Download For the questions 1 - 6, assume that the sample space contains 50

For the questions 1 - 6, assume that the sample space contains 50 elementary events
{e1,…, e50}, and that each of them are equiprobable: Pr(ei) = .02. Also assume that H =
{e1, e2, e3, e4, e5, e6, e7}. Define an event E, and calculate the relevant probabilities that
show that:
1.
2.
3.
4.
5.
6.
Pr(E) > Pr(H | E).
Pr(E) = Pr(H | E).
Pr(E) < Pr(H | E).
Pr(E | H) > Pr(H | E).
Pr(E | H) = Pr(H | E).
Pr(E | H) < Pr(H | E).
Answers:
1) Selecting E = {e7,e8,e9,e10,…..e50}
Pr(E) = 44/50 = 0.88
Pr(H/E) = Pr(H and E)/P(E) = Pr({e7})/Pr(E) = (1/50)/(44/50) = 1/44
→ Pr(E) > Pr(H | E)
2)
Let x = #(E) and y #( H and E)
Pr(E) = x/50
Pr(H/E) = Pr(H and E)/Pr(E) = (y/50)/(x/50) = y/x
→ x/50 =y/x → x2 = 50y , one solution can be y = 2 and x =10
Selecting E = {e6,e7,e8,e9,e10,e11,e12,e13,e14,e15}
Pr(E) = 10/50 = 0.2
Pr(H/E) = Pr(H and E)/P(E) = (2/50)/(10/50) = 2/10 =0.2
→ Pr(E) = Pr(H | E)
3) Selecting E = {e5,e6,e7}
Pr(E) = 3/50 = 0.06
Pr(H/E) = Pr(H and E)/P(E) = Pr({e5,e6,e7})/Pr(E) = Pr(E)/Pr(E) = 1
→ Pr(E) < Pr(H | E)
4)
Selecting E = {e7,e8,e9,e10,…..e50}
Pr(E) = 44/50 = 0.88 , Pr(H) = 7/50 = 0.14
Pr(H/E) = Pr(H and E)/P(E) = Pr({e7})/Pr(E) = (1/50)/(44/50) = 1/44
Pr(E/H) = Pr(H and E)/P(H) = Pr({e7})/Pr(H) = (1/50)/(7/50) = 1/7
→ Pr(E/H) > Pr(H | E)
5)
Let x = #(E) and y #( H and E)
Pr(E) = x/50 , P(H) = 7/50
Pr(H/E) = Pr(H and E)/Pr(E) = (y/50)/(x/50) = y/x
Pr(E/H) = Pr(H and E)/Pr(H) = (y/50)/(7/50) = y/7
→ y/x =y/7 → 7y = xy , → x = 7 and y can take any possitive value or y = 0
Selecting E = {e6,e7,e8,e9,e10,e11,e12 }
Pr(E) = 7/50 = 0.14
Pr(H/E) = Pr(H and E)/P(E) = (2/50)/(7/50) = 2/7
Pr(E/H) = Pr(H and E)/P(H) = (2/50)/(7/50) = 2/7
→ Pr(E/H) = Pr(H | E)
6)
Selecting E = {e5,e6,e7}
Pr(E) = 3/50 = 0.06 , Pr(H) = 7/50 = 0.14
Pr(H/E) = Pr(H and E)/P(E) = Pr(E)/Pr(E) = 1
Pr(E/H) = Pr(H and E)/P(H) = Pr(E)/Pr(H) = (3/50)/(7/50) = 3/7
→ Pr(E/H) < Pr(H | E)
7. For any events A and B with nonzero probabilities, Pr(A | B) ≥ Pr(A & B);
explain.
