Education: Bachelor of Science, option in Human Resource, Management
Masters, Public Health or Health Management
Skills/experience: Check resume
5 force model
Threat of new entry:
New entrants moving towards
The official language of Brazil is Portuguese; this is due to the fact that the Portuguese
settled in the area from the 16th through the 18th century.1 The strong influence of the Portuguese
continues to play a role even today, three centuries
Jaskaran singh Kharbanda
open sample project
Go to inventory value at cost highlighted it
Then analyze statistical statistics
I attached the screenshot of the answer I got.
open sample project
Go to market value highlighted it
Writing assignment # 4
Results & discussion
Based on the fast run of titration the first and second equivalence point appeared around 9.00
and 27.00 milliliters (mL), respectively. We have then performed a slow titration with approximatel
I would describe myself as outgoing and feel comfortable working in a busy environment.
Please explain why you would be an excellent candidate for the company.
Strong work ethic runs throughout my family, I was taught at a very young age to
Spring 2014 Dr. S.
Print Your Name:_ Score:_
Print your name above and also somewhere on the back of this quiz. Write your answers directly on this quiz, not on a separate sheet
of paper. You must show adequate amount of work to get
STAT 3401 SOLUTIONS FOR SECTIONS 2.10 AND 2.11
2.124 Let F be the event that a randomly selected voter favors the election issue. Let R be the event that
the voter is a Republican and let D be the event that the voter is a Democrat. The probl
STAT 3401 SOLUTIONS FOR SECTION 3.11
3.167 Y has mean = 11 and standard deviation = 9 = 3 .
a) P ( 6 < Y < 16) = P ( 6 < Y < 16 ) = P ( 6 11 < Y < 16 11) = P ( 5 < Y < 5) =
P ( Y < 5) 1 k12 , where 5 = k . So k =
and 1 k12 = 1 (5 /13)2 = 1 ( 3 )2
STAT 3401 SOLUTIONS FOR SECTION 3.8
3.121 Y is Poisson with = 2 .
e 2 2 4
a) P (Y = 4) =
b) P (Y 4) = 1 P (Y 3) . But P (Y 3) = p(0) + p(1) + p(2) + p(3) =
e 2 20 e 2 21 e 2 22 e 2 23
= e 2 (1 + 2 + 2 + 8 ) = e 2 ( 19 ) . So
STAT 3401 SOLUTIONS FOR SECTION 3.9
3.149 Suppose that the random variable Y has the mgf m(t ) = (.6et + .4)3 = ( pet + q )n . This is the
mgf of the binomial distribution with n = 3 and p = .6 (and q = .4 ). By the uniqueness property, Y
must have this d
STAT 3401 SOLUTIONS FOR SECTION 3.7
3.103 Let X be the number of non-defective machines out of the five selected machines. Then X is
hypergeometric with N = 10, r = 10 4 = 6 (non-defectives), n = 5 . So
P ( X = 5) = =
STAT 3401 SOLUTIONS FOR SECTION 3.6
3.90 Let Y be the number of employees that must be tested in order to find three that test positive
for asbestos in their lungs. Then Y is negative binomial with r = 3 , p = .40 , and q = 1 p = .60 . So
P (Y = 10) =
STAT 3401 SOLUTIONS FOR SECTION 3.5
3.67 Let Y be the number of applicants that are interviewed until the first one is found who has
advanced training in programming. Then Y is geometric with p = .30 and q = 1 p = .7 . So
P (Y = 5) = q 51p = (.7)4 (.3) =
STAT 3401 SOLUTIONS FOR SECTION 3.4
3.39 Let Y be the number of components out of the four components that operate longer than 1000
hours. Then Y is binomial with n = 4 and p = 1 .2 = .8 .
a) P (Y = 2) = (.8)2 (.2)2 = 6(.8)2 (.2)2 = .1536 .
b) The s
STAT 3401 SOLUTIONS FOR SECTION 3.2
3.1 Let A be the event that impurity A is found in the well and B, the event that impurity B is found.
The problem gives P ( A) = .40, P (B) = .50, P ( A B ) = .20 . By DeMorgans rule and the complement
rule, P ( A B) =
STAT 3401 SOLUTIONS FOR SECTION 2.9
2.110 Let D be the event that a randomly selected item is defective. Let I and II be the events that
correspond to which line the item came from. The problem gives: P ( I) = .40, P ( II) = .60, P (D | I) = .08, and
STAT 3401 SOLUTIONS FOR SECTION 2.8
2.85 Since, A and B are independent, P ( A B) = P ( A) P (B) and so it follows that
P ( A B) = P ( A) P ( A B) = P ( A) P ( A) P (B) = P ( A) (1 P (B) = P ( A) P (B) . Therefore, A and B are
independent. Another way: By
STAT 3401 SOLUTIONS FOR SECTION 2.6
2.35 Consider the 2-stage job: In stage 1, you select a flight from New York to California; in stage 2, you
select a flight from California to Hawaii. The job can be done in 6 7 = 42 ways.
2.41 Consider the 7-stage job:
STAT 3401 SOLUTIONS FOR SECTION 2.7
P ( AB) .1 1
P ( B)
P (BA) .1 1
P (B | A) =
P ( A)
First note that P ( A B) = P ( A) + P (B) P ( AB) = .5 + .3 .1 = .7 . Also note that A ( A B) = A since
P ( A ( A B)
P ( A)
A ( A B) . T
STAT 3401 SOLUTIONS FOR SECTION 2.5
2.26 Label the four cans as: 1, 2, 3, 4. Let the labels 1 and 2 represent the cans with water.
= cfw_ cfw_1,2,cfw_1,3,cfw_1, 4,cfw_2,3,cfw_2, 4,cfw_3, 4 . Each outcome represents the two cans that the expert says
STAT 3401 SOLUTIONS FOR SECTION 2.4
2.12 Let L, R, and S be the events that the vehicle turns left, turns right, and goes straight, respectively.
= cfw_L, R, S .
b) P ( It turns ) = P ( cfw_L, R ) =
a) P (E1 ) + P (E3 ) + P (E4 ) = .01 + .09
STAT 3401 SOLUTIONS FOR SUPPLEMENTARY EXERCISES
The sample space is the set of all combinations of five cards chosen from the 52 cards and the number of
such combinations is = 52 5 = 2,598,960 . There are four suits: spades, clubs, diamonds,
STAT 3401 SOLUTIONS FOR SECTION 2.3
A = cfw_FF . B = cfw_MM . C = cfw_FM , MF, MM . A B = . A B = cfw_FF, MM . A C = .
A C = S. B C = B. B C = C. C B = cfw_FM , MF .
a) AB = A B .
b) A B .
c) AB = A B .
d) ( AB ) (BA) .
2.7 First note that the sam