IB Math – High Level Math – Probability Discrete RVs
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Discrete Random Variable Practice
1.
A biased die with four faces is used in a game. A player pays 10 counters to roll the die. The table
below shows the possible scores on the die, the probability of each score and the number of counters
the player receives in return for each score.
Score
1
2
3
4
Probability
1
1
1
1
2
5
5
10
Number of counters player receives
4
5
15
n
Find the value of n in order for the player to get an expected return of 9 counters per roll.
Working:
Answer:
…………………………………………..
2.
The quality control department of a company making computer chips knows that 2% of the chips are
defective. Use the normal approximation to the binomial probability distribution, with a continuity
correction, to find the probability that, in a batch containing 1000 chips, between 20 and 30 chips
(inclusive) are defective.
(Total 4 marks)
(Total 7 marks)
3.
A supplier of copper wire looks for flaws before despatching it to customers.
It is known that the number of flaws follow a Poisson probability distribution
with a mean of 2.3 flaws per metre.
(a) Determine the probability that there are exactly 2 flaws in 1 metre of the wire.
(3)
(b)
Determine the probability that there is at least one flaw in 2 metres of the wire.
(3)
(Total 6 marks)
4.
The continuous random variable X has probability density function f (x) where
 e  ke kx , 0  x  1
fk (x) = 
otherwise
 0,
(a) Show that k = 1.
(3)
(b)
What is the probability that the random variable X has a value that lies between
1 and 1 ? Give your answer in terms of e.
4
2
(c)
Find the mean and variance of the distribution. Give your answers exactly, in terms of e.
(2)
(6)
The random variable X above represents the lifetime, in years, of a certain type of battery.
(d) Find the probability that a battery lasts more than six months.
(2)
A calculator is fitted with three of these batteries. Each battery fails independently of the other two.
Find the probability that at the end of six months
(e) none of the batteries has failed;
(2)
(f)
exactly one of the batteries has failed.
(2)
(Total 17 marks)
5.
In a game a player rolls a biased tetrahedral (four-faced) die. The probability of each possible score is
shown below.
Score
1
2
3
4
Probability
x
1
2
1
5
5
10
Find the probability of a total score of six after two rolls.
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IB Math – High Level Math – Probability Discrete RVs
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Working:
Answer:
..................................................................
(Total 3 marks)
6.
The probability distribution of a discrete random variable X is given by
x
P(X = x) = k  2  , for x = 0, 1, 2, ......
3
Find the value of k.
Working:
Answer:
..................................................................
(Total 3 marks)
7.
A satellite relies on solar cells for its power and will operate provided that at least one of the cells is
working. Cells fail independently of each other, and the probability that an individual cell fails within
one year is 0.8.
(a) For a satellite with ten solar cells, find the probability that all ten cells fail within one year.
(2)
(b)
For a satellite with ten solar cells, find the probability that the satellite is still operating at the
end of one year.
(c)
For a satellite with n solar cells, write down the probability that the satellite is still operating at
the end of one year. Hence, find the smallest number of solar cells required so that the
probability of the satellite still operating at the end of one year is at least 0.95.
(2)
(5)
(Total 9 marks)
1
of the students travel to school by bus. Five students are chosen at random. Find the
3
probability that exactly 3 of them travel to school by bus.
Working:
Answer:
…………………………………………..
8.
In a school,
9.
X is a binomial random variable, where the number of trials is 5 and the probability of success of
each trial is p. Find the values of p if P(X = 4) = 0.12.
Working:
Answer:
..........................................................................
10.
(a)
Patients arrive at random at an emergency room in a hospital at the rate of 15 per hour
throughout the day. Find the probability that 6 patients will arrive at the emergency room
between 08:00 and 08:15.
(b)
The emergency room switchboard has two operators. One operator answers calls for doctors
and the other deals with enquiries about patients. The first operator fails to answer 1% of her
calls and the second operator fails to answer 3% of his calls. On a typical day, the first and
second telephone operators receive 20 and 40 calls respectively during an afternoon session.
Using the Poisson distribution find the probability that, between them, the two operators fail to
answer two or more calls during an afternoon session.
(Total 3 marks)
(Total 3 marks)
(3)
(5)
(Total 8 marks)
11.