Proof:
Pr(A/B) = Pr(A & B)/P(A) , since 0 < P(A) ≤ 1 we have that 1/P(A) ≥1
→ Pr(A/B) =Pr(A & B)(1/Pr(A))
≥ Pr(A & B) (1) = Pr(A & B)
→ Pr(A | B) ≥ Pr(A & B);
8. Assuming you are flipping a completely fair coin (50% chance of coming up
heads, 50% chance of coming up tails, all flips independent), which is more
likely:
(i) the sequence of 6 outcomes HTHHTT, or (ii) the sequence of 6 outcomes
HHHHHH? Show your work.
i) P(HTHHTT) = P(H)P(T)P(H)P(H)P(T)P(T) = (0.5)6
ii) P(HHHHHH) = P(H)P(H)P(H)P(H)P(H)P(H) = (0.5)6
Answer: The sequences are equally likely
9. Assuming you are flipping a completely fair coin (50% chance of coming up
heads, 50% chance of coming up tails, all flips independent), which is more
likely: (i) the sequence of 6 outcomes HTHHTT, or (ii) the sequence of 5
outcomes HHHHH? Show your work.
i) P(HTHHTT) = P(H)P(T)P(H)P(H)P(T)P(T) = (0.5)6
ii) P(HHHHH) = P(H)P(H)P(H)P(H)P(H) = (0.5)5
Answer: The sequence HHHHH is more likely since
P(HHHHH) =(0.5)5 > (0.5)6= P(HTHHTT)
10. The local casino promises to triple your winnings if you win on both of two
independent roulette-type games in a single play on each. After some observation,
you observe that the chances of winning on a single play the first game is 5%, and
the chances of winning on a single play of the second is 7%. However, you
observe that the chances of winning on both in a single play of each is about 1 in a
thousand. Should you suspect foul play on the casino's part, or are they being
honest, or are they somehow themselves getting ripped off? Explain.
Solution:
P(W1) = 0.05 and P(W2)=0.07 → P(W1 and W2) = P(W1)P(W2)
=0.0035 > 1/1000 then we suspect foul play on the casino’s part
11. Suppose Pr(X|Y) = .72, Pr(Y|X) = .81, Pr(Y|not X) = .12. Find Pr(X).
Solution:
Pr(X|Y) = 0.72 → Pr( X and Y) = 0.72Pr(Y)
Pr(Y|X) = 0.81 → Pr( X and Y) = 0.81Pr(X)
Pr(Y|not X)=0.12 → Pr( notX and Y) = 0.12Pr(notX)
Also we know that Pr(notX and Y) = Pr(Y) – Pr(X and Y) =Pr(Y)-0.72Pr(Y)
Then we have 2 equations
0.72Pr(Y) = 0.81Pr(X)
0.28Pr(Y) = 0.12(1-Pr(X))
Then
8Pr(Y) = 9Pr(X)
28Pr(Y) = 12- 12Pr(X)
So
28(9/8)Pr(X) = 12-12Pr(X)
21Pr(X) = 12 – 12Pr(X)
33Pr(X) = 12
Pr(X) = 4/11
Answer: Pr(X) = 4/11
12. The glossy brochure for the Honor's program at UC Torrance brags that the
proportion of Honors students at UCT who get straight As is .68. Across all of
UCT, the proportion of students who get straight As is .35. Moreover, the Honors
program is popular with students who get straight As: 80% of them belong to it.
What proportion of UCT students are in the Honors program?
Solution:
P( A/ H) = 0.68 , P(A) = 0.35 , P(H/A) = 0.8
P(A and H) = P(A)P(H/A) = 0.35(0.8) = 0.28
P(A/H) = P(A and H)/P(H) = 0.68
Then: 0.28/P(H) = 0.68
P(H) = 0.28/0.68 = 7/17
Answer: P(H) = 7/17
13. The glossy brochure for the Honor's program at UC Diamond Bar brags that the
proportion of Honors students at UCDB who get straight As is .71. Moreover, it
later notes that 80% of the straight A students are part of the honor's program, and
that only 11% of students who don't get straight As are part of the Honor's
program. What is the proportion of UCDB students who get straight As?
