HONORS STATISTICS
Mrs. Garrett Block 2 & 3
Wednesday September 26, 2012
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Daily Agenda
1. Welcome to class
2. Please find folder and take
your seat.
3. Notes quiz Ch2 section 1
4. Week at a glance
5. Introduce Ch2 Section 1
6. Test return and grade calculations
6. Collect folders
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Density Curves
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STUDENT PRACTICE
Review exercises 2.9, 2.10, 2.11, 2.12
on page 128
2.9 Density Curves
Sketch density curves that might describe the distributions with the following shapes.
a) Symmetric, but with two peaks.
b) Single peak and skewed to the left.
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2.10 A uniform density curve
Use the areas under the density curve to answer the following questions.
USE PROPER NOTATION TO ANSWER THE QUESTIONS.
a) Why is the total area under this curve equal to 1?
b) What percent of the observations lie above 0.8?
c) What percent of the observations lie below 0.6?
d) What percent of the observations lie between 0.25 and 0.75?
e) What is the mean μ for this distribution?
f) What percent of the observations lie below 1.3?
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2.11 A uniform density curve, II
Refer to the previous exercise. Can you construct a modified box plot for the
uniform distribution? If so, do it. If not, explain why.
USE PROPER NOTATION TO ANSWER THE QUESTIONS.
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2.12 Finding means and medians.
page 128 mean, median, mode placement on distribution
(a)
(b)
(c)
Mean Median
Mode 12
Standardized Scores (z­scores)
A z­score tells us how many standard deviations away from the mean the original observation falls.
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Standardized Scores (z­scores)
NOTATION
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Review exercises on page 118 2.1
Eleanor scores 680 on the mathematics part of the SAT. The distribution of SAT scores in a reference population is symmetric and single­peaked with a mean of 500 and a
standard deviation of 100. Gerald takes the American
College Testing (ACT) mathematics test and scores 27.
ACT scores also follow a symmetric, single­peaked
distribution but with a mean of 18 and a standard deviation
of 6. Find the standardized scores for both students.
Assuming that both test measure the same kind of ability,
who has the higher score?
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2.3 Measuring Bone Density
Individuals with low bone density (osteoporosis) have a high risk of broken bones (fractures). Physicians who are concerned about low bone density in patients can refer them for specialized testing. Currently, the most common method for testing bone density is dual­
energy X­ray absorptiometry (DXA). A patient who undergoes a DXA test usually gets bone density results in grams per square centimeter (g/cm2) and in standardized units.
Judy, who is 25 years old, has her bone density measured using DXA. Her results indicate a done density in the hip of 948 g/cm2 and a standardized score of z =
­1.45. In the reference population of 25 year old women like Judy, the mean bone density in the hip is 956 g/cm2.
a) Judy has not taken a statistics class in a few years. Explain to her in simple language what the standardized score tells her about her bone density.
b) Use the information provided to calculate the standard deviation of bone density in the reference population.
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A standardized "set" of data always has the following....
Mean = 0 units
Standard deviation = 1 unit.
example Suppose a data set has a mean = 35 inches with a standard deviation = 5 inches
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20
30
25
­2
­1
0
1
2
40
45
50
5 units
-35
­3
35
3
1 unit
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OTL C2#2
Pg 118: 2.2, 2.4
Pg 122: 2.7, 2.8
Pg 129: 2.13, 2.14
All skip 2.7d
A Skips 2.2 and 2.14
Show much arithmetic work the selected problems. Use the worksheet that you received last Friday for the figure 2.13.
Follow the book instructions closely to complete 2.14.
Sketch the resulting histograms.
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2.7
a) Nationally Scott's score places him almost in the top 5.2% or
(1-.052 = .948) 94th percentile. Locally (at his school) he score
does not seem so outstanding. It is only 68th percentile.
(He has smart people at his school.)
b) Convert Scott's score to a national z-score and to a local z-score.
