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PROBABILITY and STATISTICS EXERCISES ( 03 credits)

CHAPTER 2: PROBABILITY
A. Sample space and event:
Example 1 : a/ Tossing a coin.   {H , T }

b/ Flipping 2 coins.
c/ Rolling a dice.
Example 2: Suppose we roll a fair die, with each of the numbers one, through six represented on a
face of the die, and observe which of the numbers shows
1/ An event A which is {1,2,3,4,5,6} is called ………………………
2/ An event B which is { 2} is called ………………………
3/ An event C which is { the number that shows is even} is called ………………………
4/ An event D which is { the number that shows is 7} is called ………………………
Example 3: A class has 10 students good at Maths, 7 students good at English and 3 students good
at both Maths and English.
A is the event that a student is good at Maths.
B is the event that a student is good at English.
C = A.B = ?
D = A + B = at least one subject=?
Example 4: A is the event : “My class has more than 1 girl”
AC is the event “………………………………”.
Example 5: Tossing a die. Which events are mutually exclusive?
A={1,2,4}
B={5,3}
C={2,4,6}
D={3,5,6}
Example 6: 3 people shoot a target with their gun ( each person shoots a bullet)
A is an event of person 1 hitting the target.

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B is an event of person 2 hitting the target.

C is an event of person 3 hitting the target.
D is an event of having 3 bullets on the target.
E is an event of having 0 bullet on the target.
F is an event of having 1 bullet on the target.
G is an event of having 2 bullets on the target.
H is an event of having at least 1 bullets on the target.
Show D, E,F,G,H by A, B , C and A, B, C ?

B. Counting techniques:
* Examples:
Example 1 : You have 3 shirts and 4 pants. How many outfits (1 shirt, 1 pant) will you have?
Example 2 : You have 3 T-shirts and 4 blouses. Choose 1 to wear. How many cases will you have?
Example 3: a/ How many ways 5 people can sit at a long table?
b/ How many numbers have 4 digit numbers which all have number 4,6,8,1 ? ( for
example 6814)
Example 4: a/ How many numbers have 5 digit numbers which 1 occurs 2 times, 3 occurs 2 times
and 4 occurs 1 time?
b/ How many numbers have 9 digit numbers which 3 occurs 2 times, 8 occurs 3 times and
7 occurs 4 times?
Example 5: Let a set A= {a, b, c} . Find A32  ? ; C32  ?

Example 6: A class has 30 students.

a/ How many groups have 2 students ?
b/ How many groups have 2 students who will be a monitor and vice-monitor ?
Example 7: A box has 6 red pens and 7 blue pens. A student chooses 5 pens from the box randomly.
a. How many ways can he choose 5 pens randomly?
b. How many ways can he choose 5 pens which have 2 red pens? (525)
c. How many ways can he choose 5 pens which have 1 blue pens? (105)
d. How many ways can he choose 5 pens which have at least 1 red pens?

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e. What is the probability of choosing 5 pens which have 2 red pens?

f. What is the probability of choosing 5 pens which have 1 blue pens?

Example 8: A box has 10 products, 8 of them are of type A (good product). The customer check the
products as follows: randomly take 3 products from the box. Find the probability that there are 2 type
A.

Example 9: A box has 8 oranges and 7 apples. Take randomly 5 fruits from the box. Find the
probability that there is at least 1 orange in 5 fruits.

* Exercises : (page 71: 30, 31, 34,35,38, 41*)

1.(30) A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of
zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different
wineries.
a. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways
are there to do this?
b. If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are
there to do this?
c. If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each
variety?
d. If 6 bottles are randomly selected, what is the probability that this results in two bottles of
each variety being chosen?
e. If 6 bottles are randomly selected, what is the probability that all of them are the same
variety?
Solution: a/ person 1: 8 cases
person 2: 7 cases
person 3: 6 cases

So there are 8.7.6 =336 cases A8  8.7.6

3

C82 .C102 .C122

6
b/ C30
2 2 2
c/ C8 .C10 .C12 d/ C306 b/

2. (31) Beethoven wrote 9 symphonies, and Mozart wrote 27 piano concertos. If a university radio
station announcerwishes to play first a Beethoven symphony and then a Mozart concerto, in how
many ways can this be done? Ans: 243= C91.C27
1
 9.27

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3. (34) Computer keyboard failures can be attributed to electrical defects or mechanical defects. A
repair facility currently has 25 failed keyboards, 6 of which have electrical defects
and 19 of which have mechanical defects.
a. How many ways are there to randomly select 5 of these keyboards for a thorough inspection
(without regard to order)? Ans: C255  53130

b. In how many ways can a sample of 5 keyboards be selected so that exactly two have an
electrical defect? => 3 is mechanical defects
Ans: C 62 .C193

c. If a sample of 5 keyboards is randomly selected, what is the probability that at least 4 of

these will have a mechanical defect?
Ans:  4 : at least 4 keyboards … 4 keyboards having mechanical defect s(1 : electrical..): C194 .C 61

5 keyboards having mechanical defects: C195

Let A be the event that exactly two have an electrical defect

There are A  C194 .C61  C195 cases that exactly two have an electrical defect in 5 keyboards

There are   C255 cases that take 5 keyboards randomly

C194 .C61  C195

The probability: P ( A)  5
 0.65658
C25

4. (35) A production facility employs 20 workers on the day shift, 15 workers on the swing
shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these
workers for in-depth interviews. Suppose the selection is made in such a way that any particular
group of 6 workers has the same chance of being selected as does any other
group (drawing 6 slips without replacement from among 45).
a. How many selections result in all 6 workers coming from the day shift? Ans: C206  38760

What is the probability that all 6 selected workers will be from the day shift?
Ans: Let A be the event that all 6 selected workers will be from the day shift
6 selected workers will be from the day shift: A  C206  38760

6 selected workers will be selected randomly:   C456

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C206
The probability: P( A)   0.004759
C456

b. What is the probability that all 6 selected workers will be from the same shift?
Ans: Let A be the event that all 6 selected workers will be from the same shift
6 selected workers will be from the same shift( day + swing+ graveyard):
A  C206  C156  C106

6 selected workers will be selected randomly:   C456

C206  C156  C106

The probability: P ( A)  6
 5.398 103
C45

4c. What is the probability that at least two different shifts will be represented among the
selected workers?
Hint: 6 workers: at least two different shifts2 different shifts or 3 different shifts
2 different shifts in 6 workers: (swing, day) 1 swing- 5day; 2swing -4day,…..
LONG !
Another way: : at least two different shifts (  2 shifts )

Complement is : < 2 shifts  1 shift

Ans: Let A be the event at least two different shifts will be represented among the selected workers
6 selected workers will be from the one shift( day + swing+ graveyard):

A  C206  C156  C106

6 selected workers will be selected randomly:   C456

C206  C156  C106

The probability: P ( A)  1  6
 1  5.398 103  0.994602
C45

d. What is the probability that at least one of the shifts will be unrepresented in the sample of
workers?

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5. (38) A box in a certain supply room contains four 40-W lightbulbs, five 60-W bulbs, and six 75-W
bulbs. Suppose that three bulbs are randomly selected.
a. What is the probability that exactly two of the selected bulbs are rated 75-W?
b. What is the probability that all three of the selected bulbs have the same rating?
c. What is the probability that one bulb of each type is selected?
d. Suppose now that bulbs are to be selected one by one until a 75-W bulb is found. What is the
probability that it is necessary to examine at least six bulbs?
C62 .C91 C43  C53  C63 4.5.6
Ans: a / b/ c/
C153 C153 C153

d/ let A be event of choosing 75w bulbs ; B be event of choosing 40-W lightbulbs or 60-W bulbs.
Examine at least six bulbs  (5B,1A)+(6B,1A)+(7B,1A)+(8B,1A)+(9B,1A)
Case 1: (5B,1A)
9 8 7 6 5 6 18
P  BBBBBA    . . . .  . 
 15 14 13 12 11  10 715

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Similarly: …..
Theprobability is

 9 8 7 6 5  6  9 8 7 6 5 4  6  9 8 7 6 5 4 3 6
 . . . . .   . . . . . .   . . . . . . .
 15 14 13 12 11  10  15 14 13 12 11 10  9  15 14 13 12 11 10 9  8
 9 8 7 6 5 4 3 2 6  9 8 7 6 5 4 3 2 1 6
  . . . . . . . .   . . . . . . . . .
 15 14 13 12 11 10 9 8  7  15 14 13 12 11 10 9 8 7  6
18 8 3 6 1 6
       0.041958
715 715 715 5005 5005 143

6. (41*) An ATM personal identification number (PIN) consists of four digits, each a 0, 1, 2, . . . 8,
or 9, in succession.
a. How many different possible PINs are there if there are no restrictions on the choice of
digits?
b. According to a representative at the author’s local branch of Chase Bank, there are in fact
restrictions on the choice of digits. The following choices are prohibited: (i) all four digits identical
(ii) sequences of consecutive ascending or descending digits, such as 6543 (iii) any sequence starting
with 19 (birth years are too easy to guess). So if one of the PINs in (a) is randomly selected, what is
the probability that it will be a legitimate PIN (that is, not be one of the prohibited sequences)?
c. Someone has stolen an ATM card and knows that the first and last digits of the PIN are 1
and 8, respectively. He has three tries before the card is retained by the ATM (but does not realize
that). So he randomly selects the 2nd and 3rd digits for the first try, then randomly selects a different
pair of digits for the second try, and yet another randomly selected pair of digits for the third try (the
individual knows about the restrictions described in (b) so selects only from the legitimate
possibilities). What is the probability that the individual gains access to the account?
d. Recalculate the probability in (c) if the first and last digits are 1 and 1, respectively.

7*. There are 4 counters at a shop. Six customers are waiting to be charged . Find the probability that
each counter has at least one customer waiting. Ans: 0.3808

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C. Basic formulas:
* Examples:
Example 1: In a certain residential suburb, 60% of all households get Internet service from the local
cable company, 80% get television service from that company, and 50% get both services from that
company. If a household is randomly selected, what is the probability that it gets at least one of these
two services from the company, and what is the probability that it gets exactly one of these services
from the company?
Example 2: Tossing a dice. A is an event that the number on the dice is 1. B is an event that the
number on the dice is odd number. If dice is odd number, find the probability of 1.

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Example 3: There are 20 students in a class. Given that 17 students are good at Maths, 8 students are
good at English and 5 students are good at both Maths and English . Choose a student good at Maths
in this class, find the probability that this student is also good at English.
Example 4: A box has 20 products, 2 of them are type B ( bad product). A man check one product
after another. He will stop checking if he has 2 type B.

a/ Find the probability that checking will stop at the second turn.
.
b/ Find the probability that checking will stop at the third turn.
Example 5: Shooting 3 bullets to a target independently. The probabilities of each bullet which hits
target are 0.6 ; 0.9 ; 0.7 . Find the probability of event which
a/ 3 bullets hit target. b/ 0 bullet hits target
c/ 1 bullet hits target. d/ 2 bullets hit target.
e/ at least 1 bullet hits target.
Example 6a: The probability of an event that each lottery wins is 0,2. Buy 3 tickets, find the
probability of winning 1 ticket.
Example 6b: The probability of an event that each lottery wins is 0,1. Buy 20 tickets, find the
probability of winning 4 tickets.
Example 7: The probability of each student who passes an exam is 0,9. There are 30 students taking
the exam, find the probability that 10 students pass the exam.
Example 8a: Let 𝑃(𝐴) = 0,5; 𝑃(𝐵 ) = 0,4 and 𝑃(𝐴𝐵) = 0,2.
Find the probability that at least one event occurs.
Example 8b: Let 𝑃(𝐴) = 0,5; 𝑃(𝐵) = 0,4 and 𝑃(𝐴𝐵) = 0,3
Find the probability that A and B do not occur.
Example 8c: Let 𝑃(𝐴) = 0,5; 𝑃(𝐵) = 0,4 and 𝑃(𝐴𝐵) = 0,2.
1/ Find the probability that only A occurs.
2/ If B occurs, find the probability that A occurs.
Ans: (0.3;0.5)
Example 8d: Let 𝑃(𝐴) = 0,4; 𝑃(𝐵) = 0,65 and 𝑃(𝐴𝐵) = 0,25
Find the probability that only A occurs.
Example 8e: Let A, B be disjoint and P( A)  0,3 ; P( B)  0, 4 . Which sentences below are wrong?

