Download Probability and Statistics

POPULATION STATISTICS
100%
ACCURATE SINCE THEY INCLUDE A DATA VALUE FROM
ALL UNITS
IN THE POPULATION
• IN OTHER WORDS, A CENSUS
• IF THE REGISTRAR’S OFFICE CALCULATES THE MEAN AGE OF A STUDENT FROM ALL
ENROLLED STUDENTS, THEN THAT IS THE MEAN AGE, THERE IS NO VARIATION
SAMPLE STATISTICS
NOT 100% ACCURATE SINCE THEY DON’T INCLUDE A DATA VALUE FROM
ALL UNITS IN THE POPULATION
• IN OTHER WORDS, A SAMPLE MEAN CAN VARY FROM ONE SAMPLE TO THE NEXT
• IN THIS SECTION WE WILL DISCUSS THE BEHAVIOR OF THE VARIATION OF SAMPLE
MEANS TAKEN FROM A POPULATION
OBSERVE THE FOLLOWING POPULATION
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Population of 1000 grades
Possible score on test 0 to 140
Population Statistics
μ = 77.4
median = 77.9
a) Lowest grade_____
15
σ = 22.3
Highest grade______
137
μ ± 2σ = ______________________
( 32.8 , 122 )
b) Since the mean and median are so close, the data is most
Approximately Normal
likely approximately _________________________________
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10.5
32.8
55.1
77.4
99.7
122
0.176
d) P(70 ≤ X < 80) = ______________
The top 20% of the grades were above ________________
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67
71
129
113
16
43
45
41
47
46
66
25
58
61
49
57
60
54
62
60
69
51
71
74
61
92
115
69
46
48
44
50
79
82
45
104
109
51
60
63
57
66
70
67
65
72
75
64
66
68
115
109
114
104
119
107
105
116
111
117
106
76
80
73
83
63
74
69
60
63
78
67
67
72
99
104
95
108
84
82
106
72
75
98
46
48
44
50
58
67
40
71
74
51
96
56
86
65
68
62
71
50
64
54
46
48
68
68
71
65
74
77
81
67
88
92
71
66
69
116
31
33
30
34
50
59
24
63
67
38
83
87
79
91
82
83
86
84
88
84
119
125
52
68
71
65
74
78
88
62
101
106
71
77
81
74
84
69
74
77
64
67
78
70
74
85
89
81
93
96
81
104
87
91
85
58
61
55
63
57
71
47
67
70
62
86
90
102
107
97
112
64
87
84
56
59
100
53
56
51
58
72
71
57
81
85
57
120
126
20
21
19
22
57
58
19
76
79
29
88
92
84
96
78
87
83
78
82
88
47
49
92
97
88
101
88
99
86
94
99
92
63
66
60
69
72
77
61
78
82
66
63
72
80
84
76
88
92
95
80
106
111
81
77
81
74
84
78
88
72
83
88
78
114
130
99
104
95
108
96
94
106
92
97
98
76
80
73
83
85
93
71
98
103
78
67
77
89
93
85
97
74
86
81
74
77
89
90
95
86
99
98
99
93
106
112
90
82
54
57
52
59
69
71
54
79
83
58
101
106
97
111
94
97
103
89
94
99
115
131
81
85
77
89
78
83
80
75
79
82
82
86
78
90
77
90
73
86
90
83
45
51
70
74
67
77
71
85
60
78
82
72
68
71
65
74
84
82
73
86
90
71
94
88
69
72
66
76
88
97
63
104
109
71
96
101
92
105
83
103
82
94
98
95
97
108
10.5
32.8
55.1
77.4
99.7
122
144.3
f) Make a guess. If you select 30 random grades and average them, which of the
following would be realistic for the average that you obtain? Why?