Solution:
P(A/H) = 0.71, P(H/A) = 0.8 , P(H/not A) = 0.11
Then: P(A and H) = P(H)0.71 and P(A and H) = P(A)0.8
So 0.71P(H) = 0.8P(A)
Also P(H and not A) = 0.11P(not A)
P(H) – P(A and H) = 0.11 (1-P(A))
Then:
(0.8/0.71)P(A) – 0.8P(A) = 0.11-0.11P(A)
(80/71-0.8+0.11)P(A) = 0.11
(8000/7100 – 8/10 +11/100)P(A) = 0.11
(8000 – 5680 + 781)P(A) = 7100(0.11)
3101P(A) = 781
P(A) = 781/3101
Answer: P(A) = 781/3101
14. Your friend tells you that her school, UC Costa Mesa is great, because 60% of the
students get straight As, and UCCM has an Honors program restricted to just .08
of the students. Moreover, she claims that 80% of the straight-A students are in
the honors program. Your friend is not telling the truth. Prove it.
P(A) = 0.6 P(H) = 0.08
P(H/A) = P(A and H)/P(A) ≤ P(H)/P(A) = 0.08/0.6 = 2/15
Then P(H/A) is not 0.8
15. You tell your friend that you have a 30% chance of getting your first-choice
summer internship, and a 40% chance of getting your second-choice internship
(the two internship offers are with different companies, and so are decided
independently). What are your chances of getting at least one of these internships?
Solution:
P(1st) = 0.3 , P(2nd) = 0.4
P(at least one) = 1-P( get both) = 1-P(1st and 2nd) = 1-P(1st)P(2nd) = 1-(0.3)(0.4) =
0.88
Answer: 0.88
16. Suppose events A, B, C, and D are i.i.d., with a shared probability of .1. In a given
trial, what is the probability that at least one of these events will occur? [Hint:
Use the rules of probability, and consider the event: not A & not B & not C & not
D; if this event does not occur, what must be the case?]
Solution:
P(at least A,B,C and D) = 1- P(notA,notB,notC,notD) = 1P(notA)P(notB)P(notC)P(notD) = 1-(1-0.75)4 = 0.9961 (rounded to 4 dp)
Answer: 0.9961
17. You've applied for 12 summer internships, and you guess that you have a 10%
chance of getting any one of them (where each of your applications will be judged
independently of the others. What are your chances of being offered at least one
internship? [Hint: Use the rules of probability, and consider the probability that
you won't be offered any internships at all.]
Solution:
P(at least one) = 1-P(no offers) = 1- (1-0.1)12 = 0.7176 (rounded to 4 dp)
Answer: 0.7176
18. You live in a house where there are 6 of you total. You are all taking the same
mega-MOOC course that has 14 discussion sections, and students are assigned
randomly to them. What is the probability that at least two of you in your house
will be assigned to the same discussion section? [Hint: Use the rules of
probability, and consider the probability that none of you are assigned to the same
discussion section.]
Discrete Probability Distributions
X=1
X=0
Y=1
a
c
g (=a+c)
Y=0
e (=a+b)
b
f (=c+d)
d
h (=b+d)
1. Using the above table, let a = .3, b = .4, c = .1, d = .2. Are X and Y independent?
Are they identically distributed? Are they iid?
Solution:
P(X =1 ,Y=1) = a = 0.3
P(X=1) = a + b = 0.7
P(Y=1) = a + c = 0.4
Then P(X=1)P(Y=1) = 0.28  0.3 = P(X =1 ,Y=1)
So P(X=1)P(Y=1)  P(X =1 ,Y=1) , then X anY are not independent
P(X=1)P(Y=1) then X and Y are not identically distributed
Also X anY are not iid (independent and identically distributed)
2. Using the above table, let a = .3, b = .4, c = .1, d = .2.What is Pr[X=1 | Y = 0]?
What is Pr[Y = 0 | X = 1]?