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a / P( A / B)  0 b / P( AB)  0,12 c / P( A  B)  0, 7 d / P( A B)  0,3

Example 10: Company X supplies 45% of the computers sold and is late 5% of the time.
Company Y supplies 55% of the computers sold and is late 3% of the time.
a/ What is the probability that a computer arrives late?
b/ A computer arrives late - what is the probability that it came from Company X?
Example 11: A person uses his car 40% of the time, walks 40% of the time and takes the bus 20% of
the time as he goes to work. He is late 5% of the time when he drives; he is late 10% of the time
when he walks; and he is late 20% of the time he takes the bus.

a. What is the probability he is late?

b. What is the probability he drives if he was late?
c. What is the probability he took the bus if he was late?
d. *What is the probability he walked if he is on time?

Example 12 : A box has 8 oranges, 7 apples, 5 tomatoes . For oranges, 1% are rotten, whereas 2% of
apples are rotten and 3% of tomatoes are rotten.
A fruit chosen is rotten. What is the probability that it is tomato ?
Example 13: Incidence of a rare disease: Only 1 in 1000 people is afflicted with a rare disease for
which a diagnostic test has been developed. The test is such that when an individual actually has the
disease, positive result will occur 99% of the time, whereas an individual without the disease will
show a positive test result only 2% of the time.
a/ An individual is selected randomly to test . Find the probability that the result is positive?
b/ If a randomly selected individual is tested and the result is positive, what is the probability that the
individual has the disease?
Example 14: A box has 3 bags I and 5 bags II. Each bag I has 3 blue balls and 4 red balls, while
each bag II has 5 blue balls and 7 red balls. Selecting randomly a bag from the box, then pick
randomly 2 balls from this bag.
a/ Find the probability that getting 2 blue balls. Ans 137/924
b/ Find the probability that 2 blue balls which are gotten are from bag I.
Example 15*(exercise 13 page 62) : A computer consulting firm presently has bids out on three
projects. Let Ai  {awarded project i} , for i  1, 2, 3 , and suppose that
P(A1 )  0.22, P( A2 )  0.25 , P( A3 )  0.28 ,
P( A1. A2 )  0.11 , P( A1. A3 )  0.05 , P( A2 . A3 )  0.07, P( A1. A2 . A3 )  0.01 .
Express in words each of the following events, and compute the probability of each event:

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Hint : Example : 𝑃(𝐴) = 0,3; 𝑃(𝐵 ) = 0,25; 𝑃 (𝐶 ) = 0,4 and

𝑃(𝐴𝐵) = 0,1; 𝑃(𝐴𝐶 ) = 0,2; 𝑃(𝐵𝐶 ) = 0,15; 𝑃(𝐴𝐵𝐶 ) = 0,05
1/ Tính xác suất không có biến cố nào trong 3 biến cố 𝐴, 𝐵, 𝐶 xảy ra
2/ Tính xác suất 2 biến cố A và B không xảy ra
3/ Tính xác suất có chỉ có biến cố C xảy ra trong 3 biến cố 𝐴, 𝐵, 𝐶
4/ Tính xác suất có ít nhất một biến cố A hoặc B xảy ra biết biến cố C xảy ra
ĐS:
1/ Tính xác suất không có biến cố nào trong 3 biến cố 𝐴, 𝐵, 𝐶 xảy ra
𝑃(𝐴’𝐵’𝐶’) = 1 − 𝑃(𝐴 + 𝐵 + 𝐶) = 1 − (0,3 + 0,25 + 0,4 − 0,1 − 0,2 − 0,15 + 0,05) = 1 − 0,55
= 0,45
2/ Tính xác suất 2 biến cố A và B không xảy ra
𝑃(𝐴’𝐵’) = 1 − 𝑃(𝐴 + 𝐵) = 1 − (0,3 + 0,25 − 0,1) = 0,55
3/ Tính xác suất có chỉ có biến cố C xảy ra trong 3 biến cố 𝐴, 𝐵, 𝐶
𝑃(𝐴’𝐵’𝐶) = 𝑃(𝐴’𝐵’) − 𝑃(𝐴’𝐵’𝐶’) = 0,55 − 0,45 = 0,1
4/ Tính xác suất có ít nhất một biến cố A hoặc B xảy ra biết biến cố C xảy ra
(𝐴 + 𝐵) 𝑃((𝐴 + 𝐵)𝐶) 𝑃(𝐴𝐶 + 𝐵𝐶)
𝑃 𝐶 = = = 0,75
𝑃(𝐶) 𝑃(𝐶)
𝑃(𝐴𝐶 + 𝐵𝐶) = 𝑃(𝐴𝐶) + 𝑃(𝐵𝐶) − 𝑃(𝐴𝐵𝐶) = 0,2 + 0,15 − 0,05 = 0,3

* Exercises:
8.(49) The accompanying table gives information on the type of coffee selected by someone
purchasing a single cup at a particular airport kiosk
Small Medium Large
Regular 14% 20% 26%
Decaf 20% 10% 10%
Consider randomly selecting such a coffee purchaser.

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a. What is the probability that the individual purchased a small cup? A cup of decaf coffee?
b. If we learn that the selected individual purchased a small cup, what now is the probability
that he/she chose decaf coffee, and how would you interpret this probability?
c. If we learn that the selected individual purchased decaf, what now is the probability that a
small size was selected, and how does this compare to the corresponding unconditional probability of
(a)?
P( A.B) 0.2
P(small size/ purchased decaf)  P(A/ B)    0.5
P( B) 0.2  0.1  0.1

9.(51) One box contains six red balls and four green balls, and a second box contains seven red balls
and three green balls. A ball is randomly chosen from the first box and placed in the second box.
Then a ball is randomly selected from the second box and placed in the first box.
a. What is the probability that a red ball is selected from the first box and a red ball is selected
from the second box? Ans: 6/10 * 8/11
b. At the conclusion of the selection process, what is the probability that the numbers of red
and green balls in the first box are identical to the numbers at the beginning?

10.(59) At a certain gas station, 40% of the customers use regular gas (A1), 35% use plus gas (A2),
and 25% use premium (A3). Of those customers using regular gas, only 30% fill their tanks (event B).
Of those customers using plus, 60% fill their tanks, whereas of those using premium, 50% fill their
tanks.
a. What is the probability that the next customer will request plus gas and fill the tank ? Ans:
0.21
b. What is the probability that the next customer fills the tank? Ans: 0.455
c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus?
Premium? Ans: 0.264; 0.462; 0.275
100%regular ----0,3 fill the tank
40% regular-----
P(A1)= 0,4 regular P(B/A1)= 0,3 , P(B’ not fill../A1)= 0,7
B fill the tank

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customer request P(A2)= 0,35 Plus P(B/A2)= 0,6

P(A3)= 0,25 Premium P(B/A 3)= 0,5

a/ P  A2 .B   P  A2  .P  B | A2   0,35* 0, 6  0, 21

b/ P(B)=0,4*0,3+0,35*0,6+0,25*0,5=0,455
0, 4*0,3 24
c/ P (regular | fills the tank)= P  A1 | B     0.2637362
0, 455 91

0,35*0, 6
P (Plus | fills the tank)= P  A2 | B  
0, 455

P (Premium | fills the tank)=

11. (60)Seventy percent of the light aircraft that disappear while in flight in a certain country are
subsequently discovered. Of the aircraft that are discovered, 60% have an emergency locator,
whereas 90% of the aircraft not discovered do not have such a locator. Suppose a light aircraft has
disappeared.

11a. If it has an emergency locator, what is the probability that it will not be discovered? Ans: 0.0667
b. If it does not have an emergency locator, what is the probability that it will be discovered? Ans:
0.51
P(B/A1)= 0,6 have an emergency locator

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P(A1)= 0,7 discovered

Light aircraft P ( B / A1) = 1-0,6=0,4 not have locator

P(B/A2)=1-0,9=0,1 have an emergency locator

P(A2)=1-0,7 not discovered

P ( B / A 2) = 0,9 not have locator

P( A2 ).P( B / A2 ) 0.3*0.1
11a/ P(not be dis cov ered / emergency locator )  P( A2 / B)  
P( B) 0.7*0.6  0.3*0.1

P ( A1 ).P ( B / A1 ) 0.7 * 0.4

11b/ P ( dis cov ered / not emergency locator )  P ( A1 / B )  
P( B) 0.7 *0.4  0.3*0.9

12.(62) A company that manufactures video cameras produces a basic model and a deluxe model.
Over the past year, 40% of the cameras sold have been of the basic model. Of those buying the basic
model, 30% purchase an extended warranty, whereas 50% of all deluxe purchasers do so. If you learn
that a randomly selected purchaser has an extended warranty, how likely is it that he or she has a
basic model? Ans: 0.286
100%basic---0.3 warranty
40%basic-- warranty
P(A1)= 0,4 basic P(B/A 1)= 0,3
B extended warranty
select a video camera
P(A2)=1-0,4=0,6 deluxe P(B/A 2)= 0,5

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P(extended warranty)= P(B)=0,4*0,3+0,6*0,5=0.42

P  A1.B  P  A1  .P  B | A1  0, 4*0,3
P (basic model | extended warranty)= P  A1 | B    
P  B P  B 0, 42

13.(64) The Reviews editor for a certain scientific journal decides whether the review for any
particular book should be short (1–2 pages), medium (3–4 pages), or long (5–6 pages). Data on recent
reviews indicates that 60% of them are short, 30% are medium, and the other 10% are long. Reviews
are submitted in either Word or LaTeX. For short reviews, 80% are in Word, whereas 50% of
medium reviews are in Word and 30% of long reviews are in Word. Suppose a recent review is
randomly selected.
13/ a. What is the probability that the selected review was submitted in Word format? Ans: 0.66
P(word)  0, 6*0,8  0,3*0,5  0,1*0,3  0.66
b. If the selected review was submitted in Word format, what are the posterior probabilities of
it being short, medium, or long?
P( A1 ).P( B / A1 ) 0.6*0.8
P( short / Word )  P( A1 / B)  
P( B) 0.66

14. (67)There has been a great deal of controversy over the last several years regarding what types of
surveillance are appropriate to prevent terrorism. Suppose a particular surveillance system has a 99%
chance of correctly identifying a future terrorist and a 99.9% chance of correctly identifying someone
who is not a future terrorist. If there are 1000 future terrorists in a population of 300 million, and one
of these 300 million is randomly selected, scrutinized by the system, and identified as a future

