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Title: THEORETICAL STATISTICS
Description: This book contains more than 30 solved exercises and past questions and answers from the course outline which include sets, Venn diagrams, probability theory, probability measures and axioms, Bayes theorem, probability distribution and expectation and many more.
Description: This book contains more than 30 solved exercises and past questions and answers from the course outline which include sets, Venn diagrams, probability theory, probability measures and axioms, Bayes theorem, probability distribution and expectation and many more.
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for
Mathematics and Computer Science Students
&
Earth Science Students
Theoretical Statistics
User
This book contains more than 30 solved exercises and
past questions and answers from the course outline
which include sets, Venn diagrams, probability theory,
probability measures and axioms, Bayes theorem,
probability distribution and expectation and many
more
...
Note: This is not a substitute to
the course lecturer’s notes
...
De
Signed: Management
Powered by:
DESIGNAY ACADEMY
Location:
Abuja,Lagos and Delta
08145579855
BY:
AYODEJI E
...
Conventionally, sets are denoted by capital letter A,B,C,D, etc
...
are used to denote the element of a set
...
Note: In listing the element of a set,
the order of the arrangement of these elements is immaterial and the elements are
separated by a comma
...
Then, we
write a ∈ A
...
(1)
A = {12, 14, 16, 18, 20}……………………………………
...
Examples:
1
...
A= {x: x2 + 8x + 15 = 0, x ∈ Ƶ}
Where Ƶ is the set of integers
...
B = {x: x ∈ N and x is a multiple of 5} i
...
B = {5, 10, 15, 20,…} where N is the
set of natural/counting numbers
...
Consider the set T = {2, 4, 6, 8, 10, 11, 12} then 2∈T, 5∉T & 20∉T
Definition:
Universal, Singleton and Empty Sets:
A universal set in any given concepts is the set of all elements under
consideration at a particular time
...
A set which contains only one element is called a unit set or singleton set, while a set
which have no element is called an empty set
...
Since x2+ 1=0 has no real (real number) root
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Theoretical Statistics
DESIGNAY 08145579855
Subset:
The word subset means part of set
...
Know that our definition of subset
does not exclude the possibility of tactical A equals to tactical B
Two set A & B are said to be comparable if either A is a subset of B or B is a
subset of A
...
Then B is said to be a superset of A
denoted by (B⊃A)
...
If set A is wholly contained in Y but not equal to it, X is a called a proper subset of Y
otherwise X is an improper subset of Y
...
g
...
Consider A = {1, 2, 3}, B = {1, 2, 3, 4, 5}
A is a proper subset of B i
...
A ⊂ B
Example 1:
Let A = {Parallelogram}
B = {Quadrilaterals} Then, B ⊃ A and A ⊂ B
Example 2:
Let S = {All countries in the Africa Union}
T = {Nigeria, Togo, Algeria}
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DESIGNAY 08145579855
Theoretical Statistics
Equal Sets:
Two or more set containing the same distinct element are said to be equal or
identical
...
Example 1:
A = {a, b, c, d},
A=B
B = {a, a, b, b, c, d}
since we have four distinct elements
...
Cardinality of a set:
The number of element in set B is called the cardinality of B or the order of B,
and is denoted by n(B)
...
Finite and Infinite Set:
A set that has a finite number of elements is called a finite set
...
Example 1:
A = {a, e, I, o, u} n (A) = finite
B = {1, 2, 3,…
...
E
...
A {{3,5}, {6,9}, {1}} is a family of sets
...
Any
given set Y has 2() (Y) subsets where () (Y) is the number of distinct element in set Y
otherwise known as the order of the set Y
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Theoretical Statistics
Example:
Let Y = {1, 2, 3}
() (Y) = 3
...
The collection of the 8 subsets above as a set is the power set of Y
...
It is denoted by A´ or A ᶜ
...
, 10},
A = {2, 4, 6},
B = {1, 2, 3, 5, 7}
Then, Aᶜ = {1, 3, 5, 7, 8, 9, 10} and B’ = {4, 6, 8, 9, 10}
Difference of Sets:
The difference of two set A & B, denoted by A-B or A\B is the set of element in
A that are not in B
...