Pr(X=1/Y=0) = Pr(X=1,Y=0)/Pr(Y=0) = b/(b+d) = 0.4/0.6 = 2/3
Pr(Y=0/X=1) = Pr(X=1,Y=0)/Pr(X=1) = b/(a+b) = 0.4/0.7 = 4/7
3. Using the above table, let a = .3, b = .4, c = .1, d = .2. Are X and Y independent?
Are they identically distributed? Are they iid?
See question 1 ( is the same)
4. Using the above table, assume that X and Y are iid, and that c = .15. What are a,
b, and d?
Pr(X=1)=Pr(Y=1) → a+b = a+0.15 → b = 0.15
P(X=1,Y=1) = a
Pr(X=1)Pr(Y=1) = (a+0.15)2
Since X and Y are independent: a = (a+0.15)2 → a2-0.7a+0.0225 =0
Solving it there are 2 solutions for a:
a = (0.70.4)/2 = 0.350.1
So the 2 sets of solutions are:
a
0.35+0.1
0.35-0.1
b
0.15
0.15
c
0.15
0.15
d
0.35- 0.1
0.35+ 0.1
5. Using the above table, assume that X and Y are iid, and that c = .15. What is
Pr[Y=1]?
Solution:
Pr(Y =1 ) = a + c = 0.5+0.1
6. Suppose X1 and X2 are Bernoulli trials. You learn that Pr[X1 = 1] = .6, Pr[X2 = 0]
= .4, and Pr[X1 = X2 = 1] = .3. Are X1 and X2 independent? (If you do not have
enough information to determine this, indicate why.)
Answer: X1 and X2 are not independent because Pr(X=1)Pr(Y=1) =0.24
And Pr[X1 = X2 = 1] = .3 (those probabilities s are not equal)
7. Suppose X1 and X2 are Bernoulli trials. You learn that Pr[X1 = 1] = .6, Pr[X2 = 0]
= .4, and Pr[X1 = X2 = 1] = .3. Are X1 and X2 identically distributed? (If you do
not have enough information to determine this, indicate why.)
Answer: X1 and X2 are not id because Pr(X=1)  Pr(Y=1)
(those probabilities s are not equal)
8. Suppose X1 and X2 are Bernoulli trials. You learn that Pr[X1 = 1] = .6, Pr[X2 = 0]
= .4, and Pr[X1 = X2 = 1] = .3. Are X1 and X2 i.i.d.? (If you do not have enough
information to determine this, indicate why.)
Answer: No, because from questions 6 and 7 are not independent and not id
9. Suppose Pr(X = 1) = p, and Pr(X = 0) = (1 – p). Calculate the mean of X. (Be sure
to show your work.)
Solution: Mean of X = E(X) = 1(p)+0(1-p) = p+0 = p
10. Suppose Pr(X = 1) = p, and Pr(X = 0) = (1 – p). Calculate the variance of X. (Be
sure to show your work.)
Solution:
E(X2) = 12(p)+02(1-p) = p
V(X) = E(X2)-E2(X) = p – p2 = p(1-p)
11. Suppose Pr(X = 1) = .3, and Pr(X = 0) = .7. Calculate the mean of X. (Be sure to
show your work. Do not simply plug these values into a formula.)
Solution:
Mean of X = E(X) = 1(0.3)+0(0.7) = 0.3 + 0 = 0.3
12. Suppose Pr(X = 1) = .3, and Pr(X = 0) = .7 Calculate the variance of X. (Be sure
to show your work. Do not simply plug these values into a formula.)
Solution:
E(X2) = 12(0.3)+02(0.7) = 0.3
V(X) = E(X2) – E2(X) = 0.3 -0.32 = 0.3 – 0.09 = 0.21
13. If you have a 25% chance of winning $10, a 35% chance of winning $40, and
otherwise you will lose $55, what is  for this gamble?
Solution:

 = 10(0.25)+40(0.35) +(-55)(0.4) = -5.5
14. If you have a 25% chance of winning $10, a 35% chance of winning $40, and
otherwise you will lose $55, what is  for this gamble?