94
A) 76.4 – 78.4
B) 69 - 85
C) 60 – 94.8
D) 55.1 – 99.7
E) 32.8 - 122
109
114
104
119
99
90
123
90
94
106
91
96
87
100
97
98
95
109
114
91
103
108
80
83
87
79
91
55
62
80
42
44
84
98
103
94
107
90
84
109
79
83
97
117
123
105
23
24
22
25
27
34
17
30
32
31
77
81
74
84
55
69
67
49
51
78
119
125
87
30
32
29
33
37
45
23
40
42
37
29
30
28
32
47
54
24
61
64
36
65
68
25
42
44
40
46
51
64
31
59
62
48
64
67
61
70
70
74
63
78
82
67
64
67
94
58
61
55
63
76
89
48
98
103
62
72
76
69
79
71
71
76
69
73
74
51
54
93
92
97
88
101
108
107
98
119
125
92
67
70
64
73
58
68
60
59
62
70
98
113
81
120
126
115
131
107
113
119
109
114
116
45
47
43
49
56
59
44
62
66
50
112
128
29
87
91
83
95
89
87
94
85
89
87
65
68
62
71
63
72
60
66
69
68
114
130
76
87
91
83
95
73
89
75
73
76
87
58
61
55
63
66
74
53
75
79
62
62
71
114
54
57
52
59
77
85
49
101
106
58
72
76
69
79
79
87
68
91
96
74
61
70
59
98
103
94
107
105
100
107
101
106
97
83
87
79
91
89
90
87
97
101
84
49
56
102
107
112
102
117
88
108
92
98
103
105
91
96
87
100
82
84
94
73
77
91
113
120
131
63
66
60
69
93
81
78
101
106
66
69
72
66
76
62
79
56
69
73
71
121
115
45
120
126
115
131
88
103
111
85
89
116
51
54
49
56
76
76
53
91
96
56
95
101
106
49
51
47
54
68
61
58
66
70
54
98
103
94
107
90
96
97
89
94
97
66
68
88
85
89
81
93
63
87
65
71
75
85
78
82
75
85
79
87
74
89
93
79
59
74
103
37
39
35
40
76
67
48
92
97
43
76
80
73
83
83
81
81
90
95
78
72
79
118
113
119
108
124
87
97
109
76
79
110
86
90
82
94
74
73
91
62
65
86
101
131
66
55
58
53
60
63
74
47
82
86
59
57
60
54
62
46
62
45
47
50
61
129
123
15
69
72
66
76
82
73
82
81
86
71
33
35
32
36
55
61
29
69
73
40
119
88
112
91
96
87
100
64
77
82
53
55
91
75
79
72
82
79
87
71
90
94
77
66
63
92
32
34
31
35
48
58
24
62
65
39
80
84
76
88
87
87
83
92
96
81
52
63
108
63
66
60
69
76
82
60
93
98
66
89
93
85
97
81
87
87
82
86
89
47
91
124
85
89
81
93
89
84
95
90
94
85
68
71
65
74
70
73
69
73
77
71
137
101
70
89
93
85
97
68
74
87
52
55
89
69
72
66
76
65
71
66
64
67
71
129
113
16
43
45
41
47
46
66
25
58
61
49
57
60
54
62
60
69
51
71
74
61
92
115
69
46
48
44
50
79
82
45
104
109
51
60
63
57
66
70
67
65
72
75
64
66
68
115
109
114
104
119
107
105
116
111
117
106
76
80
73
83
63
74
69
60
63
78
67
67
72
99
104
95
108
84
82
106
72
75
98
46
48
44
50
58
67
40
71
74
51
96
56
86
65
68
62
71
50
64
54
46
48
68
68
71
65
74
77
81
67
88
92
71
66
69
116
31
33
30
34
50
59
24
63
67
38
83
87
79
91
82
83
86
84
88
84
119
125
52
71
65
74
78
88
62
101
106
71
77
81
74
84
69
74
77
64
67
78
70
74
89
81
93
96
81
104
87
91
85
58
61
55
63
57
71
47
67
70
62
86
90
107
97
112
64
87
84
56
59
100
53
56
51
58
72
71
57
81
85
57
120
126
21
19
22
57
58
19
76
79
29
88
92
84
96
78
87
83
78
82
88
47
49
97
88
101
88
99
86
94
99
92
63
66
60
69
72
77
61
78
82
66
63
72
84
76
88
92
95
80
106
111
81
77
81
74
84
78
88
72
83
88
78
114
130
104
95
108
96
94
106
92
97
98
76
80
73
83
85
93
71
98
103
78
67
77
93
85
97
74
86
81
74
77
89
90
95
86
99
98
99
93
106
112
90
82
94
54
57
52
59
69
71
54
79
83
58
101
106
97
111
94
97
103
89
94
99
115
131
81
85
77
89
78
83
80
75
79
82
82
86
78
90
77
90
73
86
90
83
45
51
70
74
67
77
71
85
60
78
82
72
68
71
65
74
84
82
73
86
90
71
94
88
72
66
76
88
97
63
104
109
71
96
101
92
105
83
103
82
94
98
95
97
108
68
85
102
20
92
80
99
89
69
10.5
32.8
55.1
77.4
99.7
122
g) Collect at least 1 sample of test grades by randomly circling 30 test grades. Once
you randomly select 30 test grades, enter them into the calculator and report the
mean and standard deviation.