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terrorist, what is the probability that he/she actually is a future terrorist? Does the value of this
probability make you uneasy about using the surveillance system? Explain. Ans: 0.003289
P(B/A1)= 0.99 with B -identify terrorist-correct
P(A1)= 1000/300*10^6 future terrorists

a individual
P(B/A2)=1-0,999 with B -identify terrorist-wrong

P(A2)=1- … not future terrorists

P ( B / A 2) = 0,999 identify NOT terrorist

P( A1 ).P ( B / A1 )
P(actually future terrorist / identified future terrorist )  P ( A1 / B ) 
P( B)
1000
6
*0.99
 300.10
1000 1000
6
*0.99  (1  )*(1  0.999)
300.10 300.10 6

CHAPTER 3: DISCRETE RANDOM VARIABLES and PROBABILITY DISTRIBUTIONS

A. Discrete-type random variables:
* Examples:
Example 1: A box has 10 red pens and 4 blue pens. A boy picks up randomly 3 pens in the box. Let
X be the number of red pens in 3 pens which he picks. Determine the pmf and cdf of X
Example 2: A box has 12 apples and 6 oranges. A person takes randomly 2 fruits in the box. Let X
the number of oranges in 2 fruits.
a/ Determine the pmf of X b/ Find E(X) c/ Var(X)
d/ P(X < 1) e/ E(3X+2), E(2X-1) f/ Var (2X+1) ; Var(6-3X)
* Exercises: (49, 51, 59, 60* ,62, 64 page 80)

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Lê Thị Mai Trang 2021

15. Consider the following information about travelers on vacation. There are 20 people check work
email, 30 people use a cell phone to stay connected to work, 18 people bring a laptop with them.
Select randomly 2 people. Let X be the number of people using a cell phone in 2 people who are
selected. Find pmf of X and mean of X.
Ans:
X 0 1 2
p C 2
38
38*30 C302
C 2
68
C682 C682
38*30 C302
EX  0   2* 2 =15/17
C682 C68
16. An individual who has automobile insurance from a certain company is randomly selected. Let Y
be the number of moving violations for which the individual was cited during the last 3 years. The
pmf of Y is
Y 0 1 2 3
p(y) 0.6 0.25 0.1 0.05
a. Compute E(Y), Var(Y) , the standard deviation of Y . Ans: EY=0.6;VY=0.74
b. Suppose an individual with Y violations incurs a surcharge of $100Y 2 . Calculate the
expected amount of the surcharge. Ans: E (100Y 2 )  100 E (Y 2 )  100(0  0.25  22 *0.1  32 *0.05)  110

17. The probability that each product of Company A fails during the warranty period is 0.15. The
company will gain 100 000 VND for a product which don’t need to be warranted and loss 300 000
VND for a warranted product. The company sells 55,000 products. Find the average profit?
X -300 000 100 000
p 0.15 1-0.15
EX=(-300 000)*0.15+100 000 (1-0.15)= 40 000 ( 1 product)
Average profit = 55 000 *EX=2200*10^6
18. In Las Vegas the roulette wheel has a 0 and a 00 and then the numbers 1 to 36 marked on equal
slots; the wheel is spun an a ball stops randomly in one slot. When a player bets 1 dollar on a number,
he receives 36 dollars if the ball stops on this number, for a net gain of 35 dollars; otherwise, he loses
his dollar bet. Find the expected value for his winnings. Ans: -0.052
19. In a second version of roulette in Las Vegas, a player bets on red or black. Half of the numbers
from 1 to 36 are red, and half are black. If a player bets a dollar on black, and if the ball stops on a
black number, he gets his dollar back and another dollar. If the ball stops on a red number or on 0 or
00 he loses his dollar. Find the expected winnings for this bet. Hint: EX is the expectation or expected
value. Ans: -0.052

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Lê Thị Mai Trang 2021

B. Distributions:
* Examples:
Example 1: The probability of any single ticket winning is 0.1. Let X denote the number of winning
lottery tickets when you buy 20 tickets.
a/ Find the probability that you win 3 tickets.
b/ Find EX, Var X
Example 2: The probability of winning a free concert ticket in a raffle is 4%. You buy 10 raffle
tickets. What is the probability that:
a/ You win at least one concert ticket?
b/ You win 2 concert tickets?
c/ You win 10 concert tickets?
Example 3: A box has 50 pens. There are 10 blue pens. Pick 20 pens out. Let X be the number of
blue pens. Find P(X=6) and EX,VX.
Example 4: The probability of any single ticket winning is 0.1. Stop buying if you win for the first
time. Let X denote the number of lottery tickets you buy.
a/ What is the probability that you buy 5 tickets ?
b/ Find EX, VX
Example 5: The probability of any single ticket winning is 0.1. Stop buying if you win 3 tickets. Let
X denote the number of lottery tickets you buy .
a/ What is the probability that you buy 10 tickets ?
b/ Find EX, Var X
Example 6: An oil company conducts a geological study that indicates that an exploratory oil well
should have a 20% chance of striking oil.
a/ What is the probability that the first strike comes on the third well drilled?
b/ What is the probability that the third strike comes on the seventh well drilled?
c/ What is the mean and variance of the number of wells that must be drilled if the oil company wants
to set up three producing wells?
Example 7: The probability of any single ticket winning is 0.005. Let X denote the number of
winning lottery tickets when you buy 1,000 tickets.

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Lê Thị Mai Trang 2021

a/ Find the probability that you win 60 tickets.

b/ Find EX, Var X
Example 8: Vehicles pass through a junction on a busy road at an average rate of 300 per hour.
a/ Find the probability that none passes in a given minute
b/ What is the expectation
Example 9: A company makes electric motors. The probability an electric motor is defective is 0.01.
What is the probability that a sample of 300 electric motors will contain exactly 5 defective motors?
* Exercises:
20. (49) A company that produces fine crystal knows from experience that 10% of its goblets have
cosmetic flaws and must be classified as “seconds.”
a. Among six randomly selected goblets, how likely is it that only one is a second?

Ans: P ( X  6)  C6 *0.1* (1  0.1)  0.354294

1 5

b. Among six randomly selected goblets, what is the probability that at least two are seconds?

Ans: P ( X  2)  1  P ( X  2)  1  P ( X  0)  P ( X  1)  1  (1  0.1)  C6 * 0.1 * (1  0.1)  0.114265

6 1 1 5

c. Compute the expected cosmetic flaws for 10 goblets.

EX=n.p=10*10%=1

21. (50) A particular telephone number is used to receive both voice calls and fax messages. Suppose
that 25% of the incoming calls involve fax messages, and consider a sample of 25 incoming calls.
What is the probability that
a. At most 6 of the calls involve a fax message? Ans: P( X  6)
b. Exactly 6 of the calls involve a fax message?
c. At least 6 of the calls involve a fax message?
d. More than 6 of the calls involve a fax message?
Đs: X là số cuộc gọi có tin nhắn fax , X có pp nhị thức p=0,25 và n= 25
6 6
P ( X  6)   Cnx p x (1  p )n  x   C25x * 0.25 x *(1  0.25) 25 x 
a/ x 0 x0 0,561
25  6
b/ P ( X  6)  C25 * 0.25 * (1  0.25)  0,183
6 6

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Lê Thị Mai Trang 2021

25 25
P ( X  6)   Cnx p x (1  p )n  x   C25x * 0.25 x *(1  0.25) 25 x 
c/ x 6 x6 0,622

e/ EX=n.p=0,25*25=6,25 ;   VX  25* 0, 25* 0, 75  2,16506

22. (68). An electronics store has received a shipment of 20 table radios that have connections for an
iPod or iPhone. Twelve of these have two slots (so they can accommodate both devices), and the
other eight have a single slot. Suppose that six of the 20 radios are randomly selected to be stored
under a shelf where the radios are displayed, and the remaining ones are placed in a storeroom. Let
among the radios stored under the display shelf that have two slots.
a. What kind of a distribution does X have (name and values of all parameters)?
b. Compute P ( X  2), P ( X  2) and P ( X  2) .
c. Calculate the mean value and standard deviation of X
23. A life insurance salesman sells on the average 3 life insurance policies per week. Use Poisson’s
law to calculate the probability that in a given week he will sell
a. 2 or more policies but less than 5 policies
b. Assuming that there are 5 working days per week, what is the probability that in a given day
he will sell one policy?
24.(80) Let X be the number of material anomalies occurring in a particular region of an aircraft gas-
turbine disk. The article “Methodology for Probabilistic Life Prediction of MultipleAnomaly
Materials” (Amer. Inst. of Aeronautics and Astronautics J., 2006: 787–793) proposes a Poisson
distribution for X. Suppose that mean is 4 .
a. Compute both P ( X  4) and P ( X  4) .
b. Compute P (4  X  8) .
c. Compute P (X  8) .
d*. What is the probability that the number of anomalies exceeds its mean value by no more than one
standard deviation?
25. (83) An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the
defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the
approximate distribution of the number who carry this gene? Use this distribution to calculate the
approximate probability that
a. Between 5 and 8 (inclusive) carry the gene.
b. At least 8 carry the gene
26. Last month your company sold 10,000 new watches. Past experience indicates that the probability
that a new watch will need repair during its warranty period is 0.002. Compute the probability that no
more than 5 watches will need warranty work. Ans: 7.1908 x10^(-5)

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27. Of the people passing through an airport metal detector, 0.5% activate it. If there are 500 people
passing the detector, what is the probability that 5 of them activate the detector?
28. (75) Suppose that p  P ( male birth )  0.5 . A couple wishes to have exactly two female children in
their family. They will have children until this condition is fulfilled.
a. What is the probability that the family has x male children?
b. What is the probability that the family has four children?
c. What is the probability that the family has at most four children?
d. How many male children would you expect this family to have?

29.(76) A family decides to have children until it has three children of the same gender. Assuming
P ( B )  P (G )  0.5 , what is the pmf of X = the number of children in the family?

30.(28) A representative from the National Football League's Marketing Division randomly selects
people on a random street in Kansas City, Kansas until he finds a person who attended the last home
football game. Let p, the probability that he succeeds in finding such a person, equal 0.20. And,
let X denote the number of people he selects until he finds his first success.

a. What is the probability that the marketing representative must select 4 people to find one
who attended the last home football game?
Ans: p=0.2; r=1
P(X=4)= (1-0.2)^3 * 0.2=(0.8^3)*0.2=0.1024
b. What is the probability that the marketing representative must select more than 6 people to
find one who attended the last home football game? Ans: 0.262144
6
P ( X  6)  1  P( X  6)  1   Cnr11 *0.2r 1 *(1  0.2) n  r * 0.2
n 1
6
 1   Cn01 *0.20 *(1  0.2) n1 *0.2
n 1
6
 1   (1  0.2) n 1 *0.2  0.262144
n 1

c. How many people should we expect (that is, what is the average number) the marketing
representative needs to select before he finds one who attended the last home football game? And,
while we're at it, what is the variance? Ans : EY=r/p; VY=… with r=1, p=0.2 (negative binomial)

31.(29) Products produced by a machine has a 3% defective rate. What is the probability that the first
defective occurs in the fifth item inspected? Ans: 0.026

CHAPTER 4: CONTINUOUS RANDOM VARIABLES and PROBABILITY

DISTRIBUTIONS

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A. Continous-type random variables:

* Examples:
 k (1  x 2 ) if  1  x  1
Example 1: X is a continuous random variable with pdf: f ( x)  
0 if x  1 or x  1

a / Find k
b / Compute P ( X  1 / 2) ; P (0.5  X  0.8) ; P ( X  0.75)

k x (1  x) if 0  x  2
Example 2: X is a continuous random variable with pdf: f ( x)  
0 if x  2 or x  0

a / Find k
b / Compute P(1  X  2) ; P(X  0.5)
c / Find F ( x)

3
 (1  x ) 1  x  1
2

Example 3: X is a continuous random variable with pdf f ( x)   4

0 else

a/ Find CDF ?
b/ What is the 30th percentile of the distribution?
c/ What is the 50th percentile of the distribution?
3
 (1  x ) x 1
2

Example 4: X is a continuous random variable with pdf f ( x)   4

0 x 1


Find mean, variance and standard deviation ?