Solution:
E(X2) = 102(0.25)+402(0.35) +(-55)2(0.4) = 1795
2=E(X2)-E2(X) = 1795 – (-5.5)2 = 1764.75
15. If you have a 25% chance of winning $10, a 35% chance of winning $40, and
otherwise you will lose $55, what is  for this gamble?
Solution:
From question 14 we know that 2 = 1764.75
Then  =1764.75 = 42.009
X=
1
3
probability
.3
.4
16. Calculate the standard deviation of X
4
.2
5
.1
E(X) = 1(0.3)+3(0.4)+4(0.2)+5(0.1) = 2.8
E(X2) =1(0.3)+9(0.4)+16(0.2)+25(0.1) = 9.6
V(X) = E(X2)-E2(X) = 9.6 – 2.82 = 1.76
Sd(X) = V(X) = 1.76 = 1.3266
17. If X1, X2 and X3 are Bernoulli trials, give the sample space of joint outcomes of
these three variables.
This question is using the table above ?
18. If X1, X2 and X3 are Bernoulli trials, and Y = X1 + X2 + X3, then what are the
possible outcomes (i.e. the sample space) of Y?
This question is using the table above ?
n  n 

19. In words, explain why: 
k


.
  nk
(n,k) counts the number of ways we can select k from n objects, for each of these
selections we have n-k objects not selected.
Then (n,k) this is the same number of ways to select n-k objects from n
n  n 

20. Calculate that: 
k



  nk
(n,k) = n!/(n-k)!k!
(n,n-k) = n!/(n-(n-k))!(n-k)! = n!/k!(n-k)!
Then (n.k) = (n,n-k)
21. Give the general formula that expresses the linearity of expectations.
E(aX+BY) = aE(X)+bE(Y) with a and b real numbers
22. Calculate the mean of B.33 directly (do not use the linearity of expectations).
What is the meaning of B.33 ??
23. Calculate the standard deviation of B.33 directly (do not use the linearity of
expectations).
What is the meaning of B.33 ??
24. What is the relationship between a Bernoulli process (i.e. a collection of i.i.d.
Bernoulli trials) and the associated binomially distributed random variable?
The sum of the n Bernoulli iid with parameter p is a binomial distribution
With parameters n and p
8
25. Suppose X ~ B p , and Pr(X = 8) = .007. What is the value of p (as always, show
your work)?
Solution:
Pr(X=8) = p8 = 0.007 then p = (0.007)1/8 = 0.538 (rounded to 3 dp)
Answer: p = 0.538
26. At your job, each day 15% of the employees are randomly selected for a
screening procedure designed to ensure blind, unthinking allegiance to corporate
values. In the last 5 days, you have been selected 3 times, and you are beginning
to suspect that you are being singled out for special attention. If there really is
only a 15% chance of your being selected, then what is the probability of your
being selected 3 or more times in the last 5 days? [Comment: This is the
underlying logic of statistical testing, to be covered in 15B.]
Solution:
Let X a binomial distribution with n = 5 and p = 0.15
Pr(X≥3) = (5,3)(0.15)3(0.85)2 +(5,4)(0.15)4(0.85)1+ (5,5) (0.15)5(0.85)0
= 10(0.15)3(0.85)2 +5(0.15)4(0.85)1+ 1 (0.15)5(0.85)0 = 0.0266
Answer: 0.0266
27. On incoming international flights, the TSA now randomly inspects 22% of all
checked bags. You’ve heard rumors of laxity at LAX. As a business traveler who
frequently travels to Dubai, you have made 13 round trips there since this policy
was implemented (always flying back into LAX). However, your bags have been
opened for inspection only 2 times. If the TSA really is inspecting 22% of the
bags, what is the probability that yours would have only been inspected 2 or fewer
times? [Comment: This is the underlying logic of statistical testing, to be covered
in 15B.]