A coin is biased so that when it is tossed the probability of obtaining heads is
2
. The coin is tossed
3
1800 times. Let X be the number of heads obtained. Find
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IB Math – High Level Math – Probability Discrete RVs
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(a) the mean of X;
(b) the standard deviation of X.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 3 marks)
12.
When John throws a stone at a target, the probability that he hits the target is 0.4. He throws a stone 6
times.
(a) Find the probability that he hits the target exactly 4 times.
(b) Find the probability that he hits the target for the first time on his third throw.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
13.
Two children, Alan and Belle, each throw two fair cubical dice simultaneously. The score for each
child is the sum of the two numbers shown on their respective dice.
(a) (i)
Calculate the probability that Alan obtains a score of 9.
(ii) Calculate the probability that Alan and Belle both obtain a score of 9.
(Total 6 marks)
(2)
(b)
(i)
(ii)
Calculate the probability that Alan and Belle obtain the same score,
Deduce the probability that Alan’s score exceeds Belle’s score.
(c)
Let X denote the largest number shown on the four dice.
(4)
4
(i)
(ii)
(iii)
 x
Show that for P(X  x) =   , for x = 1, 2,... 6
6
Copy and complete the following probability distribution table.
x
1
2
3
4
5
6
P(X = x)
1
15
671
1296 1296
1296
Calculate E(X).
(7)
(Total 13 marks)
14.
The random variable X is Poisson distributed with mean  and satisfies P(X = 3) = P(X = 0) + P(X =
1).
(a) Find the value of , correct to four decimal places.
(b)
(3)
For this value of  evaluate P(2  X  4).
(3)
(Total 6 marks)
15.
Give your answers to four significant figures.
A machine produces cloth with some minor faults. The number of faults per metre is a random
variable following a Poisson distribution with a mean 3. Calculate the probability that a metre of the
cloth contains five or more faults.
16.
When a boy plays a game at a fair, the probability that he wins a prize is 0.25. He plays the game 10
times. Let X denote the total number of prizes that he wins. Assuming that the games are
independent, find
(a) E(X)
(b) P(X  2).
Working:
Answers:
(Total 4 marks)
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IB Math – High Level Math – Probability Discrete RVs
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(a) ..................................................................
(b) ..................................................................
(Total 6 marks)
17.
Give all numerical answers to this question correct to three significant figures.
Two typists were given a series of tests to complete. On average, Mr Brown made 2.7 mistakes per
test while Mr Smith made 2.5 mistakes per test. Assume that the number of mistakes made by any
typist follows a Poisson distribution.
(a) Calculate the probability that, in a particular test,
(i)
Mr Brown made two mistakes;
(ii) Mr Smith made three mistakes;
(iii) Mr Brown made two mistakes and Mr Smith made three mistakes.
(6)
(b)
In another test, Mr Brown and Mr Smith made a combined total of five mistakes. Calculate the
probability that Mr Brown made fewer mistakes than Mr Smith.
(5)
(Total 11 marks)
18.
On a television channel the news is shown at the same time each day. The probability that Alice
watches the news on a given day is 0.4. Calculate the probability that on five consecutive days, she
watches the news on at most three days.
Working:
Answer:
.........................................................................
19.
The random variable X has a Poisson distribution with mean λ.
(a) Given that P(X = 4) = P(X = 2) + P(X = 3), find the value of λ.
(Total 6 marks)
(3)
(b)
Given that λ = 3.2, find the value of
(i)
P(X  2);
(ii) P(X  3  X  2).
(5)
(Total 8 marks)
20.
The random variable X has a Poisson distribution with mean λ. Let p be the probability that X takes
the value 1 or 2.
(a) Write down an expression for p.
(b)
Sketch the graph of p for 0  λ  4.
(c)
Find the exact value of λ for which p is a maximum.
(1)
(1)
(5)
(Total 7 marks)
21.
Let X be a random variable with a Poisson distribution such that Var(X) = (E(X))2 – 6.
(a) Show that the mean of the distribution is 3.
(b)
(3)
Find P(X  3).
(1)
Let Y be another random variable, independent of X, with a Poisson distribution such that E(Y) = 2.
(c) Find P(X + Y < 4).
(2)
(d)
Let U = X + 2Y.
(i)
Find the mean and variance of U.
(ii) State with a reason whether or not U has a Poisson distribution.
(4)
(Total 10 marks)
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IB Math – High Level Math – Probability Discrete RVs Markscheme
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Discrete Random Variable Practice - MarkScheme
1.
Let X be the number of counters the player receives in return.
E(X) = p(x) × x = 9
1
 1  1
 1

