CFA Practice Question

There are 434 practice questions for this study session.

CFA Practice Question

Suppose you were told that scores on an examination were converted to standard scores with a mean of 500, range of 800, and a standard deviation of 100. A person with a score of 600 has performed better than what percentage of the persons taking the test?
A. 57 percent
B. 97.5 percent
C. 84 percent
Explanation: Z = (X - MU)/Standard Deviation
Z = (600 - 500)/100 = 1
Area between Z and Mean = 0.3413
Area to Left of Z = 0.5 + 0.3413 = 0.8413
Percentile Rank = 100*0.8413 = 84.13
Therefore, this person has performed better than 84% of the people who took the test.

User Contributed Comments 13

User Comment
DAS11 Where did they get .3413 from?
MUSK More simpler way for doing this is to look for probability value (area under the curver for z=1, in one tailed test). Value is .8413
jam99003 will we have the tables for the test? it makes sense that we should, but I've not heard anything either way.
julamo another super easy way to guess: around 68% of observations are within 1 standard deviation of the mean (400-600), and 32/2=16% of the scores will be below 400. 68+16=84% of the scores below 600...
Bibhu Very good answer julamo.
Kuki excellent julamo, i knew there had to be an easier way to do this!
sh21 P.S: No table will be given in the exam. Check cfainstitute.org, (FAQS)
NNikolaiss Or mean plus minus 1 st is 68 % area meaning 34% at one side. As mean is 50 % so 50+34=84
homersimpson Thanks a zillion, julamon!
maria15 Where did the 68 come from?
maria15 I didn't understand Julamo's shortcut. I understood analystnotes' explanation better :(
GBolt93 68 is the observations within 1 standard deviation of the mean. You have to know the benchmark deviations. I.E. 1, 2, 3 standard deviations equate to 68%, 95%, 99.8% of obersavations for a 2 tailed z-test
edrei7 Just remember the 68-95-99.7 rule for normal distributions.
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