* Exercises: (exercise 3,5,6 page 142; 11 page 148)
32. (3) The error involved in making a certain measurement is a continuous rv X with pdf:
 k .(4  x 2 ) ;  2  x  2
f ( x)  
0 otherwise

a. Find k . [Hint: Total area under the graph of f (x) is 1.]

b. Compute P ( X  0) ; P ( 1  X  1) ; P ( X  0.5 or X  0.5) .

c. Find the mean value of measurement, the variance and the standard deviation

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d. What is the 75th percentile of the distribution?

e. Find the Median of X?
33. (5) A college professor never finishes his lecture before the end of the hour and always finishes
his lectures within 2 min after the hour. Let the time that elapses between the end of the hour and the
end of the lecture and suppose the pdf of X is
 k .x 2 ; 0 x2
f ( x)  
0 otherwise

a. Find the value of k .

b. What is the probability that the lecture ends within 1 min of the end of the hour?
c. What is the probability that the lecture continues beyond the hour for between 60 and 90
sec?
d. What is the probability that the lecture continues for at least 90 sec beyond the end of the
hour?
e. Find the expectation of measurement, the variance and the standard deviation
f. What is the 60th percentile of the distribution?
g. Find the Median of X?
34.(6) The actual tracking weight of a stereo cartridge that is set to track at 3 gram on a particular
changer can be regarded as a continuous rv X with pdf :
k 1  ( x  3)2  ; 2 x4
f ( x)    
0 otherwise

a. Find the value of k.

b. What is the probability that the actual tracking weight is greater than the prescribed weight?
c. What is the probability that the actual weight is within .25 gram of the prescribed weight?
d. What is the probability that the actual weight differs from the prescribed weight by more
than .5 gram?
Solution:


a/ 

f ( x )  1 ⇔ ∫ 𝑘(1 − (𝑥 − 3) )𝑑𝑥 = 1 ⇔ 𝑘 =

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Lê Thị Mai Trang 2021

b/ 𝑃(𝑋 > 3) = (1 − (𝑥 − 3) ) 𝑑𝑥 = 0.5

.
c/ 𝑃(3 − 0.25 < 𝑋 < 3 + 0.25) = (1 − (𝑥 − 3) ) 𝑑𝑥 = ⋯
.

d/ 𝑃(|𝑋 − 3| > 0.5) = 𝑃(𝑋 − 3 > 0.5) + 𝑃(𝑋 − 3 < −0.5)

.
= (1 − (𝑥 − 3) ) 𝑑𝑥 + (1 − (𝑥 − 3) ) 𝑑𝑥 = ⋯
.

35.(13) Suppose that in a different traffic environment, the distribution of time headway has the form
k
 ; x 1
: f ( x)   x 4
0 ; x 1

a. Determine the value of k

b. Obtain the cumulative distribution function.
c. Determine the probability that headway exceeds 2 sec and also the probability that headway
is between 2 and 3 sec.
d. Obtain the mean value of headway and the standard deviation of headway.
36.(11) Let X denote the amount of time a book on two-hour reserve is actually checked out, and
suppose the cdf is
0 x0
 2
x
F ( x)   0x2
4
1 2 x

Use the cdf to obtain the following:

a. P ( X  1) ; P (0.5  X  1) ; P ( X  1.5)
b. Find f ( x ) ? [ Hint: f ( x)  F / ( x) ]
c. E( X ) ; Var X , X

B. Uniform distribution, Exponential distribution, normal distribution:

* Examples:

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Lê Thị Mai Trang 2021

Example 1: X  N (0;1). Compute :

Example 2: X  N (3; 4) . Compute :

Example 3: The probability of winning a ticket is 0.1 . Buy 100 tickets.

a/ find the probability of winning 16 tickets.
b/ find the probability of winning at most 16 tickets.
c/ find the probability of winning at least 16 tickets.
Example 4: X has the exponential distribution with mean = 2.
a/ Find variance of X b/ Calculate P ( X  4) ; P (2  X  6)

Example 5: Let T be an exponentially distributed random variable with parameter   ln 2

Find the simplest expression possible for P{T  t} as a function of t for t  0 , and find
P (T  1| T  2)

Example 6: (uniform distribution)?

*Exercises: (Exercise 2 page 142; 14 page 151; 59,60 page 170 ; 28,29,30, 32, 34, 35, 54, 55 page
162 )
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37. (2) Suppose the reaction temperature X (in oC) in a certain chemical process has a uniform
distribution with A= -5 ; B=5
a. Compute P ( X  0) .
b. Compute P ( 2.5  X  2.5) .
c. Compute P( 2  X  3) .

d. Compute mean and variance

 1  1 1
 ; a X b   ; 5  X  5
Solution: f ( x)   b  a   5  5 10
0 ; else 0 ; else

0 0 2.5
1 1 1 1
a/ P ( X  0)  

f ( x ) dx   dx 
5
10 2
b/ P ( 2.5  X  2.5)   10 dx  2
2.5

3
1 1
c/ P(2  X  3)   10 dx  2
2

a  b 5  5 (a  b) 2 (5  5) 2 25
d/ EX   0 ;VX   
2 2 12 12 3

38. (14) The article “Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants”
(Water Research,1984: 1169–1174) suggests the uniform distribution on the interval (7.5, 20) as a
model for depth (cm) of the bioturbation layer in sediment in a certain region.
a. What are the mean and variance of depth?
b. What is the pdf of depth?
c. What is the probability that observed depth is at most 10? Between 10 and 15?
d. What is the probability that the observed depth is within 1 standard deviation of the mean value?
Within 2 standard deviations?
39. The amount of time, in minutes, that a person must wait for a bus is uniformly distributed
between zero and 15 minutes.
a. What is the probability that a person waits fewer than 12.5 minutes?
b. On the average, how long must a person wait? Find the mean, µ, the standard deviation, σ.
c. Ninety percent of the time, the time a person must wait falls below what value?
40. Ace Heating and Air Conditioning Service finds that the amount of time a repairman needs to fix

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Lê Thị Mai Trang 2021

a furnace is uniformly distributed between 1.5 and four hours. Let X = the time needed to fix a
furnace. Then X ∼ U (1.5, 4).
a. Find the probability that a randomly selected furnace repair requires more than two hours.
b. Find the probability that a randomly selected furnace repair requires less than three hours.
c. Find the mean and standard deviation
41. (59) Let X be the time between two successive arrivals at the drive-up window of a local bank. If
X has an exponential distribution with   1 , compute the following:
a. The expected time between two successive arrivals
b. The standard deviation of the time between successive arrivals
c. P( X  4) d. P(2  X  5)

42. (60) Let X denote the distance (m) that an animal moves from its birth site to the first territorial
vacancy it encounters. Suppose that for banner-tailed kangaroo rats, X has an exponential distribution
with parameter   0.01386 (as suggested in the article “Competition and Dispersal from Multiple
Nests,” Ecology, 1997: 873–883).
a. What is the probability that the distance is at most 100 m? At most 200 m? Between 100
and 200 m?
b. Find the mean, µ, and the standard deviation, σ.
43. (28) Let Z be a standard normal random variable and calculate the following probabilities,
drawing pictures wherever appropriate

44. (29) In each case, determine the value of the constant c that makes the probability statement
correct.

45. (30) Find the following percentiles for the standard normal distribution. Interpolate where
appropriate.
a. 91st b. 9th c. 75th d. 25th e. 6th

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46. IQ in a particular population (as measured by a standard test) is known to be approximately

normally distributed with   100 ;   15 . What is the probability that a randomly selected
individual has an IQ of at least 125?
Ans
125  100
c/ P( X  125)  1   ( )  1   (1.67)  1  0.9525  0.0475
15
X 
with Z   N (0;1)

47. (32) Suppose the force acting on a column that helps to support a building is a normally
distributed random variable X with mean value 15.0 kips and standard deviation 1.25 kips. Compute
the following probabilities :
P ( X  15) ; P( X  17.5) ; P( X  10) ; P(14  X  18) ; P( X  15  3)

15  15
a/ P ( X  15)   ( )   (0)  0.5
1.25
17.5  15
b/ P ( X  17.5)   ( )   (2)  0.9772
1.25
10  15
c/ P ( X  10)  1   ( )  1   ( 4)  1  0  1
1.25
18  15 14  15
d/ P (14  X  18)   ( ) ( )   (2.4)   (  0.8)  0.9918  0.2119  0.7799
1.25 1.25
P (| X  15 | 3)  P (15  3  X  15  3)  P (12  X  18)
e/ 18  15 12  15 Giải lao đến 11g25
 ( ) ( )   (2.4)   ( 2.4)  0.9918  0.0082  0.9836
1.25 1.25

48. (34) The article “Reliability of Domestic-Waste Biofilm Reactors” (J. of Envir. Engr., 1995: 785–
790) suggests that substrate concentration (mg/cm3) of influent to a reactor is normally distributed
with   0.3 ;   0.06 .

a. What is the probability that the concentration exceeds .25?

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0.25  0.3
P ( X  0.25)  1   ( )  1   ( 0.83)  1  0.2033  0.7967
0.06
X 
with Z   N (0;1)

b. What is the probability that the concentration is at most .10?
0.1  0.3 X 
b / P ( X  0.1)   ( )   ( 3.33)  0.0004 with Z   N (0;1)
0.06 

49. (35) Suppose the diameter at breast height (in.) of trees of a certain type is normally distributed
with mean 8.8 and standard deviation 2.8 , as suggested in the article “Simulating a Harvester-
Forwarder Softwood Thinning” (Forest Products J., May 1997: 36–41).
a. What is the probability that the diameter of a randomly selected tree will be at least 10 in.?
Will exceed 10 in.?
b. What is the probability that the diameter of a randomly selected tree will exceed 20 in.?
c. What is the probability that the diameter of a randomly selected tree will be between 5 and
10 in.?
d*. What value c is such that the interval (8.8-c; 8.8+c) includes 98% of all diameter values?
E*. If four trees are independently selected, what is the probability that at least one has a
diameter exceeding 10 in.?
Solution:
Let X be the diameter at breast height (in.) of trees
𝑋 ∼ 𝑁(8.8; 2.8 ) ; 𝜇 = 8.8 ; 𝜎 = 2.8
a/ * at least 10
.
𝑃(𝑋 ≥ 10) = 𝑃 ≥ 𝑤𝑖𝑡ℎ 𝑍 = ∼ 𝑁(0,1)
.
= 𝑃(𝑍 ≥ 0.43) = 1 − 𝜙(0.43) = 1 − 0.6664 = 0.3336

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* exceed 10
.
𝑃(𝑋 > 10) = 𝑃 > 𝑤𝑖𝑡ℎ 𝑍 = ∼ 𝑁(0,1)
.
= 𝑃(𝑍 > 0.43) = 1 − 𝜙(0.43) = 1 − 0.6664 = 0.3336
b/ exceed 20
𝑋 − 𝜇 20 − 8.8 𝑋−𝜇
𝑃(𝑋 > 20) = 𝑃 > 𝑤𝑖𝑡ℎ 𝑍 = ∼ 𝑁(0,1)
𝜎 2.8 𝜎
= 𝑃(𝑍 > 4) = 1 − 𝜙(4) = 1 − 1 = 0
c/ between 5 and 10 in
. .
𝑃(5 ≤ 𝑋 ≤ 10) = 𝑃 ≤ ≤ 𝑤𝑖𝑡ℎ 𝑍 = ∼ 𝑁(0,1)
. .
= 𝑃(−1.36 ≤ 𝑍 ≤ 0.43) = 𝜙(0.43) − 𝜙 (−1.36) = 0.6664 − 0.0869 = 0.5795
D*/ What value c is such that the interval (8.8-c; 8.8+c) includes 98% of all diameter values?
8.8 − 𝑐 − 8.8 𝑋 − 𝜇 8.8 + 𝑐 − 8.8 𝑋−𝜇
𝑃(8.8 − 𝑐 ≤ 𝑋 ≤ 8.8 + 𝑐) = 𝑃 ≤ ≤ 𝑤𝑖𝑡ℎ 𝑍 = ∼ 𝑁(0,1)
2.8 𝜎 2.8 𝜎
𝑐
= 𝑃(−𝑐/2.8 ≤ 𝑍 ≤ 𝑐/2.8) = 𝜙(𝑐/2.8) − 𝜙 − = 0.98 (∗)
2.8
We have 𝜙(𝑐/2.8) + 𝜙 − = 1 (**) for example 𝜙(1.96) + 𝜙 (−1.96) = 1
.

