1. A criminologist developed a test to measure recidivism, where low scores indicated a lower probability of repeating an undesirable behavior. The test is normed so that it has a mean of 140 and a standard deviation of 40.
a. What is the percentile rank of a score of 172?
b. What is the Z score for a test score of 200?
c. What percentage of scores fall between 100 and 160?
d. What proportion of respondents should score above 190?
Normal chart is distributed 34% first sigma , 12.5 % second
sigma , 2.5% 3 sigma.
a. z = (x - m)/s
Given,
z = z-score
x = data point
m = mean
s = standard deviation
z = (172 - 140)/40
z = .8
.8 * 34 = 27.2
27.2 + 50 = 77.2 percentile.
b. z = (200 - 140)/40
z = 1.5
c.
let's find the z-scores of 100 and 160,
for x=100
z = (100 - 140)/40
z = -1
percentiles
50 -1(34) = 16%
for x=160
z = (160 - 140)/40
z = .5
percentiles
50 + .5(34) = 67%
hence, 67-16 = 51% falls between 100 and 160
d. Z-score of 190,
z = (190 - 140)/40
z = 1.25
percentile
50 + 34 + .25(13.5) = 87.38% will score under 190, so 100 - 87.38 =
12.63%
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