EF 507 PS-Chapter 5 FALL 2008

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1

EF 507




PS
-
Chapter
5



FALL 200
8


1.

Consider the following probability distribution. Which of the following is true?


x

0

1

2

3

4

5

6

7

P
(
x
)

0.05

0.16

0.19

0.24

0.18

0.11

0.05

0.02


A)

P
(2 <
X

< 5) = 0.42

B)

P
(
X

> 6) = 0.07

C)

P
(
X



3) = 0.64

D)

P
(
X



6) = 0.93


ANSWER:

A


QUESTIONS
2

AND
3

ARE BASED ON THE FOLLOWING INFORMATION:

Consider the following probability distribution function.



x

0

1

2

3

4

5

6

P
(
x
)

0.07

0.19

0.23

0.17

0.16

0.14

0.04


2
.


What is t
he expected value of
X
?


A)

2.74

B)

0.46

C)

1.78

D)

3.02


ANSWER:

A


3
.


What is the standard deviation of
X
?


A)

13.25

B)

3.64

C)

1.62

D)

4.13


ANSWER:

C


4.
Let
X

and
Y

be two random variables with means
X


and
Y


and variances
2
X


and
2
Y

, respectively. Which of the following statements is not always true?

A)

E
(
X

+
Y
) =
X


+
Y


B)

E
(
X

-

Y
) =
X


-

Y


C)

Var(
X

+
Y
) =
2
X

+
2
Y


D)

Var(
X



Y
) =
2
X

+
2
Y

-

2Cov(
X
,
Y
)


ANSWER:

C


QUESTIONS
5

THROUGH
8

ARE BASED ON THE FOLLOWING
INFORMATION:


2

On average, you re
ceive 2.6 pieces of junk mail a day. Assume that the number of
pieces of junk mail you receive each day follows the Poisson distribution.


5
.

What is the probability that you receive exactly three pieces of junk mail
today?

A)

0.198

B)

0.233

C)

0.218

D)

0.176


ANS
WER:

C

6
.

What is the probability of receiving more than three pieces of junk mail today?


A)

0.123

B)

0.264

C)

0.482

D)

0.242

ANSWER:

B

7

What is the standard deviation of the number of pieces of junk mail you
receive daily?

A)

1.61

B)

2.60

C)

6.76

D)

There is not sufficient inf
ormation to determine this.


ANSWER:

A

8
.

What is the expected number of pieces of junk mail you receive daily?

A)

6.76

B)

1.61

C)

3.22

D)

2.60


ANSWER:

D

9.
The following table displays the joint probability distribution of
X

and
Y
. What is
the covariance between th
e
X

and
Y
?




X




0

1

2


0

0.13

0.08

0.021

Y

1

0.16

0.14

0.03


2

0.07

0.04

0.13

A)

0.29

B)

0.23

C)

0.25

D)

0.21


ANSWER:

A



3

10.
Generally speaking, if two variables are unrelated (as one increases, the other
shows no pattern), the covariance between the variable
s will be


A)

a large positive number.

B)

a large negative number.

C)

a positive or negative number close to zero.

D)

None of the above.


ANSWER:

C


11.
Which of the following is an example of a binomial experiment?


A)

A shopping mall is interested in the income level o
f its customers and is
taking a survey to gather information.

B)

A business firm introducing a new product wants to know how many
purchases its clients will make each year.

C)

A sociologist is researching an area in an effort to determine the
proportion of house
holds with a male head of household.

D)

A study is concerned with the average number of hours worked by high
school students.


ANSWER:

C


12.
If
X

and
Y

are random variables with Var(
X
) = 7.5, Var(
Y
) = 6 and
Cov
(
X
,
Y
) = 4,
then Var(2
X
+3
Y
) is:


A)

132.

B)

37.

C)

88.

D)

33.

ANSWER:

A


13.
Which of the following identities is true If
X

and
Y

are independent random
variables?