YOUR Sample Mean = _________ YOUR Sample Standard Deviation = ______________
144.3
i) We will make a stem and leaf plot of
the means:
STEM LEAVES
10.5
32.8
55.1
77.4
99.7
122
144.3
65-<68
68-<71
71-<74
74-<77
77-<80
80-<83
83-<86
86-<89
Find the mean of the
sample means.
Definition of a SAMPLING DISTRIBUTION
The SAMPLING DISTRIBUTION of the mean is the probability distribution of all possible values of the
random variable
𝑥
computed from samples of size n from the same population.
Therefore the sampling distribution is NOT a distribution for x but for 𝑥 . The sampling distribution of 𝑥
represents the variation of the sample means of size n from a population.
THE CENTRAL LIMIT THEOREM
• THE STEM AND LEAF DISPLAY ON THE BOARD IS THE BEGINNING OF A SAMPLING
DISTRIBUTION. A COMPLETE SAMPLING DISTRIBUTION CONTAINS EVERY POSSIBLE
MEAN FROM EVERY POSSIBLE SAMPLE OF N DATA VALUES (30 IN OUR CASE). IN
EVERYDAY STATISTICS, WE DON’T CREATE SAMPLING DISTRIBUTIONS BECAUSE OF
THE FOLLOWING THEOREM ABOUT THEM:
THE CENTRAL LIMIT THEOREM
THE CENTRAL LIMIT THEOREM STATES THE FOLLOWING:
GIVEN: A RANDOM SAMPLE OF N DATA POINTS FROM A POPULATION WITH A
MEAN
𝜇
AND A STANDARD DEVIATION OF 𝜎 .
THE DISTRIBUTION DOES NOT HAVE TO BE NORMAL.
IT CAN BE UNIFORM, EXPONENTIAL, SKEWED LEFT, SKEWED RIGHT, OR
ANYTHING ELSE.
Central Limit Theorem:
Given: A random sample of n data points
from a population with a mean 𝜇 and a standard deviation of 𝜎.
91
73
60
68
85
88
72
88
71
99
60
76
Stems Leaves
0
1
2
3
4
5
6 0088
7 2136
8 5588
9 1139
10
Uniform Data
85
68
91
93
30
97
88
52 52
44 44
100
59
66
Stems Leaves
0
1
2
3 0
4 44
5 229
6 6329
7 5557
8 258
9 57
10 0
Normal Data
75
63
75
62
75
69
77
85
95
85
72
55
18
33
10
63
10
33
11
99
10
43
11
12
10
49
28
13
29
55
Stems Leaves
0
1 0112345888000
2 898992
3 00033
4 239
5 55
6 3
7 2
8
9 9
10
Exponential Data
29
14
29
28
15
22
30
18
30
42
18
30
Central Limit Theorem:
Given: A random sample of n data points from a
population with a mean 𝜇 and a standard deviation
of 𝜎.
THE CENTRAL LIMIT THEOREM
GIVEN: A RANDOM SAMPLE OF N DATA POINTS FROM A POPULATION WITH A
MEAN
𝜇 AND A STANDARD DEVIATION OF 𝜎.
