Question

probation officer caseloads have a mean of 115 and a standard deviation of 10. Caseloads sizes are normally distributed.

a. what is the probabily in proportion of percentage that a probation officer has a caseload between 90 and 105?

b. whatbis the probability in proportion or percentage that a probability officer has a caseload larger than 130?

c. one probation officer has more caseloads than 80 percent of all officers. at the least, how many caseloads does this officer have?

Answer 1

Solution :

Given that ,

mean = = 115

standard deviation = = 10

a) P(90 < x < 105) = P[(90 - 115)/ 10) < (x - ) /  < (105 - 115) / 10) ]

= P(-2.5 < z < -1.00)

= P(z < -1.00) - P(z < -2.5)

Using z table,

= 0.1587 - 0.0062

= 0.1525

b) P(x > 130) = 1 - p( x< 130)

=1- p P[(x - ) / < (130 - 115) / 10]

=1- P(z <1.5 )

Using z table,

= 1 - 0.9332

= 0.0668

c) Using standard normal table,

P(Z > z) = 80%

= 1 - P(Z < z) = 0.80  

= P(Z < z) = 1 - 0.80

= P(Z < z ) = 0.20

= P(Z < -0.8416 ) = 0.20  

z = -0.8416

Using z-score formula,

x = z * +

x = -0.8416 * 10 + 115

x = 106.58