Solution:
Let X a binomial distribution with n = 13 and p = 0.22
Pr(X≤3) = (13,0)(0.22)0(0.78)13 +(13,1)(0.22)1(0.75)12+ (13,2) (0.22)2(0.78)11
= 1(0.22)0(0.78)13 +13(0.22)1(0.85)12+ 78(0.15)2(0.85)11 = 0.7401
Answer: 0.7401
28. An elevator in SSPA breaks down once in every 5000 operations. If you use this
elevator twice a day for 4 days a week, for 30 weeks a year, for 4 years, what is
the probability that you will be trapped in it twice? At least twice?
Solution
4 years = 4(30)(4)2 operations = 960
P(break down in an operations) = 1/5000
Let X a binomial distribution with n =960 and p =1/5000
P(X =2) = (960,2)(1/5000)2(4999/5000)958 = 0.0152
P(X≥2) = 1-P(X=0)-P(X=1) =
1- (960,0)(1/5000)0(4999/5000)960-(960,1)(1/5000)1(4999/5000)959
= 1- 0.8253 – 0.1585 = 0.0162
29. When flipping a fair coin, which is more likely: 6 of 10 flips coming up heads, or
12 of 20 flips coming up heads? [Be sure to show your work.]
P(X =6) = (10,6)(1/2)6(1/2)4 = 210/210 = 0.205
P(X =12) = (20,12)(1/2)12(1/2)8 = 125970/220 =0.12
Answer: is more likely 6 of 10 flips coming up heads
30. You give a multiple choice test (37 questions, 6 possible answers per question) to
your club. Anyone who fills out the test gets a taco, so some people just guess
completely at random. On average, how many questions would you expect the
random guessers to get right?
Solution:
X = binomial distribution with n = 37 and p =1/6
Expected value is np = 37/6
31. You give a multiple choice test (37 questions, 6 possible answers per question) to
your club. Anyone who fills out the test gets a taco, so some people just guess
completely at random. You want to identify some talent, and it is very important
that you screen out the random guessers. So you decide to identify only those
persons whose score is at least 1.5 standard deviations above the score that would
be expected from sheer guessing. How many questions must a person get right for
them to be identified in this way?
Solution:
X = binomial distribution with n = 37 and p =1/6
Expected value is np = 37/6 and sd is np(1-p) = 2.267
So the person must get right more than 37/6 + 1.5(2.267) = 9.57 (rounding up 10)
Answer: 10 questions or more
32. You give a multiple choice test (37 questions, 6 possible answers per question) to
your club. Anyone who fills out the test gets a taco, so some people just guess
completely at random. You want to whinge at the guessers, so you decide to
identify those persons whose score is within 1.2 standard deviations of the score
that would be expected from sheer guessing. What are the lower and upper
boundaries of the range of scores you identify as guessing?
Solution:
X = binomial distribution with n = 37 and p =1/6
Expected value is np = 37/6 and sd is np(1-p) = 2.267
Lower limit = 37/6 – 1.2(2.267) = 3.446
Upper limit = 37/6 + 1.2(2.267) = 8.887
33. You give a multiple choice test (37 questions, 6 possible answers per question) to
your club. Anyone who fills out the test gets a taco, so some people just guess
completely at random. Tiffany claims she just guessed at the answers, but you
suspect that she is very smart and tried hard to get her low score of only 3 correct
questions. What is the probability of someone guessing at random and getting 3 or
fewer questions correct?
Solution:
X = binomial distribution with n = 37 and p =1/6
Pr(X≤3) = P(X=0)+P(X=1)+P(X=2)+P(X=3)
P(X=0)= (37,0)(1/6)0(5/6)37 = 0.0012
P(X=1)= (37,1)(1/6)1(5/6)36 = 0.0087
P(X=2)= (37,2)(1/6)2(5/6)35 = 0.0313
P(X=3)= (37,3)(1/6)3(5/6)34= 0.0731
Then P(X≤3) = 0.1143
75. Suppose you have a data set x with 30 observations, where the mean is 13 and the
variance is 7. Two new data points, whose values are 17 and 9, are added to this set.
What is the new standard deviation (assume the mean does not change)?