, 10} , A = {2, 4, 6} B = {1, 2, 3, 5, 7}
Then,
A Δ B = (A – B) υ (B – A)
= {4, 6} υ {1, 3, 5, 7}
= {1, 3, 4, 5, 6, 7}
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...
...
It is denoted by (A υ B)
...
Note: A υ B = B υ A
Example 1:
Let
S = {x: x2 + 7x + 12 = 0, x ε Ƶ}, T = {2, 4, 6, 8, 10}
X2 + 7x + 12 = 0 = (x2 + 3x) + (4x + 2)
= x(x + 3) + 4(x + 3)
= (x + 3) (x + 4)
x = - 3 or x = -4
Then S υ T = {-4, -3, 2, 4, 6, 8, 10}
Intersection of Set:
Let A and B be two given sets, the sets of elements that can be found in both
set A and set B is called the intersection of the two sets A and B
...
Clearly, A∩B = B ∩ A
Example 1:
If A = {x: x is a prime factor of 15} B = {y: y is a prime factor of 12 }
...
Since set A = {3, 5} and set B = {2, 3}
...
If A ∩ B = Ø, then there is no element common to both set A and B
...
Example 2:
Let
A = {2, 6, 10},
S = {all males in a house},
A ∩ B= {2}
...
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...
...
Idempotent laws:
(i)
AυA=A
(ii)
A∩A=A
2
...
Commutative laws:
(i) A ᴜ B = B ᴜ A
(ii) A ∩ B = B ∩ A
4
...
Complement laws:
(i) A ᴜ A’ = µ
(ii) A ∩ A’ = Ø
(iii) (A’)’ = A
(iv) µ’ = Ø
(v) Ø’ = µ
6
...
Identity laws
(i)
AᴜØ=A
(ii)
Aᴜµ= µ
(iii) A ∩ µ = A
(iv) A ∩ Ø = Ø
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...
...
Find:
(i)
(i)
(A υ B)’ (ii) B∩ (A υ C)
A∩B’
(iv) (B’∩ C)’
Solution
µ = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
A = {3, 6, 9, 12, 15}
B = {2, 4, 6, 8, 10, 12, 14}
C = {3, 5, 7, 9, 11, 13, 15}
= {3, 5, 7, 9, 11, 13, 15}
(i)
(A υ B) = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15}
(A υ B)’ = {5, 7, 11, 13}
(ii)
(A υ C) = {3, 5, 6, 7, 9, 11, 12, 13, 15}
B ∩ (A υ C) = {6, 12}
(iii)
A ∩ B’ = {3, 9, 15}
(B’∩ C) = {3, 5, 7, 9, 11, 13, 15}
(B’∩ C)’ = {2, 4, 6, 8, 10, 12, 14}
Exercise 2:
In a group of 120 students, 72 play football, 65 play basketball and 53 play handball
...
(i) Draw a venn diagram to illustrate this information
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DESIGNAY 08145579855
Theoretical Statistics
(ii) all the three games?
(iii) exactly 2 of the 3 games?
(iv) exactly 1 of the 3 games?
(v) football alone?
Solution:
Let F, B and H denote the set of students who play football, basketball and handball
respectively
...
” This means all the
students participated in the game and no one is left out
...
n(F ᴜ B ᴜ H) = n(F) + n(B) + n(H) - n(F n H) - n(F n B) - n(B n H) + n(F n B n H)
120 = 72 + 65 + 53 – 35 – 30 – 21 + x
x = 16
(iii)
This means all the students that play two games only
...
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...
...
To get football only, subtracts all students that ever played football
n(F n H’ n B’) = n(F) - n(F n H n B’) - n(F n H’ n B) - n(F n B n H)
= 72 - 14 - 19 – 16 = 23
To get basketball only, subtracts all students that ever played basketball
n(F’ n H’ n B) = n(B) - n(F n H’ n B) - n(F’ n H n B) - n(F n B n H)
= 65 – 19 – 5 – 16 = 25
To get handball only, subtracts all students that ever played handball
n(F’ n H n B’) = n(H) - n(F n H n B’) - n(F’ n H n B) - n(F n B n H)
= 53 – 14 – 5 – 16 = 18
One game only
= n(F n H’ n B’) + n(F’ n H’ n B) + n(F’ n H n B’)
= 23 + 25 + 18 = 66 students
(v)
To get football only, subtracts all students that ever played football
n(F n H’ n B’) = n(F) - n(F n H n B’) - n(F n H’ n B) - n(F n B n H)
= 72 - 14 - 19 – 16 = 23 students
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DESIGNAY 08145579855
Theoretical Statistics
The Nondiagramatic Method
In the nondiagramatic approach, we use the principle which states that if P с Q
and Q с P, then P = Q
...
i
...