   4     5     15    n  = 9
2
 5  5
  10

1

n=3
10
 n = 30
(M1)
(M1)(A1)
(A1)
(C4)
[4]
2.
Let n = number of chips = 1000,
p = the probability that a randomly chosen chip is defective = 0.02.
Hence, the mean np = (1000) (0.02) = 20 and the
variance = np(1 – p) = (1000) (0.02) (0.98) = 19.6.
(A1)(A1)
Suppose X is the normal random variable that approximates the
binomial distribution.
The X  N (20, 19.6).
(M1)(M1)
 19.5  20

30.5  20

Z 
Thus p(19.5  X  30.5) = p 
(M1)(M1)
19.6 
 19.6
= p(–0.11  Z  2.37)
= 0.5349
(A1)
Note: Line before last should be p(–0.113  Z  2.37) = 0.536.
Accept 0.535 or 0.536.
If student’s work is not shown but there is evidence that he/she used
the calculator to find the answer, accept the answer.
3.
Note: Throughout the whole question, students may be using their graphic display calculators and
should not be penalized if they do not show as much work as the marking scheme.
(a) Let X denote the number of flaws in one metre of the wire.
(2.3) 2
Then E(X) = 2.3 flaws and P(X = 2) = e–2.3
(M1)(M1)
2!
= 0.265.
(A1)
3
Note: Award (C3) for a correct answer from a graphic display
calculator.
(b) Let Y denote the number of flaws in two metres of wire.
Then Y has a Poisson distribution with mean E(Y) = 2 × 2.3 = 4.6
flaws for 2 metres.
(M1)
–4.6
Hence, P(Y  1) = 1 – P(Y = 0) = 1 – e
(M1)
= 0.990 (3 sf)
(A1)
3
–4.6
Note: Accept 1 – e .
[7]
[6]
4.
(a)
For f (x) to be a probability distribution function,

1
 (e – ke

0
 ex  e kx
kx

1
0
0
f ( x)dx  1 .
)dx = 1
(M1)
=1
(M1)
e–e +1
=1
 e = ek 
k =1
Thus f (x) = e – ex, 0  x  1
k

1
(A1)
(AG)
3
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IB Math – High Level Math – Probability Discrete RVs Markscheme
(b)

1/ 4
=
(c)

1/ 2
(e – e x )dx  ex  e x

1/ 2
1/ 4

e
e
 e 4 e
2
4
e
 e4 e
4
=
1
 (e – e
0
1
 (ex  xe
0
1
 ex 2 
= 
 
 2 0
e
= –1
2

1
0
(M1)
(A1)
)dx 
x
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x
(M1)
)dx
e
 [ xe x – e x ]10
2
xe x dx 
2
(M1)
(A1)
e

Variance = x (e – e )dx   – 1
0
2 

1
2
2
x
(M1)
1
2
 ex 3

e

x
2
= 
 e ( x  2 x  2)   – 1
 3
0  2 
(M1)
2
e2
e–
+e–1
3
4
e e2
=1+ 
3 4
=2–
(d)
(A1)
1

p(battery lasts more than 6 months) = p  x  
2

=
(e)
6

1
1/ 2
(e – e x ) dx
= [ex  e x ]11 / 2
e
= e  or 0.290(3sf)
2
p(no battery failed) = p(all lasted more than 6 months)
(M1)
(A1)
2
(M1)
3
(f)
5.
e

