Download Continuous random variables

Continuous random variables
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probability density, mean, s.d.
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Normal distribution
A random variable X is continuous if P(X = x) = 0 for all x
uniform random variable
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Suppose X is a number picked at ‘random ’ from the
interval [0, 1]
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P(X = x) = 0 for all x
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But Probability X falls in an interval is equal to the length of
the interval,and is nonzero.
Probabilities involving X can be modeled by a continuous curve
0.0
1.0
2.0
Identifies prob with area under a function
−1.0
−0.5
0.0
0.5
1.0
1.5
2.0
Continuous Probability Density
Function
The graphical form of the probability distribution for a
continuous random variable x is a smooth curve
Density curves
A density curve is a mathematical model of a distribution.
The total area under the curve, by definition, is equal to 1, or 100%.
The area under the curve for a range of values is the probability of all
observations for that range.
Histogram of a sample with the
smoothed, density curve
describing theoretically the
population.
Density curves come in any
imaginable shape.
Some are well known
mathematically and others aren’t.
Our interest is in a special type of density called Normal density
or Normal Distribution
Normal random variables
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Describes many random processes or continuous
phenomena
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Can be used to approximate discrete probability
distributions Example: binomial
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Basis for classical statistical inference
Normal density
A Normal density has two parameters µ, σ
N(µ, σ) :
f (x) = √
1
2πσ
e
(x−µ)2
2σ 2
−∞<x <∞
E(X ) = µ, V (X ) = σ 2 , s.d(X ) = σ
Normal Distribution
Normal Distribution
f(x )
x
Mean
Median
Mode
Normal Distribution
Normal distributions
x
x
Normal Distribution
Here, means are the same (m = 15)
while standard deviations are
different (s = 2, 4, and 6).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Here, means are different
(m = 10, 15, and 20) while standard
deviations are the same (s = 3)
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Normal Distribution
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How to calculate probabilities?
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Suppose we have N(5,3) and want to find P(2 < X < 7),
how do we do that?
N(5,3)table?
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Do we need a table for every possible value of mean and
s.d?
Standard Normal Distribution
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Suppose X is N(µ, σ), i.e a normal random variable with
mean µ, and s.d σ
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Define the “ standardized variable Z by
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Z =
X −µ
σ
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Then mean of Z =0 , s.d (Z ) = 1 and
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Z is N(0,1) and is called STANDARD NORMAL
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The transformation
X 7−→ Z =
X −µ
σ
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converts a N(µ, σ) to a standard normal , i.e N(0, 1)
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As a consequence
P(a < X < b) = P
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Use standard normal tables
a−µ
b−µ
<Z <
σ
σ
Standard Normal
Standard Normal Distribution
The standard normal distribution is a normal
distribution with µ = 0 and  = 1. A random variable
with a standard normal distribution, denoted by the
symbol z, is called a standard normal random variable.
Standard Normal table
Standard Normal table
The Standard Normal Table:
P(0 < z < 1.96)
Standardized Normal
Probability Table (Portion)
Z
.04
.05
s=1
.06
1.8 .4671 .4678 .4686
.4750
1.9 .4738 .4744 .4750
2.0 .4793 .4798 .4803
m= 0 1.96 z
2.1 .4838 .4842 .4846
Probabilities
Shaded area
exaggerated
Standard Normal table
The Standard Normal Table:
P(–1.26  z  1.26)
Standardized Normal Distribution
s=1
.3962
.3962
–1.26
1.26 z
m=0
Shaded area exaggerated
P(–1.26 ≤ z ≤ 1.26)
= .3962 + .3962
= .7924
Standard Normal table
The Standard Normal Table:
P(z > 1.26)
Standardized Normal Distribution
s=1
P(z > 1.26)
= .5000 – .3962
= .1038
.5000
.3962
1.26
m=0
z
Find the probability that a N(µ, σ) variable is
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within one standard deviation of the mean
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within two standard deviation of the mean
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within three standard deviation of the mean
Problems 98,107,114,203
Problem 4.98
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Let X be the support. X is Normal with µ = 67.755 and
σ = 26.871
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We will write this compactly as X is N(67.755,26.871).
