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