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AP Stats
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9/19
Below is a list of gas mileage ratings for selected passenger cars in miles per gallon.
Choose the correct histogram of the data.
53, 43, 89, 41, 85, 86, 91, 92, 95, 94, 86, 102,114, 30, 123
Section2.4 Part 3
Chebychev’s Theorem & Grouped Data
SWBAT:
Identify and analyze patterns of distributions using shape,
center and spread.
Distributions: Normal vs. Not or Unknown
•Use EMPIRICAL RULE for normal (symmetric)
distributions
•Use CHEBYCHEV’S THEOREM for ALL distributions.
CHEBYCHEV’S THEOREM
The portion of any data set lying
within k standard deviations (k>1)
of the mean is at least:
•k represents the number of standard deviations from the mean
•k = 2:
75% of the data lie within 2σ of the mean
•k = 3:
88.9%of the data lie within 3σ of the mean
In general, Chebychev’s Theorem gives the minimum %
of data that fall within the given # of standard
deviations.
CHEBYCHEV’S THEOREM - Example
The age distributions for Alaska and Florida are given. Which is
which? What conclusions can you reach using Chebychev’s
Theorem?
CHEBYCHEV’S THEOREM – Example Solution
Left Graph:
μ – 2σ is negative
μ + 2σ = 31.6 + 2(19.5) = 70.6; thus 75% of the population is
between 0 and 70.6 years old.
Right Graph:
μ – 2σ is negative
μ + 2σ = 39.2 + 2(24.8) = 88.8; thus 75% of the population is
between 0 and 88.8 years old, ie 75% of the population is under
the age of 88.8.
Because the population is higher and they are older, we known
the right graph is Florida, and the left is Alaska
We would actually get more specific data using the
histogram or a relative frequency histogram.
Standard Deviation for Grouped Data
Earlier we found sample mean and standard deviation by creating a frequency
distribution table to organize the data then using the formula for sample
standard deviation:
x
15
16
19
20
20
x–μ
15 – 18 = -3
-2
1
2
2
(x – μ )2
9
4
1
4
4
22
Standard Deviation for Grouped Data, cont’d
If data sets are larger and contain repeated values, you can create a relative
frequency distribution to group the data:
Ex: # of children in 50 households
1, 3, 1, 1, 1, 1, 2, 2, 1,0,1,1,0,0,0,1,5,
0,3,6,3,0,3,1,1,1,1,6,0,1,3,6,6,1,2,2,3,0,1,1,4,1,1,2,2,0,3,0,2,4
x
0
1
2
3
4
5
6
f
10
19
7
7
2
1
4
∑ = 50
xf
0
19
14
21
8
5
24
∑ = 91
x–‾
-1.8
-0.8
0.2
1.2
2.2
3.2
4.2
(x – ‾)2
3.24
0.64
0.04
1.44
4.84
10.24
17.64
(x – ‾)2f
32.40
12.16
0.28
10.08
9.68
10.24
70.56
∑ = 145.4
Let’s use a calculator instead!
•Enter the values of x into L1
•Enter the frequencies, f, into L2
•Select STAT
•Select CALC  1: 1-Var Stats
•Enter 2nd L1 , 2nd L2
x
0
1
2
3
4
5
6
f
10
19
7
7
2
1
4
∑ = 50
Standard Deviation for Grouped Data, cont’d
When a frequency distribution has classes, you can estimate the sample
mean and standard deviation using the midpoint of each class.
Class
x (midpt)
f
∑=
xf
∑=
x–‾
x
(x – ‾)
x 2
(x – ‾)
x 2f
∑=
Using midpoints Example
The graph shows the results of a survey of 1000 adults. Find the mean and
standard deviation.
Using midpoints Example
The graph shows the results of a survey
of 1000 adults. Find the mean and
standard deviation.
Class
0-99
100-199
200-299
300-399
400-499
500+
x (midpt)
49.5
149.5
249.5
349.5
449.5
599.5
f
xf
380
18810
230
34385
210
52395
50
17475
60
26970
70
41965
∑ = 1000 ∑ = 192,000
$192 per year
x – x‾
-142.5
-42,5
57.5
157.5
257.5
407.5
(x – x‾)2
20306.25
1806.25
3306.25
24806.25
66306.25
166056.25
(x – ‾)
x 2f
7,716,375
415,437.5
694,312.5
1,240,312.5
3,978,375
11,623,937.5
∑ = 25,668.750
$160.3 per year
You try….
In the example, we used 599.5 as the midpoint
for the class of $500+. How would the sample
mean and standard deviation change if you
used 650 to represent the class? Find the
mean and standard deviation.
Sample mean is $195.5 per year, and the sample standard deviation
is about $169.5 per year.
HOMEWORK:
P 92: 13, 19, 29, 37, 45- 51all