CONCLUSION: WHEN THE SAMPLE SIZE N IS SUFFICIENTLY LARGE ( ≥ 30), THE
SAMPLING DISTRIBUTION OF 𝑥 WILL BE APPROXIMATELY A NORMAL DISTRIBUTION
WITH
𝜇𝑥 = 𝜇
𝜎𝑥=
𝜎
𝑛
𝜇𝑥 = 𝜇
𝑥
THE CENTRAL LIMIT THEOREM
THE CENTRAL LIMIT THEOREM
THE CENTRAL LIMIT THEOREM
f(x)
Given An
Exponential
Distribution with
mean μ and standard
deviation σ
1
x
x
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
x  
x 

n
THE CENTRAL LIMIT THEOREM
GIVEN: A RANDOM SAMPLE OF N DATA POINTS FROM A POPULATION WITH A
MEAN
𝜇 AND A STANDARD DEVIATION OF 𝜎.
CONCLUSION: WHEN THE SAMPLE SIZE N IS SUFFICIENTLY LARGE (
≥ 30), THE
SAMPLING DISTRIBUTION OF WILL BE APPROXIMATELY A NORMAL DISTRIBUTION
WITH
Note: If n is less than 30 then the distribution does have to be NORMAL. This means that is n is less that 30, then
the given distribution has to be normal to assume the sampling distribution is normal.
𝜇𝑥 = 𝜇
𝜎𝑥= 𝜎
𝑛
Given A Normal
Distribution with
mean μ and standard
deviation σ
x
`
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
x  
x 

n
EXAMPLE 1
MEN’S WEIGHTS ARE
NORMALLY DISTRIBUTED WITH
A MEAN OF 𝝁 = 182.9 LBS AND A
STANDARD DEVIATION OF 𝝈 =
𝟒𝟎. 𝟖 LBS.
A) GRAPH THE GIVEN
DISTRIBUTION AND INCLUDE
THE TWO STANDARD DEVIATION
INTERVAL.
101.3
182.9
264.5
EXAMPLE 1
MEN’S WEIGHTS ARE
NORMALLY DISTRIBUTED WITH
A MEAN OF 𝝁 = 182.9 LBS AND A
STANDARD DEVIATION OF 𝝈 =
𝟒𝟎. 𝟖 LBS.
B) SINCE WE ARE DEALING WITH A
NORMAL DISTRIBUTION,
APPROXIMATELY 95% OF MEN
WOULD WEIGH BETWEEN
_______________ AND
__________________.
101.3
182.9
264.5
EXAMPLE 1
MEN’S WEIGHTS ARE
NORMALLY DISTRIBUTED WITH
A MEAN OF 𝝁 = 182.9 LBS AND A
STANDARD DEVIATION OF 𝝈 =
𝟒𝟎. 𝟖 LBS.
C) WHAT PERCENT OF MEN WEIGH
OVER 200 LBS? OR IN DIFFERENT
WORDING, WHAT IS THE
PROBABILITY THAT A RANDOMLY
SELECTED MAN WEIGHS OVER 200
LBS?
101.3
182.9
264.5
P(x>200) = normalcdf(200, 999999, 182.9, 40.8)
P(x>200) = .338 or approximately 34% of men
EXAMPLE 1
MEN’S WEIGHTS ARE
NORMALLY DISTRIBUTED WITH
A MEAN OF 𝝁 = 182.9 LBS AND A
STANDARD DEVIATION OF 𝝈 =
𝟒𝟎. 𝟖 LBS.
D) 10% OF MEN WEIGH BELOW
_______________.
a
101.3
P(x<a) = .1
182.9
264.5
a = invnorm(.1, 182.9, 40.8)
a = 130.61 10% of men weigh below 130.61 lbs
EXAMPLE 1
MEN’S WEIGHTS ARE
NORMALLY DISTRIBUTED WITH
A MEAN OF 𝝁 = 182.9 LBS AND A
STANDARD DEVIATION OF 𝝈 =
𝟒𝟎. 𝟖 LBS.
