Question

1. A criminologist developed a test to measure recidivism (i.e., a person falling back into criminal behavior), in which low scores indicated a lower probability of repeating a criminal act. The test is normed so that it has a mean of 140 and a standard deviation of 40.

e. Suppose an individual is in the 67th percentile of this test, what is his/her corresponding recidivism score?

Answer 1

Let x be a random variable representin a person's recidivism score. Let the person's recidivism score be s, then:
P(x < s) = 0.67
P(z < (s - 140)/40) = 0.67
Using normal distribution table,
P(z < 0.44) = 0.67
(s - 140)/40 = 0.44
s - 140 = 40 x 0.44 = 17.6
s = 140 + 17.6 = 157.6

Therefore, the individual's corresponding recidivism score is 157.6