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Q: The average grade for an exam is 74, and the standard deviation is 7. (A) If 12% ( Answered,   0 Comments )
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 Subject: The average grade for an exam is 74, and the standard deviation is 7. (A) If 12% Category: Miscellaneous Asked by: zebby-ga List Price: \$2.00 Posted: 16 Dec 2003 02:19 PST Expires: 15 Jan 2004 02:19 PST Question ID: 287638
 ```The average grade for an exam is 74, and the standard deviation is 7. (A) If 12% of the students in the class are given A's, and the grades are curved to follow a normal distribution, what is the lowest possible score to receive an A? (B) What is the 60th percentile score?```
 ```Hello. A way to solve problems like these is to use a standard normal probability table, such as this one from Penn State: STANDARD NORMAL PROBABILITY TABLE http://www.stat.psu.edu/~antoniou/stat250.3/ZTable.html (A) The first step is to find the "z" value for the percentile that you need. If 12% of the class are to be given A's, then you want the "z" value for the 88th percentile. Consulting the table, we see that 0.8810 corresponds to z = 1.18. Now, we can determine the necessary score by setting the z value equal to: (X - the average) ------------------- standard deviation (where X is the score to be determined) So: X - 74 1.18 = ------- 7 Using algebra, we see that: X - 74 1.18 x 7 = --------- x 7 7 so: 8.26 = X - 74 so X = 82.26 Rounding down, 82 is the score needed to receive an A. (B) We can determine the 60th percentile score exactly the same way. Again, we consult the table: http://www.stat.psu.edu/~antoniou/stat250.3/ZTable.html We see that for the value 0.6026, z=0.26 Thus, X - 74 0.26 = ------- 7 Using algebra, we see that: X - 74 0.26 x 7 = --------- x 7 7 so: 1.82 = X - 74 so X = 75.82 Rounding up, we see that 76 is the 60th percentile score. -------------------- search strategy: "standard deviation" percentile "what is the minimum" This brought up a similar problem in this document from University of Rochester: http://www.econ.rochester.edu/khan/lect7.pdf I hope this helps.```