Given A Normal
Distribution with
mean μ and standard
deviation σ
x
`
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
x  
x 

n
E) FOR A SAMPLE OF 36
RANDOMLY CHOSEN MEN, GRAPH
normally
THE SAMPLING DISTRIBUTION OF THE C.L.T. says that 𝑥 is ______________
distributed
AND LOCATE THE TWO STANDARD
𝜎
40.8
=
= 6.8
𝜇 = 182.9
DEVIATION INTERVAL.
with 𝜇𝑥 = ____________
and 𝜎𝑥 = ____________.
𝑛
36
EXAMPLE 1
MEN’S WEIGHTS ARE
NORMALLY DISTRIBUTED WITH
A MEAN OF 𝝁 = 182.9 LBS AND A
STANDARD DEVIATION OF 𝝈 =
𝟒𝟎. 𝟖 LBS.
𝜇 = 182.9, 𝜎 = 40.8
Given A Normal
Distribution with
mean μ and standard
deviation σ
101.3
x
264.5
`
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
E) FOR A SAMPLE OF 36
RANDOMLY CHOSEN MEN, GRAPH
THE SAMPLING DISTRIBUTION OF
AND LOCATE THE TWO STANDARD
DEVIATION INTERVAL.
x  
169.3
𝜇𝑥 = 182.9
𝜎𝑥 = 6.8
196.5
x 

n
EXAMPLE 1
MEN’S WEIGHTS ARE
NORMALLY DISTRIBUTED WITH
A MEAN OF 𝝁 = 182.9 LBS AND A
STANDARD DEVIATION OF 𝝈 =
𝟒𝟎. 𝟖 LBS.
𝜇 = 182.9, 𝜎 = 40.8
Given A Normal
Distribution with
mean μ and standard
deviation σ
101.3
x
264.5
`
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
• F) FIND THE PROBABILITY THAT
A RANDOMLY SELECTED MAN
WILL WEIGH BETWEEN 180 AND
190.
THEN, FIND THE PROBABILITY
THAT A SAMPLE OF 36 MEN WILL
HAVE A MEAN WEIGHT BETWEEN
180 AND 190.
•
x  
x 

n
169.3 𝜇𝑥 = 182.9 196.5
𝜎𝑥 = 6.8
P(180< x <190) = normalcdf(180, 190, 182.9, 40.8)
P(180< 𝑥 <190) = normalcdf(180, 190, 182.9, 6.8)
=0.097
=0.517
EXAMPLE 1
H) FROM PART B,
APPROXIMATELY 95% OF
MEN WEIGHT BETWEEN
__________
AND __________
101.3
264.53
BUT WE WOULD EXPECT
THE MEAN OF 36 MEN TO
FALL BETWEEN
196.5
169.3 AND ________,
________
95% OF THE TIME.
𝜇 = 182.9, 𝜎 = 40.8
Given A Normal
Distribution with
mean μ and standard
deviation σ
101.3
x
264.5
`
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
x  
169.3 𝜇𝑥 = 182.9 196.5
𝜎𝑥 = 6.8
x 

n
Example 2 Suppose that for the summer months in a certain city the high
temperature is uniformly distributed between 80 and 100 degrees Fahrenheit.
a) Graph this distribution, including the mean, c, and d.
𝑓 𝑥 =
𝜎=
1
20
= 5.8
c= 80
μ = 90
d= 100
𝑑−𝑐
12
Example 2 Suppose that for the summer months in a certain city the high
temperature is uniformly distributed between 80 and 100 degrees Fahrenheit.
b) What percent of days is a high temperature between 94 and 100 degrees
observed? _______
30%
P(94<x< 100)
𝑓 𝑥 =
1
20
= 6 x 1/20
= 0.3
94 100
Example 2 Suppose that for
the summer months in a
certain city the high
temperature is uniformly
distributed between 80 and
100 degrees Fahrenheit.
c) Graph the sampling
distribution of and locate
the two standard deviation
interval. Assume you are
dealing with 30 days of
high temperatures.