The set equation
is true
...
Note: Whenever ∉ is replaced by ∈, ‘and’ should also be replaced by ‘or’
Note: ∩= and,
ᴜ = or
Exercise 3:
x ∈ A’ or x ∈ B
Prove (A ∩ B)’ = A’ ᴜ B’
x ∉ A and x ∉ B
Solution
x ∉ (A ∩ B)
(A ∩ B)’ = A’ ᴜ B’
x ∈ (A ∩ B)’
Let x ∈ (A ∩ B)’
(A’ ᴜ B’) ⊆ (A ∩ B)’
x ∉ (A ∩ B)
Clearly,
x ∉ A and x ∉ B
(A ∩ B)’ = A’ ᴜ B’
x ∈ A’ or x ∈ B’
x ∈ (A’ ᴜ B’)
Tip:
(A ∩ B)’ ⊆ (A’ ᴜ B’)
(A – B) = (A ∩ B’)
Conversely,
(B – A) = (B ∩ A’)
Let x ∈ A’ ᴜ B’
Exercise 4:
If
µ = {2xy2; x ∈ A, y ∈ B}
A = {1, 2, 3}
B = {0, 1, 3, 4, 6}
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...
...
There is a need for ‘µ’ before B’ can be gotten
...
Late Not Late (NL) Total
Substandard(S)
4
8
12
Not Substandard (S)
5
83
88
Total
9
91
100
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...
...
1
...
(S) =
3
...
P(S and NL) =
5
...
P(S/L)
Recall that P(S/L) is the probability of S given that L has already occurred
=
(
)
( )
P (L and S) =
=
(
,
)
( )
{Conditional Probability)
P(L) =
=
7
...
85 that an order will be ready for shipment on time and it is 0
...
What
is the probability that it is ready for shipment on time?
Solution
Let
Ready for shipment on time be R
Delivered on time be D
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...
...
85, P (R∩D) = 0
...
88
Bayes Theorem
(
P (Ai\B) =
(
) (
)
(
)
(
) (
)
)
(
) (
)
Exercise 7:
Three balls are drawn from a box containing 6 red balls, 4 white balls and 5 blue
balls
...
I
...
The total number will never
change
...
P (RWB) =
(ii)
Without Replacement
A ball is picked and not replaced
...
P (RWB) =
×
=
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...
...
It is known
are from within the state where the college is located and that
of the
students are from out-of-state or live in the dormitories
...
333
0
...
333 = 0
...
556 = 0
...
P ( B’ υ A’) = = 0
...
75= 0
...
444 – P (B’∩A’)
P (B’∩ A’) = 1
...
75 = 0
...
21, the probability that his
wife will vote is 0
...
15
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DESIGNAY 08145579855
Theoretical Statistics
Solution
P (H) = Probability of Husband voting
P (W) = Probability of Wife voting
P (H∩W) = Probability of both Husband and Wife voting
Note: “At least” here means either the husband, wife, or both
P (H∩W) =?
Recall that;
P (H υ W) = P (H) + P (W) – P (H υ W)
P (H υ W) = 0
...
28 - 0
...
34
PERMUTATION AND COMBINATION
Permutation:
Permutation refers to the number of ways in which asset of objects can be arranged
in order
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Theoretical Statistics
DESIGNAY 08145579855
Exercise 10:
Solve 10P6
10
P6 = (
=
)
= 151200
Exercise 11:
If nP4 = 12 np2
...