National
= 46.9
Sx = 10.9
Scott's standard score
Nationally is much
higher than his local
score.
Locally
= 58.2
Sx = 9.4
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2.7
c) The boy's at Scott's school did much better than the National
average(46.9) . The mean of the school is 58.2 which is higher
than the national average by almost 11 points.
The standard deviation of his school is smaller (9.4) than the
National scores (10.9). Which means that they are not spread out
as much as the National Scores.
Chebyshev's
d) Using Chebyshev's "rule"
National Scores
1 z-score
0% of the data is "not predicted or guaranteed"
2 z-scores 75% of the data is predicted
46.9 ± 2(10.9) or 46.9 ± 21.8
75% of the National scores are predicted to be from 25.1 to 68.7
3 z-scores 89% of the data is predicted
46.9 ± 3(10.9) or 46.9 ± 32.7
89% of the National scores are predicted to be from 14.2 to 79.6
(note lowest score possible is 20) or 20 to 79.6
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d) Using Chebyshev's "rule"
Local School Scores
1 z-score
0% of the data is "not predicted or guaranteed"
2 z-scores 75% of the data is predicted
58.2 ± 2(9.4) or 58.2 ± 18.8
75% of the Local scores are predicted to be from 39.4 to 77
3 z-scores 89% of the data is predicted
58.2 ± 3(9.4) or 58.2 ± 28.2
89% of the National scores are predicted to be from 30 to 86.4
(note highest score possible is 80) or 30 to 80
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2.8 Larry, Larry, Larry ...
High Blood Pressure is not a good thing. The 90th
percentile interprets as only 10% of the population has
blood pressure higher than Larry's.
Higher is NOT better this time.
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2.13 page 129 Weird Density Curve
Density Curve
A
B
All area above y axis
Area = 1
a) from x = 0 to x = 0.4 shape is a trapezoid
Area = ½(2+1)(0.4) = 0.6 u2
from x = 0.4 to 0.8 shape is a rectangle
Area = (0.4)(1) = 0.4 u2
Total Area = 0.6 + 0.4 = 1
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2.13 b) and c)
b) 0.6≤x≤0.8 is a rectangle Area is 0.2(1) = 0.2
c) refer to part a) 0≤x≤0.4 is a trapezoid Area = 0.6
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2.13 d)
(0.2, ?) it appears to be 1.5.
Let's verify this with an
equation.
d) There are a few ways to do this ...
from 0.0≤x≤0.2 is a trapezoid Area = ½(2 + ?)(.2)
The equation of the line from (0,2) to the point (0.4,1)
is ... y = mx + 2. Now find the slope
so y = -2.5x + 2
if x = 0.2 then the height at 0.2 is
y = -2.5(0.2) + 2 = 1.5
FINALLY the area from 0≤x≤0.2
is ½(2 + 1.5)(0.2) = 0.35
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2.13 e) The median ?
x=?
0.5
0.5
let's review c) and d)
the area "below" x=0.2 is 0.35
the area "below" x=0.4 is 0.65
0.5 is somewhere between 0.35 and 0.65
so the median is somewhere between x = 0.2 and 0.4
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2.14 TI - Calculator Use
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OTL C2#2
Page 118: 2.2 and 2.4
Page 122: 2.7 and 2.8
Page 129: 2.13, 2.14
2.2 volunteer on board
2.4, 2.8 class discussion
2.4
Remember from 2.3 that Judy's bone density was z = -1.45
compared to her reference population.
a) Mary's bone density standardized score is z =
So compared to their reference populations, Mary has
healthier bones because her density is 0.5 standard
deviations above her reference mean and Judy's score is 1.45 standard deviations below her reference mean.
b) The standard deviation for Mary's reference population is
calculated as follows ...
The older age group has
a standard deviation of
8 g/cm2. This is a larger
standard deviation than
Judy's reference population
(5.52 g/cm2). This does
make sense because the older
women get the more variability
there will be when comparing their
bone density.
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