Question

Within three years of release, about 66% of released prisoners were rearrested. Within five years of release, about 75% of released prisoners were rearrested. Use these recidivism rates to answer the following.

Step 1 of 4:

If we take a random sample of 80 former inmates three years after they were released. What is the probability less than half have been arrested since their release?

Step 2 of 4:

What would the probability be if we doubled our sample size?

Step 3 of 4:

If we consider a sample of size 80, what proportion would have to be rearrested within 5 years to be considered unusually high? [Express your answer as a decimal rounded to 4 places.]

Step 4 of 4:

If we examine 100 prisoners after 3 years, what recidivism rate would put the sample in the 34th percentile? [Express your answer as a decimal rounded to 4 places.]

Answer 1

Answer:

Given,

p = 66% = 0.66

n = 80

mean = p = 0.66

standard deviation= sqrt(pq/n)

= sqrt(0.66(1-0.66)/80)

= 0.053

P(x < 0.5) = P(z < (0.5 - 0.66)/0.053)

= P(z < -3.02)

= 0.0012639 [since from z table]

= 0.0013

b)

p = 0.66

n = 160

mean = 0.66

standard deviation = sqrt(pq/n)

= sqrt(0.66(1-0.66)/160)

= 0.0375

P(x < 0.5) = P((x-u)s/ < (0.5 - 0.66)/0.0375)

= P(z < -4.27)

= 0.0000098 [since from z table]

c)

p = 0.75

n = 80

mean = p = 0.75

standard deviation = sqrt(pq/n)

= sqrt(0.75(1-0.75)/80)

s = 0.0484

consider,

mean+3*sd = 0.75 + 2*0.0484

= 0.8468

P(x > 0.8468) = 1 - P(x < 0.8468)

= 1 - P(z < (0.8468 - 0.75)/0.0484)

= 1 - P(z < 2)

= 1 - 0.9772498 [since from z table]

= 0.0228

d)

p = 0.66

n = 100

mean = 0.66

standard deviation = sqrt(pq/n)

= sqrt(0.66(1-0.66)/100)

s = 0.0474

P(Z < z) = 34%

since from the standard normal table, z = - 0.4125

consider,

z = (x - mu)/s

- 0.4125 = (x - 0.66)/0.0474

-0.4125*0.0474 = x - 0.66

-0.0196 = x - 0.66

x = 0.66 - 0.0196

x = 0.6404