Transcript Document

15 September 2003
15 September 2003
1.2 Describing Distributions with Numbers
Five-number summaries and boxplots
Changing the unit of measurement
1.3 The Normal Distributions
Density curves
The normal distribution
Standardized scores
Areas under a standard normal curve (Table A)
Normal quantile plots
FIVE-NUMBER SUMMARIES
and BOXPLOTS
50
◄ Maximum
45
40
35
30
◄ 75th percentile (Q3)
25
20
◄ Median (Q2)
15
◄ 25th percentile (Q1)
10
5
◄ Minimum
Calculating the variance and the standard deviation
Mean
Variance
Standard Deviation
2
6
8
5
9
6
Deviations
from Mean
-4
0
+2
-1
+3
Squared
Deviations
16
0
4
1
9
7.5
2.74
What if we add 10 to each number?
12
16
18
15
19
Mean 16
Variance
Standard Deviation
Deviations
from Mean
-4
0
+2
-1
+3
Squared
Deviations
16
0
4
1
9
7.5
2.74
What if we add 10 to each number?
12
16
18
15
19
Mean 16
Variance
Standard Deviation
Deviations
from Mean
-4
0
+2
-1
+3
Squared
Deviations
16
0
4
1
9
7.5
2.74
What if we multiply each number by 10?
20
60
80
50
90
Mean 60
Variance
Standard Deviation
Deviations
from Mean
-40
0
+20
-10
+30
Squared
Deviations
1600
0
400
100
900
750
27.4
What if we multiply each number by 10?
20
60
80
50
90
Mean 60
Variance
Standard Deviation
Deviations
from Mean
-40
0
+20
-10
+30
Squared
Deviations
1600
0
400
100
900
750
27.4
What if we multiply each number by 10?
20
60
80
50
90
Mean 60
Variance
Standard Deviation
Deviations
from Mean
-40
0
+20
-10
+30
Squared
Deviations
1600
0
400
100
900
750
27.4
CHANGING THE UNIT OF MEASUREMENT
If you add or subtract the same amount from
each value in a distribution, then
 the mean is increased or decreased by that amount
 the spread is not changed
 If you multiply or divide each value in a
distribution by the same amount, then
 the mean is multiplied or divided by that amount
 the variance is multiplied or divided by the square of that amount
 the standard deviation is multiplied or divided by that amount
STANDARDIZED SCORES
A standardized score (or z score) tells us far
above or below the mean a given number
falls, in standard deviation units.
z
deviation

standard deviation
x
x

s
STANDARDIZED SCORES
For example, suppose the grades on a quiz have
mean of 85 and standard deviation of 5 points.



If your grade is 95, your z score is +2.00, because you
are two standard deviations above the mean.
Your friend whose grade is 83 has what z score?
Another friend tells you that his z score is -1.00. What’s
his grade?
65
70
75
80
85
90
95
100
——+—————+—————+—————+—————+—————+—————+—————+——
-4
-3
-2
-1
0
+1
+2
+3
Standardized scores
2
6
8
5
9
Mean 6
Standard Deviation 2.74
-4 /
0/
2/
-1 /
3/
2.74
2.74
2.74
2.74
2.74
z score
= -1.46
= 0.00
= 0.73
= -0.36
= 1.09
What if we add 10 to each number?
12
16
18
15
19
Mean 16
Standard Deviation 2.74
-4 /
0/
2/
-1 /
3/
2.74
2.74
2.74
2.74
2.74
z score
= -1.46
= 0.00
= 0.73
= -0.36
= 1.09
What if we multiply each number by 10?
20
60
80
50
90
Mean 60
Standard Deviation 27.4
-40 / 27.4
0 / 27.4
20 / 27.4
-10 / 27.4
30 / 27.4
z score
= -1.46
= 0.00
= 0.73
= -0.36
= 1.09
STANDARDIZED SCORES
If you take a list of numbers and

add the same amount to each number,

subtract the same amount from each number,

multiply each number by the same amount, or

divide each number by the same amount,
the z scores do not change.
THE NORMAL CURVE
THE NORMAL CURVE
The normal curve for women’s heights
How many women are at least 67 inches tall?
How many women are at least 67 inches tall?
z = (67 - 64.5) / 2.5 = +1.00
The relative frequency of women at least 67
inches tall is approximately equal to the
fraction of a standard normal curve which lies
to the right of +1.00.
Table A says that 84% of the area lies to the left
of +1.00, so 16% must lie to the right.
So about 16% of women are at least 67” tall.
How many women are at least 68 inches tall?
z = (68 - 64.5) / 2.5 = +1.40
The relative frequency of women at least 68
inches tall is approximately equal to the
fraction of a standard normal curve which lies
to the right of +1.40.
Table A says that 91.92% of the area lies to the
left of +1.40, so 8.08% must lie to the right.
So about 8% of women are at least 68” tall.
How many women are at least 68 inches tall?
What SAT-verbal score is at the 90th percentile?
What SAT-verbal score is at the 90th percentile?
 Table A says
that about 90% of a
normal histogram lies left of +1.28
( X - 505 ) / 110 = 1.28
X = 110 (1.28) + 505
X = 645.8
 So approximately 90% of SAT-v scores
are less than 645.8
What fraction of SAT-v scores are less than 645.8?
z
= ( 645.8 - 505 ) / 110
= ( 140.8 ) / 110
= 1.28
 According
to Table A, 89.97% of a
normal histogram lies to the left of
+1.28, so about 90% of the SAT scores
will be less than 645.8.
1.2
Describing Distributions with Numbers
Five-number summaries and boxplots
Changing the unit of measurement
1.3
The Normal Distributions
Density curves
The normal distribution
Standardized scores
Areas under a standard normal curve (Table A)
Normal quantile plots
Sections to skip
1.3 Normal Quantile Plots
(pp 78-83)
2.1 Categorical Explanatory Variables
(pp 113-114)