=  e   or 0.0243 (3 sf)
2

2
 3 
e 
e
p(exactly one battery failed) =  1  e   e  
2 
2
 2 
 0.179 (3 sf)
(A1)
2
(M1)
(A1)
2
[17]
 P(X  x)  1
all x
Therefore,
1 2 1
 
+x=1
5 5 10
3
10
1 1
2 3 
P(scoring six after two rolls) =     2    
 10 10 
 5 10 
1
=
4
(C3)
Therefore,
x=
(A1)
(M1)
(A1)
[3]
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IB Math – High Level Math – Probability Discrete RVs Markscheme
6.
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 P(X  x) = 1
Since X is a random variable,
all x
2
Therefore, k +
3
2
2
2
k    k    k + ......... = 1
3
3
3
(M1)




1
 =1
k
2

1  
3

1
k=
3
(C3)
(M1)
(A1)
[3]
7.
(a)
(b)
(c)
10
P(all ten cells fail) = 0.8 = 0.107.
P(satellite is still operating at the end of one year)
= 1 – P (all ten cells fail within one year)
= 1 – 0.107
= 0.893.
P(satellite is still operating at the end of one year)
= 1 – 0.8n.
We require the smallest n for which 1 – 0.8n  0.95.
Thus, 0.8n  0.05
(M1)(A1)
2
(M1)
(A1)
2
(C1)
(M1)
n
5
   20
4
log 20
n
= 13.4
log1.25
Therefore, 14 solar cells are needed.
(M1)(A1)
(C1)
5
[9]
8.
Let p be the probability of choosing a student who travels to school by bus.
Let X be the random variable “the number of students who travel to
school by bus.”
1
Then X ~ B(n, p) with n = 5 and p =
3
3
2
 5  1   2 
Therefore P(X = 3) =      (using formulae and statistical tables)
 3  3   3 
40
=
or 0.165
243
(M1)
(A1)
(A1)
[3]
9.
If X ~ Bin (5, p) and P(X = 4) = 0.12 then
5 4
  p (1 – p) = 0.12
 4
5p5 – 5p4 + 0.12 = 0
p = 0.459 (3 s.f ) or 0.973 (3 s.f)
(M1)
(A1)
(G1)(C3)
[3]
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IB Math – High Level Math – Probability Discrete RVs Markscheme
10.
(a)
(b)
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Let X be the number of patients arriving at the emergency room in
15
a 15 minute period. Rate of arrival in a 15 minute period =
= 3.75. (M1)
4
(3.75) 6 –3.75
P(X = 6) =
e
(M1)
6!
= 0.0908
(A1)
OR
P(6 patients) = 0.0908
Let F1, F2 be random variables which represent the number of
failures to answer telephone calls by the first and the second
operator, respectively.
F1 ~ P0 (0.01 × 20) = P0 (0.2).
F2 ~ P0 (0.03 × 40) = P0 (1.2).
Since F1 and F2 are independent
F1 + F2 ~ P0 (0.2 + 1.2) = P0(1.4)
P(F1 + F2  2) = 1 – P(F1 + F2 = 0) – P(F1 + F2 = 1)
= 1 – e–1.4 – (1.4)e–1.4 = 0.408
OR
P(F1 + F2  2) = 0.408
(G2)
3
(A1)
(A1)
(M1)
(M1)
(A1)
(M0)(G2) 5
[8]
11.
2
n = 1800, p =
3
(a) E(X) = np = 1200
(b)
SD(X) =
(A1)(C1)
np(1  p)  1200
1
= 20
3
(M1)(A1)
(C2)
[3]
12.
(a)
6
Probability =   × (0.4)4 × (0.6)2
4
(M1)(A1)
 


432

or 0.13824 
3125

 18 
Probability = (0.6)2 × 0.4 = 0.144  or

 125 
= 0.138  accept
(b)
(A2)
(C4)
(M1)(A1) (C2)
[6]
13.
(a)
1
(= 0.111)
9
(i)
P(Alan scores 9) =
(A1)
(ii)
P(Alan scores 9 and Belle scores 9) =    
1
9
2
1