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Find P(X < 40)
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P(X < 40) = P(Z <
40−67.755
26.871 )
= P(Z < −1.03)
0.1515
−3
−1.03
0
3
Problem 4.98
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X is N(67.755,26.871)
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Find P(40 < X < 120)
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P(< 40 < X < 120) = P( 40−67.755
26.871 < Z <
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= P(1 − .03 < Z < −1.944)
10−67.755
26.871 )
Problem 4.98
0.8225
−3
−1.03
0
1.944
3
Problem 4.98
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Find P(40 < X < 120)
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P(X > 120) = P(Z <
10−67.755
26.871 )
= P(Z < −1.944)
Problem 4.98
0.0259
−3
0
1.944
3
Problem 4.98
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Find c such that P(X < c)0.25
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P(X < c) = P(Z <
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from tables P(Z < −.6744) = .25
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so
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c = 26.871 × (−.6744) + 67.755 = 49.63
c−67.755
26.871
c−67.755
26.871 )
= .25
= −.6744
Problem 4.98
0.2502
−3
−0.674 0
3
Problem 4.107
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X : Rating is N(50,15)
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Find c such that P(X > c) = .1
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P(X > c) = P(Z >
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from tables P(Z > 1.28) = .1
I c−50
15
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c−50
15 )
= 1.28
c = 15(1.28) + 50 = 69.2
= .1
Problem 4.107
0.0999
−3
0
1.282
3
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X : Rating is N(50,15)
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Find c such that P(X > c) = .7
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P(X > c) = P(Z >
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from tables P(Z > −.524) = .1
I c−50
15
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c−50
15 )
= .7
= −.524
c = 15(−.524) + 50 = 42.14
Problem 4.114
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X: quantity injected in a unit is N(10,.20
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cost of each unit fill $20
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If fill is less than 10, cost of refill$10
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sale price$230
Problem 4.114
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Find the probability that a unit is underfilled, not underfilled
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P(X < 10) = P(Z <
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Prob not underfilled = 0.5
10−10
.2 )
= P(Z < 0) = .5
Problem 4.114
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A container initially underfilled, now refilled has 10.6 units.
What is the profit?
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230 - (10 + 20(10.6)) = 8
Problem 4.114
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Now X is N(10.1,0.2)
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What is the expected profit?
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for each unit profit is 230- 20 X ( since prob of underfilling
is 0)
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Expected profit = 230- 20 E(X) =230 - 20 (10.1) =28
Problem 4.203
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X: load is N(20,000,σ), σ unknown
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P(10, 000 < X < 30, 000) = .95
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P( 10,000−20,000
<Z <
σ
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From tables P(−1.96 < Z < 1.96) = .95
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So
30,000−20,000
σ
30,000−20,000
)
σ
= 1.96 or σ =
= .95
10,000
1.96
= 5102.041
Normal Approximation to the Binomial
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Normal distribution can be used to approximate
probabilities of a Binomial distribution
p
np(1 − p)
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Recall if X is Bin(n,p) then µ = np and σ =
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It can be shown that if n is large then the probabilities of X
can be approximated by a N(np,
p
np(1 − p)
Normal Approximation to the Binomial
Normal Approximation of Binomial
Distribution
1. Useful because not all
binomial tables exist
2. Requires large sample
size
3. Gives approximate
probability only
4. Need correction for
continuity
n = 10 p = 0.50
p(x)
.3
.2
.1
.0
0
x
2
4
6
8
10
Normal Approximation to the Binomial
Why Probability
Is Approximate
p(x)
.3
.2
.1
.0
x
0
2
Binomial Probability:
Bar Height
4
6
8
10
Normal Probability: Area Under
Curve from 3.5 to 4.5
Correction for Continuity
1. A 1/2 unit adjustment to
discrete variable
2. Used when
approximating a discrete
distribution with a
continuous distribution
3. Improves accuracy
3.5
(4 – .5)
4
4.5
(4 + .5)
problem 104
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X is Bin(100,000,.01)
√
P(X < 950) ≈ P Z < 950−.5−(100,000).01
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P(Z < −1.604) = .0542
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100,000(.01)(.99)
Problem 113
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V: The number who carry visa is Bin(100,.5), E(V) = 50
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D: The number who carry Discover is Bin(100,.09) = 9
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P(V ≥ 50) = P(Z > √50−.5−50 ) = .5398
100(.5)(.5)