A criminologist developed a test to measure recidivism, where low scores indicat

Question

A criminologist developed a test to measure recidivism, where low scores indicated a lower probability of repeating the undesirable behavior. The test is normed so that it has a mean of 140 and a standard deviation of 40. What is the percentile rank of a score of 172? What is the Z score for a test score of 200? What percentage of scores falls between 100 and 160? What proportion of respondents should score above 190? Suppose an individual is in the 67th percentile in this test, what is his or her corresponding recidivism score?

Explanation / Answer

a. Given, mu=140, sigma=40. Substitute the values in following z score formula to compute the z score.

Z=(X-mu)/sigma, where, X denotes the raw score.

=(172-140)/40

=0.8

Now, look into the Z table to find area corresponding to the Z score. The area is 0.7881. Multiply it with 100 to get the percentile rank.

The percentile rank is: 78.81.

b. The Z score for X=200 is as follows:

Z=(200-140)/40

=1.5 (ans)

c. Compute Z scores for X1=100 and X2=160.

Z1=(100-140)/40=-1 and Z2=(160-140)/40=0.5

The two Z scores are of opposite signs, find areas between mean and respective Z scores and add them.

P(100<X<160)=0.3413+0.1915=0.5328 (ans)

d. P(X>190)=P[Z>(190-140)/40]=P(Z>1.25)=1-0.8944=0.1056 (ans)

e. Compute the area associated with 67th percentile.

The area is 67/100=0.67.

Subtract it from 1 to find area beyond the percentile rank. The area is 0.33. Look into the Z table, the corresponding z score is 0.44.

Substitute the values in following Z score formula to compute the raw score.

0.44=(X-140)/40

X=157.6 ~158 (ans)

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