Download Continuous Probability Distributions

Chapter 6 Continuous Probability Distributions
Outline
6.1. Continuous Uniform Distribution
a. Definition (function) & Thm 6.1 (mean, & variance)
b. Ex 6.1 Conference Room booking time (check limits of integration)
6.2. Normal Distribution
a. Introduction – special distribution (wide use): chapter 1 Empirical Rule
b. Definition (function, mean, & variance)
c. Properties of the Normal Curve
6.3. Areas under the Normal Curve
a. Integrating pdf vs Standard Normal Curve (Table Appendix A.3)
Have X (or Z)  find P(Z)
i. Ex 6.2
ii. Ex 6.4
iii. Ex 6.5
iv. ER 2
b. Using the Normal Curve in Reverse
Have P(Z)  find X = + z
i. Ex 6.3
ii. Ex 6.6
6.4. Applications of the Normal Distribution
a. Ex 6.7 Battery Storage problem (Have X (or Z)  find P(Z))
b. Ex 6.10 Gauge Limits (Have P(Z)  find X = + z)
c. Ex 6.13 A/B cut-points problem (Have P(Z)  find X = + z)
● Selected textbook problem
Q5 p157. What kind of normal distribution problem?
6.5. Normal Approximation to the Binomial Distribution
a. When to approximate (as n   : approx good when np > 5 and nq > 5)
b. Ex 6.15 Recovery from rare blood disease
● Selected textbook problem
Q10 p165. What kind of normal distribution problem? Why?
6.6. Exponential Distributions
a. Gamma Function
b. Gamma Distribution function ( & Thm 6.3 for mean & variance)
c. Exponential Distribution: A special case of the Gamma Distribution
function (& Corollary 1 of Thm 6.3 for mean & variance)
d. Relationship to Poisson Process (Waiting Time Distribution where 
=1/)
6.7. Application of the Exponential Distribution (Time to arrival or Time to Poisson
event problems)
a. Ex 6.17 Component failure time (reliability Theory - Time to failure (or
Time to Poisson event) problem)
b. Ex 6.18 Switchboard: Time until 2 telephone calls (Gamma)
6.8. Chi-squared Distribution
a. Definition (function, Mean, & variance)
b. Lab Manual Example
c. Further discussions in section 8.6
Sec. 6.4. Applications of the normal distribution.
Example 6.10. Gauges are used to reject all
components where a certain dimension is not
within the specification 5  d . It is known that this
measurement is normally distributed with mean 1.5
and standard deviation 0.2. Determine the value of
d such that the specifications "cover" 95% of
measurements.
Example 6.11. A certain machine makes
electrical resistors, resistance of which is
normally distributed with mean 40 ohms
and standard deviation 2 ohms. What
percentage of resistors will have a
resistance exceeding 43 ohms?
Example 6.13. The average grade for an
exam is 74 and standard deviation is 7. If
12% of the class are given "A" and grades
follow the normal distribution, what is the
lowest possible "A" and highest possible
"B"?
Sec. 6.5. Normal approximation to the binomial.
Objective:
To use the normal dist to approximate the Binomial dist.
Thm: If X is Binomial r.v. with mean   n p and variance
 2  n p q , then the limiting form of the distribution of
Z 
X n p
n pq
as n   is the standard normal distribution
N(0,1) { another notation n(z;0,1).
Example:
Consider X
Bin (4,0.5) then
The graph of f(x) is like
the normal curve.
p(X=x)
0
1
0.0625
0.25
2
3
0.375
0.25
4
0.0625
Bin(4,0.5)
0.4
P(x=x)
x
0.3
0.2
p(X=x)
0.1
0
0
1
2
3
4
X
Bin (10,0.5)
4
0.205078
5
0.246094
6
7
0.205078
0.117188
8
9
0.043945
0.009766
10
0.000977
Now consider X
Bin (100,0.5)
p(X=x)
X
10
0.043945
0.117188
8
2
3
0.3
0.25
0.2
0.15
0.1
0.05
0
6
0.000977
0.009766
4
0
1
Bin(10,0.5)
2
p(X=x)
0
x
P(x=x)
Now consider X
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
96
90
84
78
72
66
60
54
48
42
36
30
24
18
12
6
p(X=x)
0
P(x=x)
Bin(100,0.5)
X
consider X
Bin (100,0.1)
Bin(100,0.1)
0.14
0.12
0.08
p(X=x)
0.06
0.04
0.02
X
The approximation:
96
88
80
72
64
56
48
40
32
24
8
16
0
0
P(x=x)
0.1
Probabilities of events related to X can be approximated by a normal distribution with
mean np and variance np1  p if the conditions np  5 and n 1  p   5 are
satisfied.
Continuity Correction
It is the adjustment made to an integer-valued discrete random variable when it is
approximated by a continuous random variable. For a binomial random variable, we
inflate the events by adding or subtracting 0.5 to the event as follows:
{X
{X
{X
{X
{X
:X
:X
:X
:X
:X
 4}  { X : 3.5  X  4.5}
 4}  { X : X  3}={X : X  3.5}
 4{ { X : X  4.5}
 4}  { X : X  5}  {X : X  4.5}
 4}  { X : X  3.5} .
The continuity correction should be applied anytime a discrete random variable is
being approximated by a continuous random variable.
Example (6.15 page 163):
The probability that a patient recovers from a
rare blood disease is 0.4. If 100 people are
known to have contracted this disease, what is
the probability that less than 30 survive?
Sol. Let X= # (patient survive)
X
Bin (100,0.4)
  n p  100(0.4)  40 >5,
nq=60>5 and
  n p q  (100)(0.4)(0.6)  4.899
29  100 
p (X  30)   x 0 
(0.4) x (0.6)100  x  0.014775318

