Section 2.1 – Describing Location in a Distribution
What if we want to determine the position of an individual within a distribution?
For instance: What score on a standardized test would you have received if you are in the 80th percentile?
What percentage of people did Jenny score above on the test?
Percentile: The pth percentile of a distribution is the value with p percent of the observations less than it.
Sometimes you can determine this from a histogram. But a graph that is really helpful is the cumulative
relative frequency graph.
*Relative Cumulative Frequency Graph
 Step 1: Set up a table
 Step 2: Label & Scale Axes
 Step 3: Plot points
Example on p. 87 – 88 - Presidential age at Inauguration
Step 1: Set up a table:
Class
40-44
45-49
50-54
55-59
60-64
65-69
TOTAL
Frequency
Relative Frequency
2
7
13
12
7
3
44
2/44 = 0.045 or 4.5%
7/44 = 0.159 or 15.9%
29.5%
27.3%
15.9%
6.8%
Cumulative
Frequency
2
9
22
34
41
44
Relative Cumulative
Frequency
2/44 = 0.045 or 4.5%
9/44 = 0.205 or 20.5%
50%
77.3%
93.2%
100%
Step 2: Label & Scale Axes, Step 3: Plot points
1. Bill Clinton was 46 when he took office.
What is his relative standing?
- About 8 or 9%
2. What inauguration age corresponds to the 60th
percentile?
- About 57
3. Find the Center of the distribution.
Center means in the middle, so 50% (Median)
- About 55
4. Find the IQR of the distribution.
- IQR = Q3 – Q1 = 59 – 51 = 8
Measuring Position: z-scores
Standardized Score (z-score): If x is an observation from a distribution that has known mean and standard
deviation, the standardized score for x is
z
x  mean
standard deviation
A standardized score is often called a z-score.
Example p. 90 (data on p. 85) Finding and Interpreting z-scores
Here are the scores of all 25 students in Mr. Pryor’s statistics class on their first test:
79
77
81
83
80
86
77
90
73
79
83
85
74
83
93
89
78
84
80
82
75
77
67
72
73
x  80 and s  6.07
Find the standardized score for each of the following:
(a) Jenny, who scored an 86.
z
(b) Norman, who earned a 72.
86  80
 0.99
6.07
z
72  80
 1.32
6.07
In words: Jenny scored 0.99 standard deviations above the mean of 80 while Norman scored 1.32 standard
deviations below the mean.
Example, p. 91 – Using z-scores for comparisons
After receiving her stats test with a grade of 86, Jenny earned an 82 on her chemistry test. The chemistry test
had a mean score of 76 and standard deviation of 4. Which test did Jenny do better on in comparison to her
classmates?
Compare the z-scores!!!!
Stats z-score ≈ 0.99
Chemistry z-score =
82  76
 1.50
4
Because her Chemistry score was 1.5 standard deviations above the mean for that class compared to 0.99
standard deviations above the mean in statistics, she did better in comparison to her classmates in chemistry.
HW: p. 99 # 1, 4, 9 a & b, 12, 25, 28, 29
Due: Monday