(1)
)
n
p2 = (
)
12 np2= 12 × (
)
=
(
)
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Theoretical Statistics
DESIGNAY 08145579855
Exercise 12:
Prove that nPr = n-1Pr + r n-1Pr-1
Solution
From the Right Hand Side
= n-1Pr + r n-1Pr-1
Recall that
= npr = (
)
Then,
n-1
Pr + r n-1Pr-1 = (
=(
=(
(
)
(
)
)
)
)
(
)
(
*
)
*(
)
(
(
(
)
(
+
)
+
)
)(
)
Take LCM
=
(
)(
(
)
(
)(
)
)
Factor out (n-1)!
= (n-1)! *(
(
)
)(
)
= (n-1)! *(
)(
)
= (n-1)! *(
)(
)
=
(
(
+
+
+
)
)(
)
Recall that
(
n
pr = (
)
(
)(
)
)
Therefore,
(
n
pr =
(
)(
(
)
)
)
n
pr =
n
pr = n-1Pr + r n-1Pr-1
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...
...
Combination deals with selection and choice of objects
...
t
...
Types of Random variable
1
...
Continuous random variable
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...
, , ,
), it is a continuous random variable else,
Properties of Discrete
A probability distribution must satisfy the following properties
1
...
∑ ( ) = 1 where the summation extends ,over all values within its domain
3
...
+ P(z) for x = a, …
...
Solution
Outcomes= {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH,
TTHH, HTTT, THTT, THTT, TTHT, TTTH, TTTT}
Let the random variable for the total number of heads be X
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DESIGNAY 08145579855
Theoretical Statistics
Observe that the numerator of these fractions (1, 4, 6, 4, 1) are the binomial
coefficient
( ), ( ), ( ),( ) and ( ) i
...
4C0, 4C1, 4C2, 4C3, 4C4
From the question, we are asked to find the formula
...
Continuous
For function to be called a probability density function it must satisfy the following
properties
1
...
f(x) = 1
3
...
Find
(i)The value of constant k that makes f(x) to ba a proper pdf
(ii) Calculate P(0
...
This
determine our UM and LM
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Theoretical Statistics
DESIGNAY 08145579855
Tip 2: You only equate to 1 only when you are looking for a constant
...
The interval (0
...
=
dx
Note: Don’t equate to 1
...
=
dx
=
*
=
[
+
*
dx =
(
)
+
]=
= - 0
...
2231 = 0
...
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...
...
F(- ∞) = 0
2
...
If a < b then f(a) ≤ f(b)
Exercise 17:
If the pdf of the random variable X is given by
f(x) =
and = 0 elsewhere
...
Solution
(i)
f(x) =
Recall tip 2
⁄
[
⁄
⁄
]
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...
...
We cannot substitute the UM(x) into it
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DESIGNAY 08145579855
Theoretical Statistics
(iii)
P(X > 1)
From the que
...
Here, it is x > 1 which means x > 1 but x ≤ 4
...
22, 0
...
87, 0
...
0 respectively
...
22
P(x ≤ 1) = 0
...
87
P(x ≤ 3) = 0
...
00
At most 1 means less than or equal to 1
...
87 – 0
...
33
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...
...
0 – 0
...
13
(Iii)
P(x ≥ 1)
“At least 1” means greater than or equal to 1
= P(x ≥ 2) - P(x = 0)
= 1
...
22
= 0
...
Find
(i)
P(X > )
(ii)
(iii)
The distribution function of this random variable
...
5, namely, such that m is the median of
the distribution X
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Theoretical Statistics
= 6[*
+]
= 6[
DESIGNAY 08145579855
]
= [(3(1)2 – 2(1)3) – (3( )2 - 2( )3)
)]
= [(3 -2) – (
= [1 - (
)]
[1 - ]
=
(ii)
=
=
Recall tip3 and tip 4
=
=6*
+
= [
]
= [ (3(x)2 – 2 (x)3) – (3(0)2 – 2 (0)3)]
= (3x2 – 2x3) – (0)
= 3x2 – 2x3
= x2 (3 – 2x) for 0 < x < 1
(iii)
=
=
*
+
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...
...
5
= 0
...