 x 
Using normal approximation
p (X  30)  p (X  29)
 p (X  29.5)  p ( Z 
29.5  40
)  p ( Z  2.14)  0.0162
4.899
Example (6.16/163):
A multiple-choice quiz has 200 questions each
with 4 possible answers of which only 1 is the
correct answer. What is the probability that
sheer guess-work yields from 25 to 30 correct
answers for 80 of the 200 problems about
which the student has no knowledge?
Sol.:
Let X= #(correct answers), p=p(correct
answer)=1/4 = 0.25, n=80
X Bin (80,0.25) mean= 20 , Std= 3.873
Using normal approximation
p (25  X  30)
24.5  20
30.5  20
Z 
)
3.873
3.873
 p (1.16  Z  2.71)  .1196
 p (24.5  X  30.5)  p (
the exact answer is:
30  80 
p (25  X  30)   x  25   (0.25) x (0.75)100 x  0.119270502
 x
6.6: Exponential Distribution
Objectives:
To introduce:
 The Exponential Distribution (and the
Waiting time).
 Relationship to Poisson process.
 Applications.
Def.: if X is a continuous r.v. has an exponential
dist. with parameter  then its density function
is given by:
 x
1e  , x 0
f (x )   

 0 , O .W .
where
The mean and variance are
respectively.

 0
and  2 ,
Note:
 The Gamma Distribution is given by:
x


1

x  1e  , x  0

f (x )   ( ) 

, O .W .
 0
The mean and variance are   and  2 ,
respectively.
 If   1 then we have the special case of the
Gamma which is the Exponential
distribution.

Relation to Poisson distribution:
 The exponential distribution describe the
time between Poisson events
Time between arrivals
(Exponential)
Arrival (Poisson)
 mean for Poisson is

and

1


But for Exponential is
(compare the Exp pdf with the
Poisson pdf at x=0)
EXAMPLE 6.18. Phone calls at a particular
switchboard arrive on average of 5 calls per
minute. What is the probability that a call arrive
in one minute?
Example:
If on the average three trucks arrive per hour to be unloaded at a
warehouse. Find the probability that the time between the arrivals of
successive trucks will be less than 5 minutes.
Sol.:
The trucks arrive with average 3 per hour ( Poisson with   3/ hr )
The question is P( time between successive arrivals < 5 min)
X= time between arrivals
1
 )
 3
 p( x  5min)  P( x  5 / 60 hours)
Exp (B 
Then X
5/ 60

0
1


e
x

5/ 60
dx 

1
60
3 e 3 x dx  [ e 3 x ]5/
0
0
1  e15/ 60  0.221199
See example 6.18 page 169.
6.8: Chi square Distribution:
It is a special case of the Gamma dist. when  

2
,   2 where v is
called degrees of freedom.
Def.: if x is Chi-square dist. with v degrees of freedom ( 2 ) then

x
 1
1 

x 2 e 2, x 0


f (x )   ( ) 2 2
 2
 0
, O .W .

The mean and the variance are v and 2v, respectively.