Then from (*) and (**) we have 2𝜙 = 1 + 0.98 → 𝜙 = 0.99

. .
𝑐
→ 𝜙(𝑐/2.8) = 𝜙(2.33) → = 2.33 → 𝑐 = 6.524
2.8
E*/ If four trees are independently selected, what is the probability that at least one has a diameter
exceeding 10 in.?
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Lê Thị Mai Trang 2021

* exceed 10
.
𝑃(𝑋 > 10) = 𝑃 > 𝑤𝑖𝑡ℎ 𝑍 = ∼ 𝑁(0,1)
.
= 𝑃(𝑍 > 0.43) = 1 − 𝜙(0.43) = 1 − 0.6664 = 0.3336
Let Y be the number of trees has a diameter exceeding 10 in four trees.Then Y has the binomial
distribution with p=0.3336 ; n=4
𝑃 (𝑌 ≥ 1 ) = 𝑃 (𝑌 = 1 ) + 𝑃 (𝑌 = 2 ) + 𝑃 (𝑌 = 3 ) + 𝑃 (𝑌 = 4 )
= 𝐶 (0.3336) (1 − 0.3336) + 𝐶 (0.3336) (1 − 0.3336) + 𝐶 (0.3336) (1 − 0.3336) +
𝐶 (0.3336) (1 − 0.3336) =0.8028

ans:
49 a/0,3336 b/3.1686x10^-5 c/0.5795 d/ 6.524 e/ 0.8028

50.(54) Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can
be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X
denote the number among these that are nonconforming and can be reworked. What is the probability
that X is
a. At most 30? b. Less than 30? c. Between 15 and 25 (inclusive)?
51.(55) Suppose only 75% of all drivers in a certain state regularly wear a seat belt. A random sample
of 500 drivers is selected. What is the probability that
a. Between 360 and 400 (inclusive) of the drivers in the sample regularly wear a seat belt?
b. Fewer than 400 of those in the sample regularly wear a seat belt?

CHAPTER 6,7: ESTIMATION

*Examples:
Example 1: Measure randomly some male students of university A:
Height (m) 1.5-1.55 1.55-1.6 1.6-1.65 1.65-1.7
The number of 4 27 23 12
students
1/ Compute sample mean and sample standard deviation. Ans: 1.607575 ; 0.0425101112
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Lê Thị Mai Trang 2021

2/ Calculate the sample proportion of students whose height is greater than 1.64 m. Ans: 2/11
3/ Calculate the sample proportion of students whose height is greater than 1.62 m.
Ans:0.5303030
Example 2: Weight of 45 pigs selected randomly in a farm:

xi 35 37 39 41 43 45 47
(kg)
ni 2 6 10 11 8 5 3
(con)
1/ Compute sample mean and sample variance. Ans: 40.9555556 ; 9.725252525
2/ Calculate the sample proportion of pigs greater than 38kg. Ans: 37/45
3/ Calculate the sample proportion of pigs at least 43 kg . Ans: 16/45
Example 3: Weight of ducks chosen randomly in a farm:
Weight (kg) 1.25 1.5 1.75 2 2.25 2.5 2.75 3
The number 2 6 24 35 39 24 14 6
of ducks
Compute the sample mean and sample variance. Ans: 2,185 ; 0,143
Example 4: Data of rice productivity in region A:
Rice 5,1 5,4 5,5 5,6 5,8 6,2 6,4
productivity (t/ha)
( tons / hectare)
Area (hectare) 10 ha 20 30 15 10 10 5
Find the point estimation for the population mean of rice productivity in region A. Ans: 5.6
Example 5: Weight (kg) of some pigs chosen randomly in a farm:
Weight 65-85 85 – 95 95 – 105 105 – 115 115-135
The number 8 40 60 42 10
of pigs
Calculate a 87% CI for true average weight of the pig with the population variance 225 kg 2.
Example 6: Expenditure X (million/month) of some people randomly selected in region A:
X 3.2-3.7 3.7-4.2 4.2-4.7 4.7-5.2 5.2-5.7 5.7- 6.2 6.2-6.7
ni 23 33 55 73 45 22 18
Suppose X has normal distribution.
a/ Calculate a 98% CI for the population mean of expenditure.

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b/ Calculate a 98% upper confidence bound for the population mean of expenditure.
Example 7: Fuel consumption for distance 200 km of some cars chosen randomly:
Fuel consumption 1.9 – 2.1 2.1 – 2.3 2.3 – 2.5 2.5 – 2.7
(l)
The number of cars 5 9 8 3
a/ Calculate an 99% CI for the true average fuel consumption of the car for distance 200km.
b/Calculate an 99% upper confidence bound for the true average fuel consumption of the car for
distance 200km.
c/ Calculate an 99% lower confidence bound for the true average fuel consumption of the car for
distance 200km.
Example 8: Observe 100 workers randomly in a factory , then the sample variance of labour
productivity is 25. If you want to calculate the 99% CI for the labour productivity with precision 0.8,
how many workers do you need to observe more?
a/134 b/156 c/160 d/260
Example 9: Electric energy consumption ( X ) of a household in city A has the normal distribution.
Observe some households in city A randomly:
X(kwh/month) 65-115 115-165 165-215 215-265
The number of 24 36 75 94
households
265-315 315-365 365-415 415-465

97 125 84 75

Find the level confidence if estimate the CI for the true average electric energy consumption of a
household in city A with error 10 kwh/month.
a/ 99% b/95% c/ 90% d/ 87%
Example 10: Measure the length of some products of city A randomly:
X(cm) 53,8 53,81 53,82 53,83 53,84 53,85 53,86 53,87
Số sp 9 14 30 47 40 33 15 12
( ni )

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X has the normal distribution. Find the 98% CI for the true proportion of products greater than 53.84
cm
Example 11: Company M checks the quality of 1200 products randomly before shipment. There are
45 products not meeting the standard. Find an upper confidence bound for the proportion of
products meeting the standard at 99% the confidence level .
Example 12: Interview randomly 400 people in city X. There are 240 people supporting a new rule.
1/ At 95% confidence level, find the CI for the proportion of people supporting the new rule.
Ans: (0,5520; 0,6480)
2/ If the error of the CI for the proportion of people supporting the new rule is 0,057 , find the
confidence level.

* Exercises:
A/ Population mean: (Exercises 5a page 276 ; 12,13,14,15 page 283 ; 34,36a,37a page 293 ; 48,49c
page 297)
52.(5a) Assume that the helium porosity (in percentage) of coal samples taken from any particular seam
is normally distributed with true standard deviation .75. Compute a 95% CI for the true average porosity
of a certain seam if the average porosity for 20 specimens from the seam was 4.85
Solution:
Let  be the true average porosity of a certain seam

n  20  40 ;   0.75 is known ; x  4.85  CASE 1

  1 0.95  1
Confidence level:   0.95   ( )    0.975  z /2  1.96
2 2
 0.75
Error:   z /2  1.96  0.3287019927
n 20

CI for the true average porosity of a certain seam:   (x   ; x   )    (4.521298007; 5.178701993)

---
53.(12) A random sample of 110 lightning flashes in a certain region resulted in a sample average radar
echo duration of .81 sec and a sample standard deviation of .34 sec (“Lightning Strikes to an Airplane in a

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Thunderstorm,” J. of Aircraft, 1984: 607–611). Calculate a 99% (two-sided) confidence interval for the
true average echo duration  , and interpret the resulting interval.
Ans: (0.7263623; 0.8936377)
Solution:
Let  be the true average echo duration

n  110  40 ;  is unknown ; x  0.81 ; s  0.34  CASE 2

  1 0.99  1
Confidence level:   0.99   ( )    0.995  z /2  2.58
2 2
s 0.34
Error:   z /2  2.58  0.0836377
n 110

CI for the true average echo duration:   (x   ; x   )    (0.7263623; 0.8936377)

54.(13) The article “Gas Cooking, Kitchen Ventilation, and Exposure to Combustion Products” (Indoor
Air, 2006: 65–73) reported that for a sample of 50 kitchens with gas cooking appliances monitored during
a one-week period, the sample mean CO2 level (ppm) was 654.16, and the sample standard deviation was
164.43.
a. Calculate and interpret a 95% (two-sided) confidence interval for true average CO2 level in the
population of all homes from which the sample was selected.
Ans: (608.5823293; 699.7376707)
Solution:
Let  be the true average CO2 level in the population of all homes

n  50  40 ;  is unknown ; x  654.16 ; s  164.43  CASE 2

  1 0.95  1
Confidence level:   0.95   ( )    0.975  z /2  1.96
2 2
s
Error:   z /2  45.57767067
n

CI for the true average echo duration:   (x   ; x   )    (608.5823293; 699.7376707)

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b. Suppose the investigators had made a rough guess of 175 for the value of s before collecting data.
What sample size would be necessary to obtain an interval width of 50 ppm for a confidence level of
95%?
Ans: 189
Solution:
Type: CI for tue true average echo duration
s= 175; w=50
  1 0.95  1
Confidence level:   0.95   ( )    0.975  z /2  1.96
2 2
We have :
w  2
s s2
w  2 z /2  n  4 z 2 /2 .  189
n w2

55.(14) The article “Evaluating Tunnel Kiln Performance” (Amer.Ceramic Soc. Bull., Aug. 1997: 59–63)
gave the following summary information for fracture strengths (MPa) of ceramic bars fired in a particular
kiln: x  89.1 ; s  3.73 ; n=169>40
a. Calculate a (two-sided) confidence interval for true average fracture strength using a confidence level
of 95%. Does it appear that true average fracture strength has been precisely estimated?
Solution:
Let  be true average fracture strength

x  89.1 ; s  3.73 ; n=169>40 ;  is unknown

2
 CASE 2
  1 0.95  1
Confidence level:   0.95   ( )    0.975  z /2  1.96
2 2
s
Error:   z /2  0.5623
n

CI for the true average fracture strength:   (x   ; x   )    (88.5376;89.6623)

b. Suppose the investigators had believed a priori that the population standard deviation was about 4
MPa. Based on this supposition, how large a sample would have been required to estimate  to within .5
MPa with 95% confidence?

