Download Describing Location in a Distribution

Describing Location in
a Distribution
Text
2.1 Measures of Relative Standing
and Density Curves
Sample Data
Consider the following test scores for a small class:
79
81
80
77
73
83
74
93
78
80
75
67
77
83
86
90
79
85
83
89
84
82
77
72
73
Julia’s score is noted in red. How did she perform on
this test relative to her peers?
6| 7
7 | 2334
7 | 5777899
8 | 00123334
8 | 569
9 | 03
Her score is “above average”...
but how far above average is it?
Standardized Value
One way to describe relative position in a data set is to
tell how many standard deviations above or below the
mean the observation is.
Standardized Value: “z-score”
If the mean and standard deviation of a distribution are
known, the “z-score” of a particular observation, x, is:
x  mean
z
standard deviation
Calculating z-scores
Consider the test data and Julia’s score.
79
81
80
77
73
83
74
93
78
80
75
67
77
83
86
90
79
85
83
89
84
82
77
72
73
According to Minitab, the mean test score was 80
while the standard deviation was 6.07 points.
Julia’s score was above average. Her standardized zx  80 86  80
score is:
z

 0.99
6.07
6.07
Julia’s score was almost one full standard deviation
above the mean. What about Kevin: x=72

Calculating z-scores
79
81
80
77
73
83
74
93
78
80
75
67
77
83
86
90
79
85
83
89
84
82
77
72
6| 7
7 | 2334
7 | 5777899
8 | 00123334
8 | 569
9 | 03
73
Julia: z=(86-80)/6.07
z= 0.99 {above average = +z}
Kevin: z=(72-80)/6.07
z= -1.32 {below average = -z}
Katie: z=(80-80)/6.07
z= 0
{average z = 0}
Comparing Scores
Standardized values can be used to compare scores
from two different distributions.
Statistics Test: mean = 80, std dev = 6.07
Chemistry Test: mean = 76, std dev = 4
Jenny got an 86 in Statistics and 82 in Chemistry.
On which test did she perform better?
Statistics
86  80
z
 0.99
6.07
Chemistry
82  76
z
1.5
4
Although she had a lower score, she performed
relatively better in Chemistry.
Percentiles
Another measure of relative standing is a percentile rank.
pth percentile: Value with p % of observations below it.
median = 50th percentile {mean=50th %ile if
symmetric}
Q1 = 25th percentile
Q3 = 75th percentile
6| 7
7 | 2334
7 | 5777899
8 | 00123334
Jenny got an 86.
8 | 569
22 of the 25 scores are ≤ 86.
Jenny is in the 22/25 = 88th %ile. 9 | 03
Density Curve
In Chapter 1, you learned how to plot a dataset to
describe its shape, center, spread, etc.
Sometimes, the overall pattern of a large number
of observations is so regular that we can describe
it using a smooth curve.
Density Curve:
An idealized description of
the overall pattern of a
distribution.
Area underneath = 1,
representing 100% of
observations.
Density Curves
Density Curves come in many different shapes;
symmetric, skewed, uniform, etc.
The area of a region of a density curve represents
the % of observations that fall in that region.
The median of a density curve cuts the area in half.
The mean of a density curve is its “balance point.”
Example
•
Pretend you are rolling a die. The numbers 1,2,3,4,5,6 are the possible
outcomes. In 120 rolls, how many of each number would you expect to
roll?
•
Calculator can do a simulation:
•
Clear L1 in your calc. Use random integer generator to generate 120
random whole numbers between 1 and 6 then store in L1
•
RandInt (1, 6, 120) STO-> L1
•
Set viewing window: X (1,7) by Y (-5,25).
•
Specify a histogram using the data in L1
•
Repeat simulation several times. 2nd Enter will recall/reuse the
previous command. In theory we should expect a uniform
outcome...
Summary
We can describe the overall pattern of a distribution
using a density curve.
The area under any density curve = 1. This
represents 100% of observations.
Areas on a density curve represent % of observations
over certain regions.
An individual observation’s relative standing can be
described using a z-score or percentile rank.
x  mean
z
standard deviation
Normal Distributions
•
Normal Curves: symmetric, single-peaked, bellshaped.  and median are the same. Size of
the  will affect the spread of the normal
curve.
Example
•
Scores on the SAT verbal test in recent years
follow approximately the N (505, 110)
distribution. How high must a student score in
order to place in the top 10% of all students
taking the SAT?
•
1. State the problem and draw a picture.
Shade the area we’re looking for.
•
2. Find the Z score with the table
•
3. Convert to raw score.
Assessing Normality
•
Method 1: Construct a histogram, see if graph
is approximately bell-shaped and symmetric.
Median and Mean should be close. Then mark
off the -2, -1, +1, +2 SD points and check the
68-95-99.7 rule.
Normal Probability Plot
•
Method 2: Construct Normal Probability Plot
•
1. Arrange the observed data values from smallest to
largest. Record what percentile of the data each value
occupies (example, the smallest observation in a set of
20 is at the 5% point, the second is at 10% etc.)
•
Use Table A to find the Z’s at these same percentiles
(example -1.645 is @ 5%, -1.28 is @10%
•
Plot each data point against the corresponding Z (xvalues on the horizontal axis, z-scores on the vertical axis
is what I do, either is fine)
•
•
rkgnt
Normal w/Outliers
Right Skew
Normal
Interpretation: draw your X = Y line with a straight edge- points shouldn’t
vary too much
Constructing Probability Plot on
Calculator
79
81
80
77
73
83
74
93
78
80
75
67
77
83
86
90
79
85
83
89
84
82
77
72
•
Students in math class
•
X values on horizontal axis
73