Download Chapter 6 Some Continuous Probability Distributions

P131, 10’s
Solution
Let X be the number of defective missiles that will not
fire in one lot, then from the description of the problem ,
we have
C74k  C3k
P( X  x ) 

C104
, k  0,1,2,3.
Hence, we have the mean and variance of X as follow

3 4
14
 1.2,  2 
.
10
25
So here Chebyshev inequality reads
14
14
1
P(1.2  k
 X  1.2  k
)  1 2 ,
5
5
k
1
P131, 10’s
Solution

Which impies that the possibilities of the number of
defective missiles vary corresponding to different k. For
example, if k=1, X could be only 1 whose possibility
will be at least zero, or X will take the value of 1 at the
most possibilities; if k=2, X could be 0,1,2 whose
possibility will be at least three fourths; For example, if
k=5, X could be 0,1,2,3 whose possibility will be at least
24-25th.
14
14
1
P(1.2  k
 X  1.2  k
)  1 2 ,
5
5
k
2
Chapter 6
Some Continuous Probability Distributions

Uniform Distribution

Normal Distribution

Gamma Distribution

Exponential Distribution

Chi-Squared Distribution
3
6.1 Continuous
Uniform Distribution

Uniform Distribution The density function of
the continuous random variable X on the
interval [A, B] is
f(x; A, B) = 1/(B – A),
= 0,
AxB
elsewhere
P(a  X  b) = (b – a) /(B – A)
4

Theorem6.1 The mean and variance of the
uniform distribution
A B

2
and
( B  A)
 
12
2
2
The probability that X falls in an interval
of fixed length within [A, B] is constant.
5
Example 6.1, page 143
Suppose that a large conference room for a certain
company can be reserved for no more than 4 hours.
However, the use of the conference room is such that both
long and short conferences occur quite often. In fact, it can
be assumed that length of X of a conference has a uniform
distribution on the interval [0, 4]

What is probability density function?
f(x)=1/4, 0  x  4
=0. elsewhere

What is probability that any given conference lasts at least 3
hours?
P( X  3)  
4
1
3 4
dx  14
6
Example
6.2 Normal Distribution

The normal distribution is perhaps the most important
distribution in statistical applications. (Central Limit
Theorem)

Many phenomena that occur in real world can be described
approximately by normal distributions or normal curves.

For examples:
Heights, weights,
Life time of a particular product,
Average scores of the graduates from a university
Measurements of manufactured parts,
Errors in scientific measurements,
The amount of rainfall in a city.
7
Frequency distribution table
height
111
113
115
117
119
121
123
125
127
129
131
133
135
frequency
1
3
9
9
15
18
21
14
10
4
3
2
1
N=110
red cells
420440460480500520540560580600620640-
frequency
2
4
7
16
20
25
24
22
16
2
5
1
N=144
8
Frequency distribution of red blood cells
30
Frequency
25
20
15
10
5
0
420- 440- 460- 480- 500- 520- 540- 560- 580- 600- 620- 640-
Red blood cells
9
Frequency distribution of height
Frequency
25
20
15
10
5
0
111 113 115 117 119 121 123 125 127 129 131 133 135
Height
10
The characteristics of frequency distribution
When
n→∞,
smooth
line
y=f(x)
y=f(x) ---density function
(密度函数)
11
Definition of normal distribution A continuous
random variable X having the bell-shaped distribution
is called a normal random variable.
The
mathematical equation for the probability distribution
of the normal variable depends upon the two
parameters,  and , its mean and standard deviation.
We denote the values of the density function of X by
n(x;  , )
n(x;  , ) =
1
 (1/ 2)[( x   ) /  ]2
e
2 
12
The Properties of a Normal Curve





The mode( 众 数 ), which is the point on the
horizontal axis where curve is a maximum, occurs
at x = .
The curve is symmetric about a vertical axis x = .
The curve has its points of inflection (拐点) at x = 
 , is concave downward if  -  < x<  + , and
concave upward otherwise.
The normal curve approaches the horizontal axis
asymptotically as we proceed in entire direction
away from the mean.
The total area under the curve and above the
horizontal axis is equal to 1.
13
The graph of a normal distribution
Bell-shaped
density
function
y =f (x)
y
μ-3σ
μ+3σ
μ
x
σ< 1: steep
μ<0: left
μ>0: right
σ>1: flat
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6.3 Area Under the Normal Curve
0≤F(x)≤1
-∞
x
+∞
The area under a density curve
bounded by the two ordinates x
= x1 and x = x2 equals the
probability that the random
variable X assumes a value
between x1 and x2.
P(x1 < X < x2)
=

x2
n(x;,)dx
x1
-∞
x1
x2
+∞
=

x2
x1
1
2 
e
 ( x2)
2
2
dx
15
The difficulty in calculation of area
F ( x)  
x

1
e
2 

( x )2
2 2
dx,
  x  
Calculation for F(x) :
1) needs integration knowledge;
2) takes time.
How to simplify the calculation of F(x)?
16
Simplify to find out the value of F(x)
If X～N (μ,σ2), then
F ( x)  
x