 81 
(= 0.0123)
(A1)
2
(b)
(i)
2
2
 1   2 
 6 
 2 
 +   +…+   +…+  
 36   36 
 36 
 36 
2
2
P(Same score) = 
2
 1 
+ 
 36 
73
=
(= 0.113)
648
(M1)
(A1)
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IB Math – High Level Math – Probability Discrete RVs Markscheme
(ii)
P(A>B) =
=
(c)
(i)
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73 
1 
1 –

2  648 
(M1)
575
(= 0.444)
1296
(A1)
P(One number  x) =
x
(with some explanation)
6
 x
6
(R1)
4
P(X  x) = P(All four numbers  x) =  
(M1)(AG)
4
(ii)
 x
 x – 1
P(X = x) = P(X  x) – P(X  x – l) =   – 

6
 6 
x
P(X = x)
4
4
1
2
3
4
5
6
1
1296
15
1296
65
1296
175
1296
369
1296
671
1296
(A1)(A1)(A1)
Note: Award (A3) if table is not completed but calculation of E(X)
in part (iii) is correct.
(iii)
1
15
671
+2×
+…+6×
1296
1296
1296
6797
=
(= 5.24)
1296
E(X) = 1 ×
(M1)
(A1)
7
[13]
14.
(a)
The given condition implies that
3
6
(b)
e–  = e–  +  e– 
(M1)
 3 – 6 – 6 = 0
   2.8473
P(2  X  4) = P(X = 2) + P(X = 3) + P(X = 4)
P(X = 2) =
 2e – 
= 0.235, P(X = 3) =
2
4 –
e
P(X = 4) =
= 0.159
24
 3e – 
6
(A1)
(G1)
3
(M1)
= 0.223,
(G1)
Hence P(2  X  4) = 0.617
OR
P(2  X  4) = P(X  4) – P(X  1)
= 0.8402 – 0.2231
= 0.617
(A1)
(M1)
(G1)
(G1)
3
[6]
15.
Note: Accept answers to an accuracy of at least 4
significant figures – do not apply AP.
P(X = 0) = 0.04979, P(X = 1) = 0.14936, P(X = 2) = 0.22404,
P(X = 3) = 0.22404, P(X = 4) = 0.16803
Sum = 0.81526
OR
P(X  4) = 0.8153 (accept 0.8152)
Hence, P(X  5) = l – 0.8153 = 0.1847 (accept 0.1848)
OR
P(X  5) = 1 – P(X  4) = 0.1847
(M1)
(A1)
(G2)
(M1)(A1)
(G4)
[4]
C:\Users\Bob\Documents\Dropbox\IB Curriculum\HL\5 StatProb\TestsQuizzesPractice\HLProb.DiscreteRV.Practice.docx on 03/04/2017 at 11:09 AMPage 5 of 7
IB Math – High Level Math – Probability Discrete RVs Markscheme
16.
(a)
(b)
Alei - Desert Academy
X is B(10, 0.25) (seen or implied)
so E(X) = 10 × 0.25 = 2.5
(R1)
(M1)(A1)
10 
P(X  2) = (0.75)10 +   (0.75)9(0.25) +
1
10 
  (0.75)8(0.25)2
2
(C3)
(M1)(A1)
= 0.526
(A1)
(C3)
[6]
17.
B ~ P(2.7), S ~ P(2.5)
e –2.72.7 
= 0.245
2
3
e –2.52.5
P(S = 3) =
= 0.214
6
2
(a)
(i)
(ii)
P(B = 2) =
(M1)(A1)
(M1)(A1)
(iii)
(b)
The two events are independent.
P((B = 2)  (S = 3)) = P(B = 2) × P(S = 3)
= 0.214 × 0.245
= 0.0524
–5.2
5
e (5.2)
P(B + S = 5) =
 0.175
120
(M1)
(A1)
6
(A1)
e –2.72.7 
e –2.52.5
P((B = 2)  (S = 3)) =
×
 0.245 × 0.214 = 0.0524
2
6
1
e –2.72.7 
e -2.5 (2.5) 4
P((B = 1)  (S = 4)) =
×
 0.181 × 0.133
24
1
2
= 0.0242
P((B = 0)  (S = 5)) =
e
–2.7
2.7 
3
0
1
×
e
2.5
–2.5
120
(A1)
5
 0.067 × 0.067
= 0.0045
P(B < S) =
(A1)
0.0524  0.0242  0.0045 0.0811

