ch2
description
Transcript of ch2
ch2ch2their distributionstheir distributions
Random variables andRandom variables and
Random variablesRandom variables
DistributionsDistributions
Some Important Discrete 2.3
Discrete Probability Distributions
Probability Distributions
nxxx ,,, 21
nin
xpxXP ii ,,2,1,1
)()(
)1
;(~n
xfX i
Uniform
We have a finite set of outcomes
which has the same probability of occurring (equally
likely outcomes).
X is said to have a Uniform distribution and we
each of
Write
So
)1
;(n
xf ibulb,
Example 2.2When a light bulb is selected at random from a boxthat contains a 40-watt bulb, a 60-watt bulb, a 75-watt
and a 100-watt bulb. Find
Solution:
.6,5,4,3,2,1,6
1)6;( xxf
Example 2.3
When a die is tossed, S={1,2,3,4,5,6}.
P(each element of the sample space) = 1/6.
Therefore, we have a uniform distribution, with
Bernoulli trial
A Bernoulli trial is an experiment which has two
Let p = P(success), q = P (failure ) (q=1-p).
‘success’ and ‘failure’.possible outcomes:
0)(1)( failureXandsuccessX
0,1
1,)()(
xpq
xpxpxXPThe pmf of X is
Xkp
0p1
1p
or
伯努利资料
Binomial
each of which must result in either a ‘success’ with
Consider a sequence of n independent Bernoulli trials
probability of p or a ‘failure’ with probability q=1-p.
Let X= the total number of successes in these n trial}.,1,0{ nRX so that
nxppCxXP xnxxn ,,1,0,)1()(
X is said to have a Binomial distribution with parameters
P( the total number of x successes ) =
n and p and we write X~Bin(n, p) or X~b(x;n, p)
1,0)1()( xppxXP xnx
),1( pb
X is said to have a Binomial distribution with parameters
n and p and we write X~Bin(n, p) or X~b(x;n, p)
Special case,
when n=1,we have
We write
B(n,p)1n
b(1, p)
二项分布的图形
The probability that a certain kind of component will survive a given shock test is 3/4. Find the probability that exactly 2 of the next 4 components tested survive.
Example 2.4
Example 2.5
The probability that a patient recovers from a rare blood
disease is 0.4. this disease,
survive,
If 15 people are known to have contracted
what is the probability that (a) at least 10 (b) from 3 to 8 survive, and (c) exactly 5 survive?
Solution:
Solution:
from a shipment.
(b) Suppose that the retailer receives 10 shipments in a
A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that the defective rate of the device is 3%.
Example 2.6
(a) The inspector of the retailer randomly picks 20 items
will be at least one defective item among these 20?What is the probability that there
shipments containing at least one defective device?shipment. month and the inspector randomly tests 20 devices per
What is the probability that there will be 3
Solution :
,2,1,0,!
)( xx
exXP
x
Poisson
The pmf of a random variable X which has a Poisson
0distribution with parameter is given by
).;(~ xpXand we write 泊松资料
电话呼唤次数 交通事故次数商场接待的顾客数
地震 火山爆发 特大洪水
泊松分布的图形
单击图形播放 / 暂停 ESC 键退出
二项分布 泊松分布)( nnp
During a laboratory experiment the average number of radioactive particles passing through a counter in 1 millisecond is 4. What is the probability that 6 articles enter the counter in a given millisecond?
Example 2.7
Solution :
handle at most 15 tankers per day.
Example 2.8
Ten is the average number of oil tankers arriving each day a certain port city. The facilities at the port can
What is the probability that on a given day tankershave to be turned away? Solution :
Solution: }100,75,60,40{S
each element of the sample space occurs with
probability 1/4.
Therefore, we have a uniform distribution:
,4
1)4;( xf .100,75,60,40x
4
3,4;2b 222
4 )4
31()
4
3( C 4
2
4
3
!2!2
!4
Assuming that the tests are independent andSolution:
p=3/4 for each of the 4 tests, we obtain
128
27
)10( XP
9662.01
)83( XP
9
0
)4.0,15;(1x
xb)10(1 XP
0338.0
0.0271-0.9050
8
3
)4.0,15;(x
xb
8
0
2
0
)4.0,15;()4.0,15;(x x
xbxb
Solution :Let X = the number of people that survive.
(a)
(b)
8779.0
Solution :
(a) Denote by X the number of defective
Devices among the 20.Then this X follows a b (x; 20,0.03). Hence
0200 )03.01(03.01 )1( XP
4562.0
)03.0,20;0(1 b)0(1 XP
(b) Assuming the independence from shipment to shipment and denoting by Y .
Y=the number of shipments containing at least one
defective.Then Y~b(y;10,0.4562).
.1602.03103310 )4562.01(4562.0 C)3( YP
Therefore,
)4;6(p
4
7851.08893.0
!6
464
e
Solution :Using the Poisson distribution with x=6 and
, We find from Table 1 that
.1042.0
6
0
5
0
)4;()4;(x x
xpxp
Solution :Let X be the number of tankers arriving eachday.
Then,
)15( XP
0487.09513.01
15
0
)10;(1x
xp)15(1 XP
using Table, we have
Jacob Bernoulli
Born: 27 Dec 1654 in Basel, Switzerland
Died: 16 Aug 1705 in Basel, Switzerland
伯努利资料
泊松资料
Born: 21 June 1781 in Pithiviers, France
Died: 25 April 1840 in Sceaux (near Paris), Fr
ance
Siméon Poisson