φ( z ) 

z

～ N ( 0, 1 )

1
e
2 
( x )2
1
e
2
2 2

z2
2
dx,
  x  
dz
calculate φ(z), and list them in a table
(see p670-671， Table A.3).
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Definition of Standard Normal Distribution
The distribution of a normal random variable with
mean zero and variance 1 is called a standard normal
distribution.
parameters: μ= 0 , σ=1
(others are same)
18
The relationship between normal
distribution and standard normal distribution
If X ～N
(μ,σ2),
Z
then
b
P ( a  X  b)  
a

z2

z1
X 

1
2 
1
2
～N (0,1),
e
e
 ( x2)
2
2
2
 z2
dz
dx
( z1 
a

, z2 
b

 P( z1  Z  z2 )
Table A. 3 indicates the area under the
standard curve corresponding to P(Z ≤ x) .
19
)
The relationship between normal distribution
and standard normal distribution
If X ～N (μ,σ2), then
～ N ( 0, 1 )
f (x)
μ-3σ
μ
Φ(z)
X
μ+3σ
Normal distribution
-3
0
Z
3
Standard normal distribution
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Example
Example a：
If z ～N(0,1), find out the probability from the
standard normal distribution table.
1） P (z<-1)= φ(-1) =
0.1587
2） P (z<1) = 1-φ(-1) =
1-0.1587=0.8413
3） P (-1<z<1)=1-2φ(-1) =
1-2*0.1587=0.6826
4） P (|z|>1) =2 φ(-1) =
2*0.1587=0.3174
-1 0 1
Z
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Example b：Estimating probability
If 1000 body weights were drawn from a normal distribution with
meanμ=70kg, sd σ=10kg . What is the portion of being less than 80kg?
Solution: Let
x   Then,
Z 

depends
on
table
0.8413
Φ(z)
70 80
X
0 1
f (z)
Z
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Example c: Estimating probability
If 1000 body weights were drawn from a normal distribution with
meanμ=70kg, sd σ=10kg . What is the portion of being large than 80kg?
Solution: Let
x   Then,
Z 

Φ(z)
70 80
X
-1 0 1
f (z)
Z23
Example d: Estimating probability
If 1000 body weights were drawn from a normal distribution with
meanμ=70kg, sd σ=10kg . What is the portion of being large than
50kg and less than 80kg?
Solution: Let
Z 
x

Then,
Φ(z)
50
70 80
X
-2 -1 0 1
f (z)
Z24
Normal Probability
3 Rule
a
1
2
3
P(|X-|<a) 0.6826 0.9545 0.9973
X
+3

异常区域
正常区域
t
－3
异常区域
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6.4 Applications of
the Normal Distribution
Example 6.13, page 155 The average grade for an
exam is 74, and the standard deviation is 7. Assume
the distribution of the grades of the exam is
approximately normal,

What percent of students scored above 85?
P(X>85)=P((X-74)/7 >(85-74)/7)=P(Z>1.571)

If 12% of students are given A’s, what is the lowest
possible A?
P(X>x)=12%
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Example
6.5 Normal Approximation
to the Binomial
Theorem 6.2 If X is a binomial random
variable with mean  = np and variance  2 =
np(1-p), then limiting form of the distribution
of
X  np
Z
npq
as n  , is the standard
normal distribution n(z; 0, 1).
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Note

The normal distribution with mean  = np and
variance  2 = np(1-p), provides:
(a) A very accurate approximation to the binomial
distribution when n is large and p is not extremely
close to 0 or 1.
(b) A good approximation to the binomial
distribution even when n is small and p is reasonably
close to 0.5.