Solution:
We know that Sum(xi)/30 = 13 and [Sum(xi-13)2]/29 = 7
let X =[Sum(xi-13)2] = 29(7) = 203
The new variance is [X+(17-13)2+(9-13)2]/31 = (203+16+16)/31= 235/31
New sd is (235/31)
Answer: 235/31
76. Suppose you have a data set x with 30 observations, where the mean is 13 and the
variance is 7. Two new data points, whose values are 17 and 9, are added to this set.
What is the new variance (assume the mean does not change)?
Answer: From question 75) the new variance is 235/31
xi  x
. Express z in terms of the other quantities.
s
106. Suppose zi 
Solution:
z-bar = Sum(zi) = Sum[(xi-x-bar)/s] = [Sum(xi)-n(x-bar)]/s = (Sum(xi)-Sum(xi))/s = 0
Answer: 0
107. Suppose zi 
xi  x
. Express s in terms of the other quantities.
s
Answer: s = (xi-(x-bar))/zi
110. If we make the transformation yi abxi , calculate the z-score of yi in terms of xi,
x , and sx, to show that it is simply the standardization of xi. You do not need to calculate
values of y or sy. You can simply insert these values (expressed in terms of the
corresponding x-values.
Solution:
y-bar = a + b(x-bar)
sy =bsx
z-score of yi is: [yi-(y-bar)]/sy
Then z-score of yi is [a+bxi –(a+b(x-bar))]/bsx = b(xi-(x-bar))/bsx = (xi-(x-bar))/sx
Then z-score of yi = (xi-(x-bar))/sx (the standardization of xi)
111. If you standardize a data set, what is its new mean?
Answer: 0
112. If you standardize a data set, what is its new standard deviation?
Answer: 1
113. If you standardize a data set, what is its new m3?
Which is m3 ??
115. Calculate the standardization of the value 3 as it occurs in the data set {1, 2, -4, 3},
and again as it occurs in the data set {3, 5, 8, 7, 2}
Solution:
The first set has a mean of 0.5 and sd of 3.109
Standardization of 3 in the first set is (3-0.5)/3.109 = 0.804
The second set has a mean of 5 and sd of 2.55
Standardization of 3 in the first set is (3-5)/2.55 = -0.784
116. If the standardized value of x is -1.4 from a data set whose mean is 7 and sd is 4,
what is the original (unstandardized) value of x?
Solution
(x – 7)/4 = -1.4
x-7 = -5.6
x = 1.4
Answer: 1.4
117. Suppose x1 = 3, in a data set with 12 observations, where the mean is 6 and the
standard deviation is 2. A new data point x13 = 9 is added and now the mean is 6.5 and
the standard deviation is 3. What is the new z-score for x1?
Answer: Z = (3-6.5)/3 = -3.5/3 = -1.167
120. In a given trial or experiment, how many of the possible outcomes in the sample
space can occur at one time? How many must occur at one time?
Answer: Only one
125. Which, if any, of the purported assignment of probabilities to the outcomes in a
sample space are possible? Why or why not?
Table
Outcomes in the sample space
1
E
E2
E3
E4
E5
E6
E7
E8
1
Pr1
.2
.2
.1
.05
.05
.1
.3
0
Pr2
.3
.1
.1
.2
.1
.1
.1
.1
Pr3
.3
.05
.04
.05
.1
.1
.1
.1
Pr4
.2
.2
.3
.1
–.1
.1
.1
.1
Answer:
Pr2 is not valid because the sum of the p(x) values is not 1
Pr3 is not valid because the sum of the p(x) values is not 1
Pr4 is not valid because there are some p(x) values < 0
Pr1 is valid because the p(x) values are >0 and its sum is 1
127. For the sample space {a, b, c}, list all the possible events.
Answer: {, {a},{b},{c},{a,b},{a,c},{b,c},{a,b,c}}
129. Explain why Axiom 3 (of the probability axioms) has the restriction it does (you can
draw a picture as part of your explanation, if you like).
I am not sure about this one