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Solution (đã sửa)

𝜎 = 4 𝑖𝑠 𝑘𝑛𝑜𝑤𝑛; 𝜀 = 0.5 ; 𝛾 = 0.95 → 𝐶𝐴𝑆𝐸 1

𝜎
𝜀 = 0.5 → 𝑧 ⁄ . = 0.5
√𝑛
4
→ 1,96 ∗ = 0.5 → 𝑛 ≈ 246
√𝑛

56.(15) Determine the confidence level for each of the following large-sample one-sided confidence
bounds:

57.(34) A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress
of 8.48 MPa and a sample standard deviation of .79 MPa (“Characterization of Bearing Strength Factors
in Pegged Timber Connections,” J. of Structural Engr., 1997: 326–332).
a. Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of
all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress?
bỏ b. Calculate and interpret a 95% lower prediction bound for the proportional limit stress of a single
joint of this type.
58.(36 a) The n  26 observations on escape time give a sample mean and sample standard deviation of
370.69 and 24.36, respectively. Calculate an upper confidence bound for population mean escape time
using a confidence level of 95%.
Solution
Let  be population mean escape time
n  26  40 ;  2 is unknown  CASE 3
x  370.69 ; s  24.36 ; one  sided

Confidence level:   0.95    1    1  0.95  0.05  t ;n 1  t0.05 ;25  1.708

s
Error:   t ;n 1  8.159780501
n

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CI for the population mean escape time:   x      378.8497805

59.(37) A study of the ability of individuals to walk in a straight line (“Can We Really Walk Straight?”
Amer. J. of Physical
Anthro., 1992: 19–27) reported the accompanying data on cadence (strides per second) for a sample of
n  20 randomly selected healthy men

A normal probability plot gives substantial support to the assumption that the population distribution of
cadence is approximately normal. Calculate and interpret a 95% confidence interval for population mean
cadence.
Solution
𝑛 = 20 < 40 ; 𝜎 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 → 𝐶𝐴𝑆𝐸 3
𝑥̅ = 0.9255 ; 𝑠 = 0.8097777

CI : (
60.(48) A triathlon consisting of swimming, cycling, and running is one of the more strenuous amateur
sporting events. The article “Cardiovascular and Thermal Response of Triathlon Performance” (Medicine
and Science in Sports and Exercise, 1988: 385–389) reports on a research study involving nine male
triathletes. Maximum heart rate (beats/min) was recorded during performance of each of the three events.
For swimming, the sample mean and sample standard deviation were 188.0 and 7.2, respectively.
Assuming that the heart-rate distribution is (approximately) normal, construct a 98% CI for true mean
heart rate of triathletes while swimming.
61.(49) For each of 18 preserved cores from oil-wet carbonate reservoirs, the amount of residual gas
saturation after a solvent injection was measured at water flood-out. Observations, in percentage of pore
volume, were
23.5 31.5 34.0 46.7 45.6 32.5
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41.4 37.2 42.5 46.9 51.5 36.4

44.5 35.7 33.5 39.3 22.0 51.2
(See “Relative Permeability Studies of Gas-Water Flow Following Solvent Injection in Carbonate
Rocks,” Soc. Of Petroleum Engineers J., 1976: 23–30.) Calculate a 98% CI for the true average amount
of residual gas saturation.
Solution
Input: nhập tất cả các giá trị vào máy với tần số là 1

1. Let  be the true average amount of residual gas saturation.

𝑛 = 18 < 40 ; 𝜎 𝑖𝑠 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 → 𝐶𝐴𝑆𝐸 3
𝑥̅ = 38.66111111 ; 𝑠 = 8.473220021
Confidence level: 𝛾 = 0.98 → 𝛼 = 1 − 𝛾 = 1 − 0.98 = 0.02 → 𝑡 ; =𝑡 . ; = 2.567

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Error: 𝜀 = 𝑡 / ; =⋯
√

CI for the population mean escape time: 𝑥̅ − 𝜀 <𝜇 < 𝑥̅ + 𝜀 →…

62.(56b) Chronic exposure to asbestos fiber is a well-known health hazard. The article “The Acute
Effects of Chrysotile Asbestos Exposure on Lung Function” (Environ. Research,1978: 360–372) reports
results of a study based on a sample of construction workers who had been exposed to asbestos over a
prolonged period. Among the data given in the article were the following (ordered) values of pulmonary
compliance (cm3/cm H2O) for each of 16 subjects 8 months after the exposure period (pulmonary
compliance is a measure of lung elasticity, or how effectively the lungs are able to inhale and exhale):
167.9 180.8 184.8 189.8 194.8 200.2
201.9 206.9 207.2 208.4 226.3 227.7
228.5 232.4 239.8 258.6
Compute a 95% CI for the true average pulmonary compliance after such exposure.

B. Population proportion: (Exercise 19,20,21,22,23,25a page 284 ; 51a,54,56b page 297)

63.(19) The article “Limited Yield Estimation for Visual Defect Sources” (IEEE Trans. on
Semiconductor Manuf., 1997: 17–23) reported that, in a study of a particular wafer inspection process,
356 dies were examined by an inspection probe and 201 of these passed the probe. Assuming a stable
process, calculate a 95% (two-sided) confidence interval for the proportion of all dies that pass the probe.
Solution:
Let p be the proportion of all dies that pass the probe
n  356 ; m  201

m 201
Sample proportion: 
p 
n 356
  1 0.95  1
Confidence level:   0.95   ( )    0.975  z /2  1.96
2 2

p (1 p )

Error:   z /2 =0.0515044
n

CI for the true average echo duration: p  (p  ; p   )  p  (0.5131023; 0.6161111)

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64.(20) The Associated Press (October 9, 2002) reported that in a survey of 4722 American youngsters
aged 6 to 19, 15% were seriously overweight (a body mass index of at least 30; this index is a measure of
weight relative to height). Calculate and interpret a confidence interval using a 99% confidence level for
the proportion of all American youngsters who are seriously overweight.
Ans: (0.5131; 0.6161)
Solution:
Let p be the proportion of all American youngsters who are seriously overweight.
n  4722 ;

Sample proportion: p  15%

  1 0.99  1
Confidence level:   0.99   ( )    0.995  z /2  2.58
2 2

p (1 p )

Error:   z /2 =0.0515044
n

CI for the true average echo duration: p  (p  ; p   )  p  (0.5131023; 0.6161111)

65.(21) In a sample of 1000 randomly selected consumers who had opportunities to send in a rebate claim
form after purchasing a product, 250 of these people said they never did so (“Rebates: Get What You
Deserve,” Consumer Reports,May 2009: 7). Reasons cited for their behavior included too many steps in
the process, amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and
doubts about receiving the money. Calculate an upper confidence bound at the 95% confidence level for
the true proportion of such consumers who never apply for a rebate. Based on this bound, is there
compelling evidence that the true proportion of such consumers is smaller than 1/3? Explain your
reasoning.
66.(22) The technology underlying hip replacements has changed as these operations have become more
popular (over 250,000 in the United States in 2008). Starting in 2003, highly durable ceramic hips were
marketed. Unfortunately, for too many patients the increased durability has been counterbalanced by an
increased incidence of squeaking. The May 11, 2008, issue of the New York Times reported that in one
study of 143 individuals who received ceramic hips between 2003 and 2005, 10 of the hips developed
squeaking.

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Calculate a lower confidence bound at the 95% confidence level for the true proportion of such hips that
develop squeaking.
67.(23) The Pew Forum on Religion and Public Life reported on Dec. 9, 2009, that in a survey of 2003
American adults, 25% said they believed in astrology.
a. Calculate and interpret a confidence interval at the 99% confidence level for the proportion of all adult
Americans who believe in astrology.
b. What sample size would be required for the width of a 99% CI to be at most .05 irrespective of the

value of p ?
Solution
a/ Let p be the true proportion of of all adult Americans who believe in astrology.
𝑛 = 2003 ;
Sample proportion: 𝑝̂ = 25%
  1 0.99  1
Confidence level:   0.99   ( )    0.995  z /2  2.58
2 2

p (1 p )

Error:   z /2 =…
n
CI for p: 𝑝 ∈ (𝑝̂ − 𝜀; 𝑝̂ + 𝜀) →⋯
68.(51) An April 2009 survey of 2253 American adults conducted by the Pew Research Center’s Internet
& American Life Project revealed that 1262 of the respondents had at some point used wireless means for
online access. Calculate and interpret a 95% CI for the proportion of all American adults who at the time
of the survey had used wireless means for online access.
69.(54) It is important that face masks used by firefighters be able to withstand high temperatures because
firefighters commonly work in temperatures of 200–500°F. In a test of one type of mask, 11 of 55 masks
had lenses pop out at 250°. Construct a 90% CI for the true proportion of masks of this type whose lenses
would pop out at 250°.

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CHAPTER 8,9: TESTS OF HYPOTHESES BASED ON A SINGLE SAMPLE

* Examples:
Example 1: A random sample of 50 hens was selected, the weight of each hen determined: sample
average weight x  3,2 kg , sample variance s2  0,25 . The farm used nutrient rich food for hens.
Suppose the investigators had believed a priori that true average weight of a hen would be greater
than 2,8kg. Does the data contradict this prior belief? Assuming normality, test the appropriate
hypotheses using a significance level of .05.
Example 2: A random sample of 25 hens was selected, the weight of each hen determined: sample
average weight x  3,2 kg , sample variance s2  0,25 . Suppose the investigators had believed a
priori that true average weight of a hen would be 3.3 kg. Does the data contradict this prior belief?
Assuming normality, test the appropriate hypotheses using a significance level of .05.
Example 3: Last year, the population mean of electric energy consumption in city A was
150kwh/month. After a public energy savings awareness campaign, obserse on electric energy
consumption of some households.
X 100- 120- 140- 160-
(KWH/month) 120 140 160 180
Households 5 11 9 3
Assuming that X is normally distributed, does the data suggest that the population mean of electric
energy consumption decrease? Carry out a test using a significance level of .05. Ans: t=-3,37
Example 4: In a restaurant customer satisfaction survey of 500 randomly selected customers, there
are 400 customers satisfied with the services.
Does this data provide compelling evidence for concluding that 90% of customers satisfied with the
services? State and test the appropriate hypotheses using a significance level of .05.
Example 5: Select randomly 400 people in city B and determine that 22 people are illiterate. Last
year, there were many classes to help illiterate people Does this data provide compelling evidence for
concluding that less than 5% of people in city B are illiterate ? State and test the appropriate
hypotheses using a significance level of .02. Ans: z=0,46
Example 6: A random sample of 710 products of factory Y reveals that 30 were low quality. Using
new machines, does this suggest that the actual percentage of low quality less than 8%? Carry out a
test of the appropriate hypotheses using a significance level of .02. Ans: z  3, 707

Example 7: Observe a sample of 75 morning shifts, the sample mean of products was 806
products/shift and the sample standard deviation was 185. Observe a sample of 100 evening shifts,