A)

E
(2
X
+3
Y
) =
E
(
X
) +
E(Y
) + 5

B)

Var(2
X
+3
Y
) =2Var(
X
) + 3Var(
Y
)

C)

Var(2
X
+3
Y
) = 4Var(
X
) + 9Var(
Y
)

D)

E
(2
X
+3
Y
) = 5
E
(
X
+
Y
)

ANSWER:

C


14.
Consider the following probab
ility distribution function. Compute the mean and
standard deviation of
x
.


x

0

1

2

3

4

5

6

7

P
(
x
)

0.05

0.16

0.19

0.24

0.18

0.11

0.05

0.02




ANSWER:



( )
X
x
xP x

 

2.97




2 2 2 2
( ) 11.49 (2.97) 2.6691
X X
x
x P x
 
    

; hence
1.6337
x




4

15.

In a box of 16 chocolates, there are four chocolates with coconut filling. What is
the probability of choosing four chocolates, one or more of which have
coconut filling?




ANSWER:




Total number of combinations of four chocolates out of 20=
20
4
 
 
 
= 4845.

Number of ways of choosing four chocolates none of which have
coconut=
16 4
1820.
4 0
  

  
  

Let
X
= Number of chocolates which have coconut filling.

Since
P
(
X
= 0) =1820 / 4845 = 0.376, then
P
(
X

1) =

1
-
P
(
X
= 0) =1
-

0.376 =
0.624.


QUESTIONS 16
AND
17 ARE BASED ON THE FOLLOWING INFORMATION:

The following table displays the joint probability distribution of
X

and
Y
.





X




0

1

2


1

0.09

0.08

0.23

Y

2

0.16

0.14

0.05


3

0.07

0.16

0.
02


16.

Find a
nd interpret the conditional probability function of
X
, given
Y

= 1.



ANSWER:


The conditional probability function of
X
, given
Y
=1 is given by
P
(
X
=
x
|
Y
=1).

Since
P
(
X
=
x
|
Y
= 1) =
P
(
X
=
x
Y
=1) /
P
(
Y
=1). Hence,

P
(
X
= 0 |

Y
=1) = 0.09 / 0.40 = 0.225,

P
(
X
= 1 |
Y
= 1) = 0.08 / 0.40 = 0.20, and

P
(
X
= 2 |
Y
= 1) = 0.23 / 0.40 = 0.575.

This means that if we know
Y
= 1, then we know that the probability
X

equals
zero is 0.225, the probability
X
equals one is 0.20, and that the p
robability
X

equals two is 0.575.

17.

Are
X

and
Y

statistically independent? Explain



ANSWER:


Since,
P
(
X
= 0) = 0.32,
P
(
Y
= 1) = 0.4, and
P
(
X
= 0
Y
= 1) = 0.09


P
(
X
=
0)

P
(
Y
= 1), then
X

and
Y

are not statistically independent.



5

QUESTIONS
18

THROUGH
2
8 ARE BASED ON THE FOLLOWING
INFORMATION:

The following table displays the joint probability distribution of two discrete random
variables
X

and
Y
. What is the covariance between

X

and
Y
?




X




1

2

3


0

0.10

0.12

0.06

Y

1

0.05

0.10

0.11


2

0.02

0.16

0.28


1
8.

Determine the marginal probability distribution for
X
.


ANSWER:



x

1

2

3

P
(
x
)

0.17

0.38

0.45


19.

Compute the expected value for
X
.


ANSWER:





2.28
X
x P x

  


2
0.

Compute the standard deviation for
X
.


ANSWER:







2
2 2 2
5.74 2.28 0.5416
X X
x P x
 
    

. Hence,
X

=0.7359.

2
1.

Determine the marginal probability distribution for
Y
.


ANSWER:



y

0

1

2

P
(
y
)

0.28

0.26

0.46


2
2.

Compute the expect
ed value for
Y
.


ANSWER:





1.18
Y
y P y

 



2
3.

Compute the standard deviation for
Y
.