5
m =
Exercise 20:
If the distribution function of the random variable X is given by f(x) = 1 – (1 + x)
for x>0 and = 0 elsewhere
...
f(x) = 1 – (1 + x)
for x>0
(i)
= P(x ≤ 2) (less than 2)
= [ –(
)
= [1 – (1 + x)
= [1- 3
= [1 - 3
]
]2 - [1 – (1 + x)
]0
] - [1 – 1(1)]
– 0]
= 1-3
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...
...
Note: Distribution ≠ pdf (Probability Density Function)
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DESIGNAY 08145579855
Theoretical Statistics
du = 1
dv = ( )
(
)(
(
)
)
Substitute into eqn (1)
)
= (-t
=[
]
= [x
-0
=[x
– 0(1)]
]
= x
Exercise 21;
A random variable X has the following pdf
f(x) = (
(i)
(ii)
(iii)
), for 2 < x < 4, = 0 elsewhere
Find K
Determine the distribution function;
Calculate P[2 < X < 3]
Solution
(i)
k
+
k*
k[
]
k[(3(4) + (4)2) – (3(2) + (2)2] = 1
k[(12 + 16) – (6 + 4)] = 1
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DESIGNAY 08145579855
Theoretical Statistics
Mathematical Expectation
For Discrete
E(x) = ∑
( )
For Continuous
( )
E(x) =
Exercise 22:
If the probability density of the random variable is given by
f(x) =
Find using the method of moment generating function:
(i)
(ii)
(iii)
The moment generating function
The mean of the distribution
The variance of the distribution
Solution
(i)
( )
[
]
( )
=
=
(
=
= *
= *
(
)
(
+
(
)
= *
=
)
(
)
+
)+
)
=
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...
...
(1)
(
=
)
Differentiate eqn (1)
(
)
=(
) ……………………………(2)
µ=
(
)
µ = (
)
(iii)
Take the second derivative of eqn (2)
( ) =
)
(
=
=
(
)
(
)
( )
( )
=1
Exercise 23:
Find the expected value of the random variable X whose probability density is given
by f(x) = (x + 1) for 2 < x < 4, = 0 elsewhere
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DESIGNAY 08145579855
Theoretical Statistics
=
*
+
=
*(
)
(
=
*(
)
(
=
*
+
=
* +
=
=
)+
)+
Exercise 24:
The random variable X takes on the value 0, 1, 2 and 3 with respective probabilities
,
of
,
Find E(X) and E( )
Use the results of part (i) to find E[(3X + 2)2]
...
(
)
(
)
(
)
= 0 + 0
...
768 + 1
...
4
E(
( )
)=∑
=(
=(
)
)
(
(
)
)
(
(
)
)
(
(
)
)
= 0 + 0
...
536 + 4
...
4
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...
...
4) + 12(2
...
96
Exercise 25:
Find µ,
and
for 0 < x < 2,
Solution
for a random variable which has the probability density f(x) =
and = 0 elsewhere
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Theoretical Statistics
DESIGNAY 08145579855
* +
*
+
*
+
( ⁄ )
=
=
Exercise 26:
A continuous random variable X has the probability density given by
f(x) = 2
, and = 0 elsewhere
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DESIGNAY 08145579855
Theoretical Statistics
*
=
+ =
*
+
µ=
(ii)
Let’s Solve for
=
=
=
-----------(1)
Using Integration by part to solve (1), we have;
*(
)
+
*
+
When x =
=*
+ = 0
When x =
=*
+ =
= *
(
)+
(
(
)=
⁄ )
⁄
=
⁄
⁄
( ⁄ )
⁄
⁄
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...
...
Find
(i)
(ii)
( )
The moment generating function
E(X)
(iii) V(X)
Solution
(i)
( )
for x > 0
( )
(
)
( )
=
=
=
(
=
(
)
)
+
= 2*
*
=
(
)
(
)
(
)
*
=
(
)
+
+ =
(
)
( ) =
(ii)
( ) =
= 2(
)
(
)
Differentiate
µ =
( )=
µ =
(
µ= ( )
)=
=
(
)
= ⁄
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...
5
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DESIGNAY 08145579855
Theoretical Statistics
Solution
λ = 3
...
0302
(ii)
x = 1
P(x, λ) =
P(x, λ) =
= 0
...
e
...
3208]
P(x ≥ 3) = 0
...