𝜇 = 90, 𝜎 = 5.8
1
20
80
100
𝜎𝑥 =
87.8
𝜇𝑥 = 90
𝜎𝑥 = 1.1
92.2
= 1.1
5.8
30
Example 2 Suppose that for
the summer months in a
certain city the high
temperature is uniformly
distributed between 80 and
100 degrees Fahrenheit.
d) What is the probability
that the AVERAGE TEMP
of 30 randomly selected
days is between 94 and 100
degrees? Why? What
about 91 to 93 degrees?
𝜇 = 90, 𝜎 = 5.8
1
20
80
100
P(94< 𝑥 <100) =
normalcdf(94, 100, 90, 1.1)
= 0.0001
87.8
𝜇𝑥 = 90
𝜎𝑥 = 1.1
92.2
Example 2 Suppose that for
the summer months in a
certain city the high
temperature is uniformly
distributed between 80 and
100 degrees Fahrenheit.
d) What is the probability
that the AVERAGE TEMP
of 30 randomly selected
days between 94 and 100
degrees? Why? What
about 91 to 93 degrees?
𝜇 = 90, 𝜎 = 5.8
1
20
80
100
P(91< 𝑥 <93) =
normalcdf(91, 93, 90, 1.1)
= 0.18
87.8
𝜇𝑥 = 90
𝜎𝑥 = 1.1
92.2
Example 2
e) Give an interval that
you expect the average
high temp of 30 days to fall
in 95% of the time.
𝜇 = 90, 𝜎 = 5.8
1
20
80
100
(87.8, 92.2)
________________
f) Give an interval that
you expect a high temp to
fall in 100% of the time.
(80, 100)
__________________
87.8
𝜇𝑥 = 90
𝜎𝑥 = 1.1
92.2
EXAMPLE
3
The time that a bus in late (x) for one particular stop is exponentially distributed with mean 3
and standard deviation _____.
3
a) Graph
this
distribution
0
3
9
EXAMPLE 3
The time that a bus in late (x) for one particular stop is exponentially distributed with mean 3
and standard deviation 3.
b) Which of the following statements describes the sampling distribution of for 40 random late times.
A) The sampling distribution of is exponentially distributed with a mean of 3 and a standard deviation of 3
B) The sampling distribution of is normally distribution with a mean of 3 and a standard deviation of 3
C) The sampling distribution of is normally distribution with a mean of 3 and a standard deviation of 0.47
D)The sampling distribution of is uniformly distributed with a mean of 3 and a standard deviation of 3
f(x)
EXAMPLE 3
𝜇 = 3, 𝜎 =3
1
c) Graph the sampling distribution and
locate the two standard deviation
interval.
Given An
Exponential
Distribution with
mean μ and standard
deviation σ
x
x
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
x  
2.06
𝜇𝑥 = 3
𝜎𝑥 =0.47
3.94
x 

n
EXAMPLE 3
f(x)
d) Suppose someone who
rides the bus is running
late. It takes them 2.5
minutes to get to the bus
stop and they are already
1minute late. What is the
probability that they catch
the bus? Which
distribution would you
use, the one for X or 𝑥 ?
2.06
𝜇 = 3, 𝜎 =3
1
Given An
Exponential
Distribution with
mean μ and standard
deviation σ
P(x > 3.5)
x
x
= 𝑒 −3.5/3
A sampling distribution or the distribution of
the sample mean x is approximately normally
=
.311with
distributed
x  
𝜇𝑥 = 3
𝜎𝑥 =0.47
3.94
x 

n
EXAMPLE 3
f(x)
𝜇 = 3, 𝜎 =3
e) Find the probability
that the average of 40 late
times is greater than 3.5
minutes.
1
Given An
Exponential
Distribution with
mean μ and standard
deviation σ
P( 𝑥 > 3.5) =
x
normalcdf(3.5, ∞, 3, 0.47)
x
= 0.14
A sampling distribution or the distribution of
the sample mean x is approximately normally
distributed with
x  
2.06
𝜇𝑥 = 3
𝜎𝑥 =0.47
3.94
x 

n