ANSWER:







2
2 2 2
2.10 1.18 0.7076
Y Y
y P y
 
    

. Hence,
0.8412
Y


.

2
4.

Compute the covariance between
X

and
Y
.


ANSWER:



Cov(
X
,
Y
)=


,
X Y
x y
xyP x y



=2.94
-
(2.28)(1.18)=0.2496


6

2
5.

Compute the correlation between
X

and
Y
.


ANSWER:





= Corr(
X
,
Y
) =






Cov,
0.2496
0.4032
0.7359 0.8412
X Y
x y

 

2
6.

Compute the mean for the linear function
W
=2
X
+
Y
.


ANSWER:



2
W X Y
  
 
=2(2.28)+1.18 = 5.74

2
7.

Compute the variance for the linear function
W
= 2
X
+
Y
.


ANSWER:







2 2
2 2 2
2 1
W X Y
  
 
+2(2)(1)Cov(
X
,
Y
) = 4(0.5416) + 0.7076 + 4(0.2496) =
3.8724


2
8.

Are
X
and
Y

statistically independent? Explain.


ANSWER:


The variables
X

and
Y

are not statistically independent since

P
(
X
=1) = 0.17,
P
(
Y
= 0) = 0.28, and
P
(
X
=1 and
Y
= 0) = 0.10

P
(
X
= 1)

P
(
Y
=
0).

QUESTIONS
29

THROUGH
33

ARE BASED ON THE FOLLOWING
INFORMATION:

From past ex
perience, it is known 90% of one
-
year
-
old children can distinguish their
mother's voice from the voice of a similar sounding female. A random sample of 20
one
-
year
-
olds is given this voice recognition test.


29
.

Find the probability at least 3 children do
not recognize their mother's voice.



ANSWER:


P(
X


3 )

=
0.323

(Here
p

= 0.10)


30
.

Find the probability all 20 children recognize their mother's voice.



ANSWER:


P(
X

= 20) = 0.122

(Here
p
= 0.90)


31
.

Let the random variable
X

denote the number of children who do not
recognize their mother's voice. Find the mean of
X
.



ANSWER:




=
nP

= 2


32
.

Let the random variable
X

denote the number of children who do not
recognize their mother's voice. Find the
variance of
X
.



ANSWER:


7


2
(1 )
nP P

 

=
1.8


33
.

Find the probability that at most 4 children do not recognize their mother’s
voice?



ANSWER:


P(
X



4) = 0.957

(Here
p

= 0.10)



QUESTIONS
34

THROUGH
37

ARE BASED O
N THE FOLLOWING
INFORMATION:

The quality of computer disks is measured by sending the disks through a certifier
which counts the number of missing pulses. A certain brand of computer disks
averages 0.1 missing pulse per disk. Let the random variable
X

den
ote the number of
missing pulses.


34
.

What is the distribution of
X
?




ANSWER:


X
has a Poisson distribution with mean 0.1.


35
.

Find the probability the next inspected disk will have no missing pulse.



ANSWER:



P
(
X

= 0) = 0.905


36
.

Find the probabili
ty the next disk inspected will have more than one missing
pulse.



ANSWER:


P
(
X
> 1) = 0.005


37
.

Find the probability neither of the next two disks inspected will contain any
missing pulse.



ANSWER:


We can assume the disks are independent.


P
(
X

= 0 on

first disk and
x

= 0 on second disk) =
P
(
X

= 0).
P
(
X

= 0) =
2
(0.905)

= 0.819


QUESTIONS
38

THROUGH
41

ARE BASED ON THE FOLLOWING
INFORMATION:

The number of people arriving at a bicycle repair shop follows a Poisson distribution
with an a
verage of 5 arrivals per hour. Let
x
represent number of people arriving per
hour)


38
.

What is the probability that seven people arrive at the bike repair shop in a one
hour period of time?



ANSWER:


8

P
(
X
= 7)

=
e

5
7
5
7
!