7967
(ii) P(0 ≤ Z ≤ b) = 0
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Theoretical Statistics
DESIGNAY 08145579855
(i)
P(z
7967
b=
0
...
83
(ii)
P(0 ≤ z ≤ b) = 0
...
4236 = P(z ≤ b) – 0
...
4236 + 0
...
9236
z < b = 0
...
9236
b = 1
...
Solution
p=
, x =6, n= 15
Binomial Distribution
P(x, n, p) = nCxPx(1 – p)n-x
P(6, 15, 0
...
4)6(1 – 0
...
4) = 5005 (4
...
4) = 5005 (0
...
01008)
P(6, 15, 0
...
2066
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...
S/N
1
2
3
4
5
6
7
8
9
10
11
12
Total
X
65
63
67
64
68
62
70
66
68
67
69
71
800
Y
68
66
68
65
69
66
68
65
71
67
68
70
811
∑
(
4420
4158
4556
4160
4692
4092
4762
4290
4828
4489
4692
4970
54109
(∑ )(∑ )
∑
=
4225
3969
4489
4096
4624
3844
4900
4356
4624
4489
4761
5041
53418
(∑ )
)
(
(
)(
) (
)
)
=
y = a + bx
a=
∑
a=
∑
(
)
a = 6
...
33
a = 34
...
253 + 0
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Theoretical Statistics
y = 34
...
5(61)
DESIGNAY 08145579855
y = 64
...
253 + 30
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Theoretical Statistics
DESIGNAY 08145579855
Exercise 36:
Given that f(x) =
(i)
(ii)
Find:
The inverse of f(x)
The value of
(
i
...
)
( )
Solution
I,
f(x) =
y=
x=√
( )
Ii,
The value
f(x) =
(
)
=
= ±3
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...
eqn(1)
Let’s solve for
=
=
=6*
+
= 6 *(
)
= 6 *(
)
=6*
)=
)+
+
+
= E(
(
=
From eqn(1)
( )
=
=
=
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...
3
1(b) Solution:
The elements of the sets are as follows:
µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {3, 5, 7, 9},
B = {2, 4, 6, 8, 10},
C = {3, 4, 5, 6}
(i)
For Left Hand Side (LHS); we have
(A ∩ B) = Ø
(A ∩ B)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = µ
For the Right Hand Side (RHS); we have
A’ = {1, 2, 4, 6, 8, 10},
B’ = {1, 3, 5, 7, 9},
A’ ᴜ B’ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = µ
The results of the LHS and RHS is the same, hence (A ∩ B)’ = A’ ᴜ B’
(ii)
For Left Hand Side (LHS); we have
(B ∩ C) = {4, 6}
A ᴜ (B ∩ C) = {3, 4, 5, 6, 7, 9}
For the Right Hand Side (RHS); we have
(A ᴜ B) = {2, 3, 4, 5, 6, 7, 8, 9, 10}
(A ᴜ C) = {3, 4, 5, 6, 7, 9}
(A ᴜ B) ∩ (A ᴜ C) = {3, 4, 5, 6, 7, 9}
The results of the LHS and RHS is the same, hence
A ᴜ (B ∩ C) = (A ᴜ B) ∩ (A ᴜ C)
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...
...
0532 – 0
...
23622 – 0
...
2062 – 0
...
0193
(ii)
P(x = 5)
= 12C5 (
= 0
...
06872 + 0
...
2835 + 0
...
13287
= 0
...
= 0
...
23622 + 0
...
0532 + 0
...
003322 + 0
...
00005767 + 0
...
00000019660 + 0
...
7252
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...
...
9987 – 0
...
84
(ii)
The number of students scoring between 40 and 60 marks is =
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...
Ogundele S
...
B
...
Oyetunde (2005) : Fundamentals of Pure Mathematics
3
...
R
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Title: THEORETICAL STATISTICS
Description: This book contains more than 30 solved exercises and past questions and answers from the course outline which include sets, Venn diagrams, probability theory, probability measures and axioms, Bayes theorem, probability distribution and expectation and many more.
Description: This book contains more than 30 solved exercises and past questions and answers from the course outline which include sets, Venn diagrams, probability theory, probability measures and axioms, Bayes theorem, probability distribution and expectation and many more.