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 10, page139.

Date:  February 11, 2004

Question:

Suppose the following grouped frequency distribution represents test anxiety measures. 

The upper 15% of the individuals form the highly anxious group,
the next 25% form the moderately anxious group,
the next 40% form the somewhat anxious group,
and the bottom 20% form the non-anxious group.

Find the percentiles associate with each classification.

 

Interval Frequency Cumulative Frequency
60-64 8 155
55-59 11 147
50-54 15 136
45-49 23 121
40-44 36 98
35-39 26 62
30-34 19 36
25-29 13 17
20-24 4 4

Answer:

Step 1:  Compute the percentiles you want to find:

If the upper 15% of the individuals form the highly anxious group,
we must search for the 100-15 = 85th percentile.


If the next 25% form the moderately anxious group,
we must search for the 100-15-25 = 60th percentile.


If the next 40% form the somewhat anxious group,
we must search for the 100-15-25-40 = 20th percentile.

If the bottom 20% form the non-anxious group,
we must search for the 0th percentile.

Step 2:

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

 

Xpercentile = LLsubf + [ ( pN - cfsubj-1 )] [j]
                                          fsubj

 

Step 3:  compute pN for each desired percentile

p N pN percentile
.85 155 131.75 85th
.60 155 93 60th
.20 155 31 20th
.0 155 0 0th

(Keep in mind that the interval j includes 1/2 point below and 1/2 point above the differences in each interval.)

Fit the pN into the cumulative frequency column of the table, beneath the cf which is larger and above the cf which is small. 

In the procedure below,
   - subtract the cumulative frequency below pN in the table;
   - divide by the frequency above pN in the table;
   - use the lower limit of interval above pN in the table, and make it a real lower limit by subtracting 0.5
   - for j, add 1 to the interval in the table.
 

For the 85th percentile:  49.5 + [(131.75 - 121)/15] [5] = 53.08  for the highly anxious

For the 60th percentile:  39.5 + [(93 - 62)/36] [5] = 43.81 for the moderately anxious

For the 20th percentile:  29.5 + [(31 - 17)/19][5] = 33.18 for the somewhat anxious

For the 0th percentile: 19.5 + [(0 - 0)/4][5] = 19.5 for the non-anxious



filename: StatHW1Ch5Prob10page139TomMeyer.doc

Tom Meyer

Thomas Meyer