOUR website! a) Probability (P) ( X > 125 ) = 1 - P ( X < 125 )
:
z-score(125) = (125 - 100)/12 = 2.0833
:
using table of z-scores, the probability associated with 2.0833 is 0.9812
:
(P) ( X > 125 ) = 1 - 0.9812 = 0.0188
:
b) because we know the standard deviation of the population, we can use the normal distribution in our calculations
:
z-score(90) = ( 90 - 100 ) / 12 = −0.8333
:
P ( X < 90 ) = 0.2033
:
300 * 0.2033 = 60.99 approximately 61
:
