Bài giảng Probability & Statistics - Lecture 6: Continuous probability - Bùi Dương Hải
Example Example 6.2. Return ($mil) of project A is normality with mean of 8 and variance of 9. Calculate the probability: (a) Return of A higher than 10 (b) Loss money (c) Return of A between 5 and 12 Return of project B is normality with mean of 10 and variance of 25. A and B are independent. Calculate the probability that: (c) Both gain positive return (d) Total return of A and B greater than 20
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Lecture 6. CONTINUOUS PROBABILITY
Continuous Random Variable
Density Function
Parameter
Uniform Distribution
Normal Distribution
Cutoff point
[1] Chapter 6. pp. 255 - 294
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6.1. Continuous Random Variable
Continuous Random Variable: uncountable values
Available value is one interval: = ( , )
Maybe: = −∞; = +∞
Probability that one point: = = 0
Consider Probability at one interval: ( <
< )
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6.2. Density Function
Discrete Continuous
∑ = 1 ∫
= 1
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X
Prob.
X ( , )
Density ( )
f(x)p
Density Function
≥ 0
∫
= 1
< < = ∫
Cutoff point level denoted by : > =
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f(x)
a b
6.3. Parameter
Expected Value:
= = ∫
Variance:
= ∫ −
= ∫
−
Standard Deviation
= ( )
Cutoff point level , denoted by :
> =
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Example
Example 6.1. Waiting time (hour), with density
function
=
2 ∈ [0,1]
0 ∉ [0,1]
(a) Prob. of waiting more than a half of hour?
(b) Prob. of waiting from 20 to 40 minutes?
(c) The average and variance of waiting time?
(d) Cutoff point level 10%?
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Example
(a) > 0.5 = ∫ 2 .
.
(b)
< <
= ∫ 2 .
/
/
(c) = ∫ .2 .
= ∫ .2 .
−
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f(x)
0.5
f(x)
1/3 2/3
6.4. Uniform Distribution
~ ( , ) if
=
∈ [ , ]
0 ∉ [ , ]
=
; =
< < =
Ex. Temperature is Uniform Distribution in the interval of
(20, 30)oC. What is the probability that temperature is
between 23 and 28 degree?
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a c d b
6.5. Normal Distribution
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0
0.05
0.1
0.15
0.2
0.25
1 2 3 4 5 6 7 8 9
0
0.05
0.1
0.15
0.2
0.25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
0
0.005
0.01
0.015
0.02
0.025
1 6 111621263136414651566166717681869196
0
0.005
0.01
0.015
0.02
0.025
0 20 40 60 80 100
, = 0.5 : = 10; 20; 100 Normality
Normal Distribution
Density Function: =
Denoted: ~ ( , )
=
=
=
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f(x)
μ μ’
1
σ 2π
Normal Distribution
Carl Friedrich Gauss
(1777-1855) in 1809
~ 3,1
~ 6,1
~ (8,0.5
)
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0
0.1
0.2
0.3
0.4
0.5
0.6
0 1 2 3 4 5 6 7 8 9 10
Standardized Normal Variable
~ ,
=
~ (0,1)
Table 1
< 1 = 0.8413
< 1.25 =
> 2 =
−1 < < 1.3 =
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0
0.1
0.2
0.3
0.4
0.5
-4 -3 -2 -1 0 1 2 3 4
-4
-3
.5 -3
-2
.5 -2
-1
.5 -1
-0
.5 0 0
.5 1
1
.5 2
2
.5 3
3
.5 4
Probability formula
~ ,
< =
−
<
−
= <
−
Ex. ~ 100,16
< 104 =
> 92 =
94 < < 102 =
Probability that X differ from the mean not more than
standard deviation =
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Example
Example 6.2. Return ($mil) of project A is normality with
mean of 8 and variance of 9. Calculate the probability:
(a) Return of A higher than 10
(b) Loss money
(c) Return of A between 5 and 12
Return of project B is normality with mean of 10 and
variance of 25. A and B are independent. Calculate the
probability that:
(c) Both gain positive return
(d) Total return of A and B greater than 20
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3-sigma Rule
− < < + = 68.26%
− 2 < < + 2 = 95.44%
− 3 < < + 3 = 99.75%
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Cutoff point
Cutoff point level , or “critical value”
Denoted:
> =
> 1.96 = 0.025 . = 1.96
> 1.64 = 0.0505 . = 1.64
> 1.65 = 0.0495 . = 1.65
Keys: . = . ; . = .
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6.6. Binomial vs Normal
Binomial: ~ ( , ) with ≥ 100
approximate: ( , )
With: = ; = (1− )
Example 6.3. Probability that visitor buy good in the
shopping mall is 0.3. In 400 visitors, what is the
probability
(a) There are at least 100 buyers
(b) Number of buyers is from 90 to 150
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6.7. Cutoff Point
Normal Distribution:
Student Distribution:
df: Degree of freedom
Table 2 (p.976)
. = 1.833; . = 2.086
≈
Chi-square Distribution:
Table 3 (p.979)
.
= 3.94 ; .
= 24.996
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Key Concepts
Continuous variable
Density function
Normal distribution
[1] Chapter 6:
(270) 3, 5
(281) 11, 12, 17, 19, 23, 24, 31
(292) 41, 44, 49
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Exercise