Often, we may use normal approximation to evaluate
binomial probabilities if both np and n(1–p) are
greater or equal to 5.
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The relationship between normal distribution and
binominal distribution
0.4
n=5 p=0.3
0.35
0.3
n=10 p=0.3
0.25
0.3
0.2
0.25
0.2
0.15
0.15
0.1
0.1
0.05
0.05
0
0 1 2 3 4 5
0
0
2
4
6
8
10
29
n=30 p=0.3
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
10
12
14
16
18
20
30
Binomial,Poisson and Normal distribution
Binomial
n>30
np<5
np>5
n(1-p)>5
n>50
Poisson
λ>20
Normal
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Exercise 14, page 165
A commonly used practice of airline companies is to
sell more tickets than actual seats to a particular flight
because customers who buy ticket do not always
show up for the flight. Suppose that the percentage
of no shows at flight time is 2%. For a particular
flight with 197 seats, a total of 200 tickets were sold.
What is the probability that the airline overbooked
this flight?
32
Solution:
X : the number of customers who will not show up for the
flight among those 200 who bought the tickets
X has a binomial distribution b(x; 200, 0.02) We need to find
P(X  2).
Since n = 200 is quite large, we use normal approximation.
 = np = (200)(0.02) = 4
2= np(1-p) = (200)(0.02)(0.98) =3.92,  = 1.98
x  np
z=
npq
24
=
= -1.01
1.98
P(X  2) = P(Z  – 1.01)  0.1562.
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Example 2

Let X be a random variable whose distribution function is
continuous and strictly increasing. Show that the random
variable Y given by Y=FX(X) has a uniform distribution on [0,1].
Proof: for y [0,1] , we have
P(Y  y)  P( FX ( X )  y)  P( X  FX1 ( y))
 FX ( FX1 ( y))  y
Consequently, Y ~ U [0,1]
Note that FX ( X ) ~ U [0,1] for all random variables. However,
the proof is more complicated when FX is not monotonic or
continuous.
34
Example 3

According to electrical circuit theory, the voltage drop across a
resistor is related to the current flowing through the resistor by the
equation V=IR, where R is the resistance level measured in ohms, I
is the current in amperes, and V is the voltage in volts. In
alternating current systems, the direction and magnitude of the
current change in a cyclical pattern. Hence, if the resistance level is
held constant, the voltage will also vary in a cyclical pattern.
Consequently, measurements of the voltage will have a distribution
that is symmetric about the mean.
Suppose that the measured voltage in a certain electrical circuit
has a normal distribution with mean 120 and standard deviation 2
and five measurements of the voltage are taken. Determine the
probability that two of the measurements lie outside the range 11835
122.
Solution

Let X1, X2, X3, X4, X5 be the voltage measurements. By
assumption, Xj ~ N(120, 22) for each j. Hence,
118  120
122  120
P(118  X j  122)  P(
Z
)
2
2
 P(1  Z  1)  0.6826
Consequently, the probability that the jth measurement lies on
outside the range 118-122 is 0.3174.
Now let N be the number of voltage measurements that lie
outside the range 118-122. Then N~B(5, 0.3174). Hence,
P( N  2)  C52 (0.3174) 2 (0.6826)3  0.3204
36
Example 4

Measurements in science and engineering are always subject to
error. Consider a scientist who is trying to determine the value of
a particular physical constant whose true but unknown value is k.
Suppose that the scientist’s measurement errors are normally
distributed with mean 0 and standard deviation 1; that is, the
scientist’s measurements are of the form k+E, where E ~ N(0, 1).
Intuition suggests that the scientist can improve her estimate of
the physical constant by taking a large number of independent
measurements and averaging the results.
Suppose that n independent measurements are made and let
E1,…, En be the respective measurement errors. Then the estimate
of the physical constant that the scientist obtains by averaging the
37
measurements is
1
{(k  E1 )  (k  E2 )  (k  En )}
n
_
1
1
  nk  ( E1    En )  k  En
n
n
_
_
Where En  ( E1    En ) / n . Hence, En is the error in the scientist’s
estimate of the physical constant.
_
1
E[ En ]  E[ E1    En ]  0
n
_
1
1
D( En )  2 [ D( E1 )    D( En )] 
n
n
_
N (0, 1n )
_
Hence, En ~
. Consequently, the distribution of En becomes
more concentrated around zero as n increases. This means that the
scientist’s estimate of the physical constant improves with the
number of observation taken, which is consistent with your
38
intuition in the case.
Example 5
• Suppose that the scientist of the previous example would like her
estimate to be within 0.01 of the true value with probability 95%.
How many measurements must be taken?
Continuing with the notation of the previous example, the
scientist’s requirement is
_
P(0.01  En  0.01)  0.95
_
Since En 
1
n
Z ,where Z~N(0,1), this requirement is equivalent to
P(0.01 n  Z  0.01 n )  0.95
From tables of the standard normal distribution,(1.96)  0.975 .Hence,
the latter requirement is equivalent to 0.01 n  1.96 That is,
n  38,416
Consequently, if the scientists wishes the estimate to be within
0.01 of the true value with probability 95%, she must take at least
39
38,416 measurements!