0.175
0.175
= 0.464 (or 0.463)
(M1)
(A1)
5
[11]
18.
METHOD 1
X is Binomial
n = 5 p = 0.4
P(X  3) = 1 – P(X = 4) – P(X = 5)
= 1 – 0.0768 – 0.01024
= 0.91296... (0.913 to 3 sf)
(C6)
METHOD 2
P(X  3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.07776 + 0.2592 + 0.3456 + 0.2304
= 0.91296... (0.913 to 3 sf)
(C6)
(A1)(A1)
(M1)
(A1)(A1)
(A1)
(M1)
(A2)
(A1)
[6]
19.
(a)
(b)




e 
e 


4!
2!
3!
λ2 – 4λ – 12 = 0  λ = 6
(i)
P(X ≥ 2) = 1 – e–3.2 – e–3.2 × 3.2 = 0.829
OR
P(X ≥ 2) = 0.829
e
4
2
3
(M1)
(A1)(A1)
3
(M1)(A1)
(G2)
C:\Users\Bob\Documents\Dropbox\IB Curriculum\HL\5 StatProb\TestsQuizzesPractice\HLProb.DiscreteRV.Practice.docx on 03/04/2017 at 11:09 AMPage 6 of 7
IB Math – High Level Math – Probability Discrete RVs Markscheme
(ii)
P(X ≤ 3X  2) =
Alei - Desert Academy
P(2  X  3)
P(X  2)
(M1)
e 3.2  3.2 2 e 3.2  3.2 3

2
6
=
3.2
1  4.2e
= 0.520
OR
P(X ≤ 3X ≥ 2) = 0.520
(A1)
(A1)
(G3)
5
[8]
20.
(a)
p = λe–λ +

2
2
e–
(A1)
1
(b)

0
(c)
4

(A1)
1
Note: Award (A1) for a maximum point in [0, 4]; sketch need not be
accurate.
dp
2
= e–λ + λ(–e–) + e– × λ +
(–e–)
(M1)(A1)
d
2
 2 

= e–λ 1 
(A1)
2 

 2 
dp
 =0
pmax when
= 0  1 
(M1)
d
2 

 λ2 = 2
  = 2 (do not accept 1.41)
(A1) 5
Note: If no working shown, award (A2) for an answer of 1.41
obviously obtained from a GDC.
[7]
21.
(a)
(b)
(c)
(d)
Note: In this question do not penalize answers given to more than three
significant figures.
Let  = E(X) = Var(X). Then λ = λ2 – 6
(M1)
1  25
Therefore  
(M1)
2
and since λ must be positive λ = 3.
(A1)
Then P(X  3) = 0.647
(A1)
(N1)
E(X + Y) = 3 + 2 = 5. Since X and Y are independent X + Y
has a Poisson distribution with mean = 5.
(M1)
Hence P(X + Y  4) = 0.265.
(A1)
(N1)
Note: Award (N0) if P (X + Y  4) is given instead.
(i)
E(U) = E(X) + 2E(Y) = 7
(A1)
Var(U) = Var(X) + 4Var(Y) = 11
(A1)
(ii) U does not have a Poisson distribution
(A1)
because Var(U)  E(U).
(R1)
4
3
1
2
[10]
C:\Users\Bob\Documents\Dropbox\IB Curriculum\HL\5 StatProb\TestsQuizzesPractice\HLProb.DiscreteRV.Practice.docx on 03/04/2017 at 11:09 AMPage 7 of 7