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the sample mean of products was 723 products/shift and the sample standard deviation was 164.
Does it appear that true average products of morning shift is different the evening shift? Test the
appropriate hypotheses at significance level .01
Example 8: Observe a sample of 5 fields not using fertilizers, the sample mean corn yeild was 44.8
ton/ha and the sample standard deviation was 10.825. Observe a sample of 5 fields using fertilizers,
the sample mean corn yeild was 46.6 ton/ha and the sample standard deviation was 13.01153. Does
it appear that the true average corn yeild of field not using fertilizers is less than field using
fertilizers ? Test the appropriate hypotheses at significance level .05
Example 9: Among 529 helmets of company A, there are 95 helmets not meeting quality standards.
Among 400 helmets of company B, there are 95 helmets not meeting quality standards. Carry out a
test of hypotheses at the 3% significance level to decide whether the quality standards of helmets of
company A differs from the helmets of company B.
Example 10: Among 500 male students, there are 45 good students. Among 400 female students,
there are 50 good students. Does it appear that the proportion of good students among male students
is greater than among female students? Test the appropriate hypotheses at significance level .02
Z=-1,698

* Exercises:
A. Test about population mean  : (Page 321: exercises 19a, 20, 22b, 23, 24, 26, 28, 29a, 31, 32)
70.(19a) The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was
determined, resulting in x  94.32 . Assume that the distribution of the melting point is normal with
  1.2 . Test H 0 :   95 versus H a :   95 using a two-tailed level .01 test.
71.(20) Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price
of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase
arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what
is advertised. A random sample of 50 bulbs was selected, the lifetime of each bulb determined, and the
appropriate hypotheses were tested using Minitab, resulting in the accompanying output.
Variable
N Mean StDev SEMean
lifetime
50 738.44 38.20 5.40
What conclusion would be appropriate for a significance level of .05?
71/ Solution: n=50 >40 ,  2 is unknown  case 2

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 : the true average lifetime

H 0 :   750
H a :   750

  0, 05   (z )  1    1  0.05  0.95  z  1.65

𝑥̅ − 750
𝑧= . √𝑛 = −2.1398309
𝑠
Because −2.1398309 < −1.65 : reject H0, accept Ha:   750 . The claim is right ( the true average lifetime
is less than 750)

72.(22 b) The article “The Foreman’s View of Quality Control” (Quality Engr., 1990: 257–280) described
an investigation into the coating weights for large pipes resulting from a galvanized coating process.
Production standards call for a true average weight of 200 lb per pipe. The accompanying descriptive
summary and boxplot are from Minitab.
Variable N Mean x Median TrMean StDev s SEMean
ctg wt 30<40 206.73 206.00 206.81 6.35 1.16
A normal probability plot of the data was quite straight. Use the descriptive output to test the appropriate
hypotheses with   0.02
72/ Solution: n=30 <40 ,  2 is unknown  case 3
 : true average weight

H 0 :   200
H a :   200

  0, 02  t /2,n 1  t0,01;29  2.462

x  200
t . n = 5.80499
s
Because t  t /2,n 1 : reject H0, accept Ha:   200 . The claim is wrong

(note: x  206.73  0  200    200 )

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73.(23) Observations on escape time (sec) for oil workers in a simulated exercise, n=26 , from which the
sample mean and sample standard deviation are 370.69 and 24.36, respectively. Suppose the investigators
had believed a priori that true average escape time would be at most 6 min. Does the data contradict this
prior belief? Assuming normality, test the appropriate hypotheses using a significance level of .05.

74.(24) Reconsider the sample observations on stabilized viscosity of asphalt specimens: (2781, 2900,
3013, 2856, and 2888). Suppose that for a particular application it is required that true average viscosity
be 3000. Does this requirement appear to have been satisfied? Test the hypotheses with   0.05 .
74 / Solution: n=5<40 ; x  2887.6 ; s  84.02559134 ;  2 is unknown  case 3
 true average viscosity
H 0 :   3000
H a :   3000

  0, 05  t / 2, n 1  t0,025;4  2.776

x  200
t . n = -2.9911606
s
t  2.9911606  2.776 : reject H0, accept Ha:   3000 . The claim is wrong

(note: x  2887.6  0  3000    3000 ) Break : 10: 30

75.(26) To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45
specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is
then measured, yielding a sample average penetration of x  52.7 and a sample standard deviation of
s  4.8 . The conduits were manufactured with the specification that true average penetration be at most
50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been
met. What would you conclude?   0.03
Ans: z=3,7733647

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76.(28) Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing
good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery
with minimal aftereffects so that horses can be left unattended. The article “A Field Trial of Ketamine
Anesthesia in the Horse” (Equine Vet. J., 1984: 176–179) reports that for a sample of horses to which
ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down)
time was 18.86 min and the sample standard deviation was 8.6 min. Does this data suggest that true
average lateral recumbency time under these conditions is less than 20 min? Test the appropriate
hypotheses at level of significance .10
77.(29) The article “Uncertainty Estimation in Railway Track LifeCycle Cost” (J. of Rail and Rapid
Transit, 2009) presented the following data on time to repair (min) a rail break in the high rail on a curved
track of a certain railway line.
159 120 480 149 270 547 340 43 228 202 240 218
A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the
population distribution of repair time is at least approximately normal. The sample mean and sample
standard deviation are 249.7 and 145.1, respectively. Is there compelling evidence for concluding that true
average repair time exceeds 200 min? Carry out a test of hypotheses using a significance level of .05
Solution: (đếm được 12 giá trị)
 : a true average repair time
𝐻 : 𝜇 = 200 (𝑤𝑖𝑡ℎ 𝜇 = 200)
𝐻 : 𝜇 > 200 (𝑜𝑛𝑒 − 𝑡𝑎𝑖𝑙𝑒𝑑)
n=12 <40 ; 𝑥̅ = 249.7; s=145.1  case 3
𝑡 , = 𝑡 . , = 1.796
249.7 − 200
𝑡= √12 = 1.1865
145.1
1.1865 < 1.796 𝑡ℎ𝑒𝑛 𝑎𝑐𝑐𝑒𝑝𝑡 Ho: 𝜇 = 200
 true average repair time is 200 min
78.(31) A well-designed and safe workplace can contribute greatly to increased productivity. It is
especially important that workers not be asked to perform tasks, such as lifting, that exceed their
capabilities. The accompanying data on maximum weight of lift (MAWL, in kg) for a frequency of four
lifts/min was reported in the article “The Effects of Speed, Frequency, and Load on Measured Hand
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Forces for a Floor-to-Knuckle Lifting Task” (Ergonomics, 1992: 833–843); subjects were randomly
selected from the population of healthy males ages 18–30. Assuming that MAWL is normally distributed,
does the data suggest that the population mean MAWL exceeds 25? Carry out a test using a significance
level of .05.
25.8 36.6 26.3 21.8 27.2
79.(32) The recommended daily dietary allowance for zinc among males older than age 50 years is 15
mg/day. The article “Nutrient Intakes and Dietary Patterns of Older Americans: A National Study” (J. of
Gerontology, 1992: M145–150) reports the following summary data on intake for a sample of males age
65–74 years: n  115 ; x  11.3 ; s  6.43 . Does this data indicate that average daily zinc intake in the
population of all males ages 65–74 falls below the recommended allowance?   0.02
B. Test concerning a population proportion p : (37a,38ab,39,42ab page 327)
80.(37a) A common characterization of obese individuals is that their body mass index is at least 30
[BMI=weight/(height)2, where height is in meters and weight is in kilograms]. The article “The Impact of
Obesity on Illness Absence and Productivity in an Industrial Population of Petrochemical Workers”
(Annals of Epidemiology, 2008: 8–14) reported that in a sample of female workers, 262 had BMIs of less
than 25, 159 had BMIs that were at least 25 but less than 30, and 120 had BMIs exceeding 30. Is there
compelling evidence for concluding that more than 20% of the individuals in the sampled population are
obese? State and test appropriate hypotheses using the rejection region approach with a significance level
of .05
80 / Solution: p: true proportion of the obese individuals
H 0 : p  0, 2
more than
H a : p  0, 2

  0, 05   ( z )  1    0,95  z  1.65

n=262+159+120= 541 , m=120

120
 0.2) 541
( p  p0 ) n
(
z  541  1,268
p0 (1 p0 ) 0.2(1 0.2)

z  z : accept H0: p  0.2 . : True proportion of the obese individuals equals 20%. The claim is wrong.

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81.(38) A manufacturer of nickel-hydrogen batteries randomly selects n=100 nickel plates for test cells,
cycles them a specified number of times, and determines that m=14 of the plates have blistered.
Does this provide compelling evidence for concluding that different 10% of all plates blister under such
circumstances? State and test the appropriate hypotheses using a significance level of .05.

81 / Solution: p: true proportion of the plates have blistered

H 0 : p  0,1
something other than
H a : p  0,1


  0, 05   ( z / 2 )  1   0,975  z /2  1.96
2
14
 0.1) 100
( p  p0 ) n
(
z  100 = 1.3333333
p0 (1 p0 ) 0.1(1 0.1)

z  z / 2 : accept H0: p  0.1 . The claim is wrong.

82.(39) A random sample of n=150 recent donations at a certain blood bank reveals that m=82 were type
A blood. Does this suggest that the actual percentage of type A donations differs from p 0 =40%, the
percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a
significance level of .01.
82 / Solution: p: percentage of the population having type A blood

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H 0 : p  04,
H a : p  0, 4


  0, 01   ( z /2 )  1   0,995  z /2  2.58
2
82
 0.4) 150
( p  p0 ) n
(
z  150 = 3.66667
p0 (1 p0 ) 0.4(1 0.4)

z  z /2 : reject H0, accept Ha: p  0.4 . The claim is right.

82
(note: p ^   0,546  p0  0, 4  p  0, 4 ) .
150
Fhqutex.hcmute.edu.vn  chapter 6,7 solution estimation
83.(42ab) With domestic sources of building supplies running low several years ago, roughly 60,000
homes were built with imported Chinese drywall. According to the article “Report Links Chinese Drywall
to Home Problems” (New York Times, Nov. 24, 2009), federal investigators identified a strong association
between chemicals in the drywall and electrical problems, and there is also strong evidence of respiratory
difficulties due to the emission of hydrogen sulfide gas. An extensive examination of 51 homes found that
41 had such problems. Suppose these 51 were randomly sampled from the population of all homes having
Chinese drywall.
a. Does the data provide strong evidence for concluding that more than 50% of all homes with Chinese
drywall have electrical/environmental problems? Carry out a test of hypotheses using   0.01 .
b. Calculate a lower confidence bound using a confidence level of 99% for the percentage of all such
homes that have electrical/environmental problems.

B. P-value:

C. Tests for a diffeference between two population means: (2b, 3, 6a, 7, 8a page 354 ; 19,28,32 page
362)
84.(2) The National Health Statistics Reports dated Oct. 22, 2008, included the following information on
the heights (in.) for non-Hispanic white females:

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Age Sample size Sample mean

20-39 866 64.9
60 and older 934 63.1
Let 1 denote the population mean height for those aged 20–39 and 2 denote the population mean height

for those aged 60 and older. Interpret the hypotheses H 0 : 1  2  1 and H a : 1  2  1 , and then carry
out a test of these hypotheses at significance level .001 using the rejection region approach
85.(3) Let 1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let 2
denote the true average
tread life for an economy brand of the same size. Test H 0 : 1  2  5000 versus H a : 1  2  5000 at level

.01, using the following data: m  45, x  42500, s1  2200 , n  45 , y  36800, s2  1500 .
86.(6) An experiment to compare the tension bond strength of poly-mer latex modified mortar (Portland
cement mortar to which
polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in
x  18.12 kgf / cm 2 for the modified mortar (m=40) and y  16.87 kgf / cm 2 for the unmodified mortar

(n=32). Let 1 and 2 be the true average tension bond strengths for the modified and unmodified
mortars, respectively. Assume that the bond strength distributions are both normal.
Assuming that  1  1.6 and  2  1.4 , test H 0 : 1  2  0 versus H a : 1  2  0 at level .01 .
87.(7) Is there any systematic tendency for part-time college faculty to hold their students to different
standards than do full-time faculty? The article “Are There Instructional Differences Between Full-Time
and Part-Time Faculty?” (College Teaching, 2009: 23–26) reported that for a sample of 125 courses
taught by full-time faculty, the mean course GPA was 2.7186 and the standard deviation was .63342,
whereas for a sample of 88 courses taught by part-timers, the mean and standard deviation were 2.8639
and .49241, respectively. Does it appear that true average course GPA for part-time faculty differs from
that for faculty teaching full-time? Test the appropriate hypotheses at significance level .01
88.(8a) Tensile-strength tests were carried out on two different grades of wire rod (“Fluidized Bed
Patenting of Wire Rods,” Wire J.,
June 1977: 56–61), resulting in the accompanying data.

