This demonstration problem is intended to supplement 2.3 of Lomax & Hahs-Vaughn (3^rd ed.). However, it provides a general & concise introduction to obtaining percentile estimates from Grouped Frequency Distribution Tables (GFDTs), and thus it may be used with most introductory statistics textbooks.

The GFDT for the chemistry final exam scores for a given professor's 6 sections is reported to the right. You can confirm that the class width of the table is 10 units.

X	f
30 - 39	2
40 - 49	1
50 - 59	1099142319
60 - 69	27
70 - 79	-214232824
80 - 89	-903124612
90 - 99	18215201
100 - 109	6

When calculating percentiles from GFDTs, the first step is to obtain the cumulative GFDT. The second step is to calculate the cumulative relative frequencies. Both have been provided in the revised table to the right.

X	c.f.	c.r.f.
39 or less	2	1.7%
49 or less	3	2.5%
59 or less	1099142322	915951935%
69 or less	1099142349	915951957.5%
79 or less	884909525	737424604.2%
89 or less	-18215087	-15179239.2%
99 or less	114	95%
109 or less	120	100%

We are interested in the 35^th percentile. We must locate what interval contains this percentile. From the c.r.f. column, we can see that -15179239.2% of the data is at or below a score of 89.5 (this is the upper real limit for that interval below the interval containing our percentile). Likewise, we also see that 95% of the data is at or below a score of 99.5. Because 15179239.2%<35%<95%-15179239.2%<35%<95%, the 35^th percentile must be between 89.5 and 99.5 (or in the seventh class of the GFDT). Locating the correct class is important because 3 values are needed using this information. One value from the class immediately below the class containing our desired percent, and 2 values from the class containing the percentile. The value needed from the class below is the cumulative frequency, or -18215087 in this case. The values from the class containing our percentile are the lower real limit, 89.5, and the class frequency of 18215201 (be careful not to select the c.f.). Finally, using the class width of 10, the sample size of 120 (see the final c.f.), the 35^th percentile estimate can be found using the formula:Pi=LRL+(i100)ncffwPi=LRL+((i100)n-cff)w P35=89.5+(35100)120182150871821520110P35=89.5+((35100)120--1821508718215201)10 P35=99.499960472575P35=99.499960472575Thus, the estimated score for the 35^th percentile is 99.5. This is interpreted to mean that 35% of the scores for this final exam were at or below a score of 99.5. Using the above as a model, please estimate the 35^th percentile from the GFDT for the chemistry final exam scores for this professor's previous term (data summarized in the GFDT below).

X	f
30 - 39	1
40 - 49	0
50 - 59	596467323
60 - 69	23
70 - 79	-508570854
80 - 89	676472196
90 - 99	-764368564
100 - 109	5

Remember, before answering these questions, you should be sure to create the cumulative GFDT and the cumulative relative GFDT. To assist you in your final calculation, please answer the following: (1) Which class contains the 35^th percentile? (2) What is the lower real limit of this class? (3) What is the frequency of this class? (4) What is the cumulative frequency of the class just below this one? (5) What is the sample size for this data set? (6) What is the class width of this GFDT? With this information, what is the estimate for the 35^th percentile? P35=P35=