To:  sbaker@odu.edu

From:  Thomas Meyer
tmeyer@ph.vccs.edu
276 656-0283
Patrick Henry Community College

Subject:  Statistics - w/  Dr.Spencer Baker - Homework Assignment #1, Ch 5 - Problem 08, page138.

Date:  February 11, 2004

Question:

Suppose the following scores represent performance on a quiz.  The instructor wants to assign grades such that the upper 15% receive As,
the next 25% receive Bs,
the next 35% receive Cs,
the next 20% receive Ds,
and the bottom 5% receive Fs.

Find the percentiles associated with each letter grade 
20, 20, 19,19,19,19, 18,18,18,18,18, 16,16,16, 15,15, 14,14, 11.

Answer:

Step 1:  Compute the percentiles you want to find:

If the grades such that the upper 15% receive As, we must search for the 100-15 = 85th percentile.
If the next 25% receive Bs, we must search for the 100-15-25 = 60th percentile.
the next 35% receive Cs, we must search for the 100-15-25-35 = 25th percentile.
the next 20% receive Ds, we must search for the 100-15-25-35-20 = 5th percentile.
and the bottom 5% receive Fs.

Step 2:  Tabulate the data you received into the following columns and rows:

Scores and Percentile Ranks of 19 students

Use text page 119, formula 5.1 to find a percentile.

xpercentile = the percentile you are searching for, such as the 85th

Step 3:  compute pN

p stands for xpercentile, SUCH AS .85
N the number of people in the study (or the highest cumulative frequency cf) = 19
pN =  (.85)(19) = 16.15. 

Note that 16.15 will be a point in the distribution affecting persons in row 2.
Using the cumulative frequency column 16.15 lies between 17 in row 2 and 13 in row 3.

LLsubf = the lower real limit of the interval that includes the X-value in the row containing the cumulative frequency  greater than the product of Xpercentile and N where N is the largest cumulative frequency.  19 is the X-value and its lower real limit is 18.5

Said differently find Row 2's X-score and write down it lower real limit, in this case the X-score is 19 and the lower real limit of 19 is 18.5

cfsubj-1 = the cumulative frequency (in row 3) less than pN (that is, less than 16.15) = 13 in row 3

fsubJ = the frequency in the interval that includes X-value = 4 in row 2
 

j = interval length = 1

Therefore the 85 percentile = 18.5  + [ (16.15 - 13 )in row 3       = 18.5 + (3.15)  =  18.5 + .7875  =  19.2875
                                                                  4  in row 2                                4     

So 19.2875 is the 85th percentile and is the cutoff score for grade A.

xpercentile = the percentile you are searching for, such as the 60th

Step 3:  compute pN

p stands for xpercentile, SUCH AS .60
N the number of people in the study (or the highest cumulative frequency cf) = 19
pN =  (.60)(19) = 11.4. 

Note that 11.15 will be a point in the distribution affecting persons in row 2.
Using the cumulative frequency column 11.4 lies between 13 in row 3 and 8 in row 4.

LLsubf = the lower real limit of the interval that includes the X-value in the row containing the cumulative frequency  greater than the product of Xpercentile and N where N is the largest cumulative frequency.  18 is the X-value and its lower real limit is 17.5

cfsubj-1 = the cumulative frequency less than pN (that is, less than 11.4) = 8 in row 4

fsubJ = the frequency in the interval that includes X-value = 5 in row 3
 

j = interval length = 1

Therefore the 85 percentile = 17.5  + [ (11.4 - 8 )in row 4       = 17.5 + (3.4)  =  17.5 + .68  =  18.18
                                                                  5  in row 2                            5     

So 18.18 is the 60th percentile and is the cutoff score for grade B.
                                 

xpercentile = the percentile you are searching for, such as the 25th

Step 3:  compute pN

p stands for xpercentile, SUCH AS .60
N the number of people in the study (or the highest cumulative frequency cf) = 19
pN =  (.25)(19) = 4.75. 

Note that 4.75 will be a point in the distribution affecting persons in row 6.
Using the cumulative frequency column 4.75 lies between 5 in row 6 and 3 in row 7.

LLsubf = the lower real limit of the interval that includes the X-value in the row containing the cumulative frequency  greater than the product of Xpercentile and N where N is the largest cumulative frequency.  15 is the X-value and its lower real limit is 14.5

cfsubj-1 = the cumulative frequency less than pN (that is, less than 4.75) = 3 in row 7

fsubJ = the frequency in the interval that includes X-value = 2 in row 6
 

j = interval length = 1

Therefore the 25 percentile = 14.5  + [ (4.75 - 3 )in row 7       = 14.5 + (1.75)  =  14.5 + .875  =  15.375
                                                                  2  in row 6                          2     

So 15.374 is the 25th percentile and is the cutoff score for grade C.
                                 

xpercentile = the percentile you are searching for, such as the 5th

Step 3:  compute pN

p stands for xpercentile, SUCH AS .05
N the number of people in the study (or the highest cumulative frequency cf) = 19
pN =  (.05)(19) = .95. 

Note that .95 will be a point in the distribution affecting persons in row 8.
Using the cumulative frequency column .94 lies beneath 1 in row 8.

LLsubf = the lower real limit of the interval that includes the X-value in the row containing the cumulative frequency  greater than the product of Xpercentile and N where N is the largest cumulative frequency.  11 is the X-value and its lower real limit is 10.5

cfsubj-1 = the cumulative frequency less than pN (that is, less than .95) = 0

fsubJ = the frequency in the interval that includes X-value = 1 in row 8
 

j = interval length = 1

Therefore the 5th percentile = 10.5  + [ (.95 - 0 )                   = 10.5 + (.95)  =  10.5 + .95  =  11.45
                                                                  1  in row 8                            1     

So 11.45 is the 5th percentile and is the cutoff score for grade D.
                                 

filename: StatHW1Ch5Prob08page138TomMeyer.doc

Tom Meyer

Thomas Meyer