Question

The monthly untily bills in a city are normally distributed with a mean of $100 with standard deviation of $15. find the probability that a randomly selected utility bill is less than $65? between $80 and $90? and more than $100?

Answer 1

Solution,

Given that ,

mean = = 100

standard deviation = =15

1) P(x < 65 ) = P[(x - ) / < ( 65 - 100) / 15 ]

= P(z < -2.33 )

Using z table

= 0.0099

2) P( 80 < x < 90 ) = P[( 80 - 100)/ 15 ) < (x - ) /  < ( 90 - 100 ) / 15 ) ]

= P( -1.33 < z < -0.67 )

= P(z < -0.67 ) - P(z < - 1.33 )

Using z table

= 0.2514 - 0.0918

= 0.1596

3) P(x > 100 ) = 1 - P(x < 100 )

= 1 - P[(x - ) / < ( 100 - 100 ) / 15 ]

= 1 - P(z < 0 )

Using z table   

= 1 - 0.5000

= 0.5000