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Grade Sample size Sample Mean Sample SD

(kg/mm2)
AISI 1064 m=129 x  107.6 s1  1.3

AISI 1078 n=129 y  123.6 s2  2

Does the data provide compelling evidence for concluding that true average strength for the 1078 grade
exceeds that for the 1064 grade by more than 10 kg/mm 2? Test the appropriate hypotheses ,   0.02
CHECK z=-28.5686667
z(alpha)=2.06 The claim is wrong

89.(19) Suppose 1 and 2 are true mean stopping distances at 50 mph for cars of a certain type equipped
with two different types of braking systems. Use the two-sample t test at significance level .01 to test
H 0 : 1  2  10 versus H 0 : 1  2  10 for the following data:

m  6 , x  115.7 , s1  5.03 , n  6 , y  129.3 , s2  5.38 .

90.(28) As the population ages, there is increasing concern about accident-related injuries to the elderly.
The article “Age and Gender Differences in Single-Step Recovery from a Forward Fall” (J. of
Gerontology, 1999: M44–M50) reported on an experiment in which the maximum lean angle—the
furthest a subject is able to lean and still recover in one step— was determined for both a sample of
younger females (21–29 years) and a sample of older females (67–81 years). The following observations
are consistent with summary
data given in the article:
YF: 29, 34, 33, 27, 28, 32, 31, 34, 32, 27
OF: 18, 15, 23, 13, 12
Does the data suggest that true average maximum lean angle for older females is more than 10 degrees
smaller than it is for younger females? State and test the relevant hypotheses at significance level .10
91.(32) The degenerative disease osteoarthritis most frequently affects weight-bearing joints such as the
knee. The article “Evidence of Mechanical Load Redistribution at the Knee Joint in the Elderly when
Ascending Stairs and Ramps” (Annals of Biomed. Engr., 2008: 467–476) presented the following
summary data on stance duration (ms) for samples of both older and younger adults.

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Age Sample Size Sample Mean Sample SD

Older 28 801 117
Younger 16 780 72
Assume that both stance duration distributions are normal.
a. Calculate and interpret a 99% CI for true average stance duration among elderly individuals.
b. Carry out a test of hypotheses at significance level .05 to decide whether true average stance duration is
larger among elderly individuals than among younger individuals.

D. Inference concerning a diffeference between population proportion: (49, 51, 53a page 380 )
92.(49) Is someone who switches brands because of a financial inducement less likely to remain loyal
than someone who switches without inducement? Let p1 and p2 denote the true proportions of switchers
to a certain brand with and without inducement, respectively, who subsequently make a repeat purchase.
Test H 0 : p1  p2  0 versus H 0 : p1  p2  0 using   0.01 and the following data:
m = 300 number of success = 30
n = 200 number of success = 180
(Similar data is given in “Impact of Deals and Deal Retraction on Brand Switching,” J. of Marketing,
1980: 62–70.)

93.(51) It is thought that the front cover and the nature of the first question on mail surveys influence the
response rate. The article “The Impact of Cover Design and First Questions on Response Rates for a Mail

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Survey of Skydivers” (Leisure Sciences, 1991: 67–76) tested this theory by experimenting with different
cover designs. One cover was plain; the other used a picture of a skydiver. The researchers speculated that
the return rate would be lower for the plain cover.   0.1
Cover Number Sent Number Returned
Plain 207 104
Skydiver 213 109
93 /Solution: p1 : number returned for plain; p2 : number returned for skydiver
H 0 : p1  p2
H a : p1  p2

 (z )  1    0.9  z  1.29

 104 109  x y 104 109 p1  p 2

p1  ;
p2  ; p 
With 207 213 n1  n2 207  213 z   =-0.191036
q  1 p
 p. q  1  1 
1
 n n 
2

Z= -0.191036 >𝑧 = −1.29: accept H 0 : p1  p2 . The claim is wrong

94.(53) Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been
anecdotal reports of gastrointestinal problems associated with olestra consumption, a randomized, double-
blind, placebo-controlled experiment was carried out to compare olestra potato chips to regular potato

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chips with respect to GI symptoms (“Gastrointestinal Symptoms Following Consumption of Olestra or

Regular Triglyceride Potato Chips,” J. of the Amer. Med. Assoc., 1998: 150–152). Among 529 individuals
in the TG control group, 17.6% experienced an adverse GI event, whereas among the 563 individuals in
the olestra treatment group, 15.8% experienced such an event. Carry out a test of hypotheses at the 5%
significance level to decide whether H 0 : p1  p2 the incidence rate of GI problems for those who consume
olestra chips according to the experimental regimen differs from the incidence rate for the TG control
treatment
94/ Solution: p1 : GI event of TG control group; p2 : GI event of the olestra treatment group
H 0 : p1  p2
H a : p1  p2

 (z /2 )  1  ( / 2)  0.975  z /2  1.96

 x x y y x y 0.176*529  0.158*563 1
p1  17.6%   ;
p2  15.8%   ; p   
With n1 529 n2 563 n1  n2 529  563 6
q  1 p


p1  p 2
z   =0.7976446 . Because z  z / 2 : accept H 0 : p1  p2
p. q  1  1 
 n n 
1 2

GI problems for those who consume olestra chips is equal to for the TG control treatment

E. Analysis of paired data:

95.A survey about the income and expenses (million/month) of some households in city A has the data:
Income 15 18 19 21 23 27 29 19 17 24 22 28 35 38 40
Expenses 12 15 15 17 21 25 22 18 17 21 18 21 30 25 26
Difference 3 3 4 4 2 2 7 1 0 3 4 7 5 13 14
a/ Find the 95% CI for the true mean difference between income and expenses of the household
b/ Is the true mean difference between income and expenses of the household something other than zero
? With significance level is 0.02
Solution

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 Lê Thị Mai Trang 2021

Let  be the true mean difference between income and expenses of the household
n  15  40 ;  2 is unknown  CASE 3
x  4.8 ; s  4.0213714 ; two  sided

Confidence level:   0.95    1    1  0.95  0.05  t /2;n 1  t0.025 ;14  2.145

s
Error:   t /2;n1  2.22718
n

CI for the population mean: x      x    2.5728    7.027182

b/ Is the true mean difference between income and expenses of the household something other than zero ?
With significance level is 0.02
H0 :   0
Ha :   0

  0.02  t /2;n 1  t0.01 ;14  2.624

x 0
z . n  4.622880
s
Because 4.622880  2.624 : reject H0, accept Ha:   0 .

the true mean difference between income and expenses of the household something other than zero
96. (example 9.10) Adding computerized medical images to a database promises to provide great
resources for physicians. However, there are other methods of obtaining such information, so the issue of
efficiency of access needs to be investigated. The article “The Comparative Effectiveness of Conventional
and Digital Image Libraries”(J. of Audiovisual Media in Medicine, 2001: 8–15) reported on an experiment
in which 13 computer-proficient medical professionals were timed both while retrieving an image from a
library of slides and while retrieving the same image from a computer database with a Web front end.

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REVIEW CHAPTER 12: SIMPLE LINEAR REGRESSION

*Examples:
Example 1: Observe a sample (X,Y)
X 1 3 4 6 8 9 11 14
Y 1 2 4 4 5 7 8 9
Find the linear regression equation of X and Y? When X=12, find Y
Example 2: Observe randomly some trees. Let X be the radius and Y be the height of these trees.
Y (m) 3 4 5 6 7 8
X (cm)
21 2 5
23 3 11
25 8 15 10
27 4 17 3
29 7 12
Determine the correlation coefficient for the above set of results and the equation of the linear regression
of Y on X ?
ANS
a=-4,911554

b=0.424155
r = 0,8148

* Exercises:
97. Let X be the price of a product in 10 regions. Let Y be the number of products sold per month.
X 34 35 36 36 35 37 38 40 39 40
Y 6 5,9 5,7 5,7 6,2 6,7 5,6 5,5 5,4 5,2

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Determine the correlation coefficient for the above set of results and the equation of the linear
regression of Y on X ?
98.
X 1 3 4 6 8 9 11 14
Y 1 2 4 4 5 7 8 9
Find the equation of the linear regression of Y on X ? Use your regression line to estimate the value of Y
when X=12? Ans: y = 0,6364 x + 0,5455 ; y = 8,1823
99. Let X (cm) be the height and Y (kg) be the weight of some students chosen randomly:
X 155 156 158 159 159 160 160 162 164 165
Y 48 47 48 49 48 50 51 51 53 54
Determine the correlation coefficient for the above set of results and the equation of the linear regression
of Y on X ?
Ans: r= 0.9383340525 ; y = -59.65720525+0.6855895197 x
100. Toughness and fibrousness of asparagus are major determinants of quality. This was the focus of a
study reported in “Post-Harvest Glyphosphate Application Reduces Toughening, Fiber Content, and
Lignification of Stored Asparagus Spears” (J. of the Amer. Soc. of Hort. Science, 1988: 569–572). The
article reported the accompanying data (read from a graph) on x shear force (kg) and y percent fiber dry
weight

x 48 57 60 72 81 85 109 121 137 148 149 184 185 187

y 2.1 2.28 2.34 2.53 2.28 2.62 2.5 2.66 2.8 3.01 2.98 3.34 3.49 3.26
Determine the correlation coefficient for the above set of results and the equation of the least squares
regression line for predicting the percent fiber dry weight. Use your regression line to estimate the
value of the percent fiber dry weight when shear force is 132 kg?
Ans: r = 0.9623895711: Strong relationship
Slope: B=0.008349892; y-Intercept: A= 1.759866075
The “best-fit” linear regression model is: y=1.759866075+0.008349892x;
If x = 132, the point estimate for the percent fiber dry weight is y=2.862051838

*****GOOD LUCK ******

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Tham khảo

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Tiến hành quan sát tốc độ cháy của hai loại chất nổ lỏng được dùng làm nhiên liệu trong tàu vũ trụ ở
trung tâm nghiên cứu. Với loại chất nổ lỏng thứ nhất khi quan sát 50 mẫu thử thì tốc độ cháy trung
bình của mẫu là 19 cm/s và độ lệch chuẩn mẫu là 3cm/s. Với loại chất nổ lỏng thứ hai khi quan sát 55
mẫu thử thì tốc độ cháy trung bình của mẫu là 23 cm/s và độ lệch chuẩn mẫu là 2.5 cm/s. Với mức ý
nghĩa 3% hãy kiểm định giả thuyết hai chất nổ này có cùng tốc độ cháy.

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