=
0.1044 or
P
(
X


7)
-

P
(
X


6) = 0.867
-

0.762 = 0.105


39
.

What is the probability that at most seven people arrive at the bike repair shop
in a one hour period of time?



ANSWER:


P
(
X


7) = 0.867


40
.

What is the probability that more than seven people

arrive at the bike repair
shop in a one hour period of time?



ANSWER:


P
(
X
> 7) = 1
-

P(
X


7) = 1
-

0.867 = 0.133


41
.

What is the probability that between 4 and 9 people, inclusively, arrive at the
bike repair shop in a one hou
r period of time?




ANSWER:


P
(4


X



9) =
P
(
X


9)
-

P(
X



3) = 0.968
-

0.265 = 0.703


QUESTIONS
42

THROUGH
49

ARE BASED ON THE FOLLOWING
INFORMATION:

A small grocery store has two checkout lines available to its customers: a regular
checkout line and an express checkout line. Customers with 5 or fewer items are
expected to use the express line. Let
X

and
Y

be the number of customers in the
regular c
heckout line and the express checkout line, respectively. Note that these
numbers include the customers being served, if any. The joint probability distribution
is given in the table below.









42
.

Find the marginal distribution of
X
. What does this distribution tell you?


ANSWER:

The marginal distribution of
X

is:

( 0) 0.28, ( 1) 0.22, ( 2) 0.26, ( 3) 0.24
P X P X P X P X
       


This distribution indicates the likelihood of observing a particular number of
customers in the regular checkout line.


43
.

Find the marginal distribution of
Y
. What does this distribution tell you?


ANSWER:

The marginal distributi
on of Y is:


Y

= 0

Y

=

1

Y

=

2

Y

3

X

= 0

0.06

0.0
4

0.03

0.15

X

=
1

0.09

0.06

0.03

0.04

X

=
2

0.08

0.05

0.01

0.12

X

3

0.07

0.05

0.03

0.09


9

( 0) 0.30, ( 1) 0.20, ( 2) 0.10, ( 3) 0.40
P Y P Y P Y P Y
       

This distribution indicates the likelihood of observing a particular number of
customers in the express checkout line.


44
.

Calculate the conditional distribution of
X

given
Y
.


ANSWER:

The Conditional distributi
on of
X

given
Y
is:



Y = 0

Y

=

1

Y

=

2

Y

3

X

= 0

0.200

0.200

0.300

0.375

X

=
1

0.300

0.300

0.300

0.100

X

=
2

0.267

0.250

0.100

0.300

X

3

0.233

0.250

0.300

0.225


1.00

1.00

1.00

1.00


45
.

What is the
practical benefit of knowing the conditional distribution in
Question
44
?


ANSWER:

If we find that the probability that customers are waiting in the
regular line when the express line is empty is relatively large, we might permit
some customers in the regu
lar line to switch to the express line when it is
empty. Conversely, if we learn that the probability that no customers are
waiting in the regular line when the express line is busy is relatively large, we
might then encourage express line customers to sw
itch to the idle regular line.
The idea here is to reduce the average waiting time of the customers.


46
.

What is the probability that no one is waiting or being served in the regular
checkout line?


ANSWER:

P
(Regular line is empty) =
P
(
X
= 0) = 0.28


47
.

What is the probability that no one is waiting or being served in the express
checkout line?


ANSWER:

P
(Express line is empty) =
P
(
Y
= 0) = 0.30


48
.

What is the probability that no more than two customers are waiting in both
lines combined?


ANSWER:

(Number in regular line + Number in expr
ess line 2)
P




=
( 0,0) ( 0,1) ( 0,2) ( 1,0)
P X Y P X Y P X Y P X Y
           


( 1,1) ( 2,0) 0.36
P X Y P X Y
     



49
.

On average, how many customers would you expect to see in each of these
two lines at the grocery store?


ANSWER:

Expected nu
mber of customers in regular line =
E
(
X
) = 1.60


Expected number of customers in express line =
E
(
Y)

= 1.60