covenant university · 3 list of courses mat411: topology mat412: differential equations ii mat413:...
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COVENANT UNIVERSITY
ALPHA SEMESTER TUTORIAL KIT (VOL. 2)
P R O G R A M M E : M AT H E M AT I C S
400 LEVEL
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DISCLAIMER
The contents of this document are intended for practice and learning purposes at the undergraduate
level. The materials are from different sources including the internet and the contributors do not
in any way claim authorship or ownership of them. The materials are also not to be used for any
commercial purpose.
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LIST OF COURSES
MAT411: Topology
MAT412: Differential Equations II
MAT413:
MAT415: Design and Analysis Experiment
MAT417: Functional Analysis
MAT418:
GEC410: Probability and Statistics
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COVENANT UNIVERSITY CANAANLAND, KM. 10, IDIROKO ROAD
P. M. B. 1023, OTA, OGUN STATE. NIGERIA COLLEGE: COLLEGE OF SCIENCE AND TECHNOLOGY DEPARTMENT: MATHEMATICS SESSION: 2014/2015 EXAMINATION SEMESTER: ALPHA COURSE TITLE: TOPOLOGY COURSE CODE: MAT 411 INSTRUCTION: Attempt any four questions. TIME: 3 hours
1. (a) Define the followings: (i) a topological space and (ii) Given that
A is a subset of a topological space X, define the relative topology
on A.
(b) Let X be a none-empty set and 𝜏= {Ø, G⊂X: 𝐺𝑐 is countable}.
Show that 𝜏 is a topology on X.
(c) Let {𝜏 :i∈I} be a collection of topologies on a non-empty set X.
Show that the intersection ∩ 𝜏𝑖 is also a topology on X.
2. (a) Let (X, 𝜏) be a topological space. Define a base 𝛽 for 𝜏.
(b) Let 𝛽= {(𝑎, 𝑏]: 𝑎, 𝑏𝜀 𝑅 𝑎𝑛𝑑 𝑎 < 𝑏} be a collection of half-open
intervals of R. Show that 𝛽 form a base for a topology on R.
(c) Let X be a non-empty set (Y, 𝜏) a topological space. If f:X→Y is a
function, we define 𝜏∗ be the collection of inverses of open
subsets of Y. i.e. 𝜏∗= {𝑓−1(G): G𝜀𝜏}, show that 𝜏∗ is a topology on
X.
3. (a) Prove that every second countable space is also first
countable.
(b) If (X,𝜏) is a first countable topological space and (A,𝜏𝐴) a
subspace. Show that (A,𝜏𝐴) is also first countable.
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(c) If X has a countable subbase, then show that X is second
countable.
4. (a) Let (X,𝜏) and (Y,𝜏∗) be topological spaces. Prove that a function
𝑓: 𝑋 → 𝑌 is continuous if and only if the inverse image of each
closed subset of Y is a closed subset of X.
(b) Show that any two open intervals of R are homeomorphic.
(c) Let 𝑓: 𝑋 → 𝑌 be a bijective function, show that f is a
homeomorphism if and only if f is continuous and closed.
5. (a) Prove that a topological space X is a 𝑇1-space if and only if
every singleton subset of X is closed.
(b) (i) Define a Hausdorff space.
(ii) Let X be an infinite set and 𝜏 a co-finite topology on X.
Show that X is a 𝑇1-space but not an Hausdorff space.
(c) Prove that if X is a normal topological space and F is a closed
subset of X, then F is normal.
6. (a) Let 𝑋 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒}. Determine whether or not each of the following classes of
subsets
of 𝑋 is a topology on 𝑋.
(i) 𝜏1 = {𝑋, ∅, {𝑎}, {𝑎, 𝑏}, {𝑎, 𝑐}}
(ii) 𝜏2 = {𝑋, ∅, {𝑎, 𝑏, 𝑐}, {𝑎, 𝑏, 𝑑}, {𝑎, 𝑏, 𝑐, 𝑑}}
(iii) 𝜏3 = {𝑋, ∅, {𝑎}, {𝑎, 𝑏}, {𝑎, 𝑐, 𝑑}, {𝑎, 𝑏, 𝑐, 𝑑}}
(b) Let 𝑋 be a discrete space and let 𝛽 be the class of all singleton subsets of 𝑋, i.e. 𝛽 =
{{𝑝}: 𝑝 ∈ 𝑋}. Show that any class 𝛽′ of subsets of 𝑋 is a base for 𝑋 if and only if it is a
superclass of 𝛽.
(c) Show that the usual topology 𝑈 on the real line ℝ is coarser than the upper limit
topology 𝜏 on ℝ which has as a base the class of open-closed intervals (𝑎, 𝑏].
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COURSE: MAT411
SESSION: 2014/2015
MARKING GUIDE
Question 1
(a) (i) Let X be a non-empty set and 𝜏 a collection of subsets of X, 𝜏 is called a topology on X
if it satisfies the following axioms:
(p1) 𝑋, ∅ ∈ 𝜏.
(p2) Union of arbitrary class of subsets in 𝜏 is also in 𝜏.
(p3) Intersection of every finite class of sets in 𝜏 is a set in 𝜏.
(a) (ii) The topology 𝜏𝐴 on A is called is called relative topology on A if A is a subset of X
and (𝑋, 𝜏𝐴) is a topological space. This relative topology can be denoted by (𝐴, 𝜏𝐴),
where 𝜏𝐴 = {𝐺 ∩ 𝐴: 𝐺 ∈ 𝜏}.
(b) (i) ∅ ∈ 𝜏 (given)
Since 𝑋𝑐 = 𝑋 − 𝑋 = ∅. Since ∅ has no element, then it is countable. Then 𝑋 ∈ 𝜏.
(ii) Let {𝐺𝑖: 𝑖 ∈ 𝐼} be collection of sets in 𝜏, i.e. 𝐺𝑖𝑐 is countable for each 𝑖 ∈ 𝐼. To
show that ⋃ 𝐺𝑖 ∈ 𝜏𝑖 , we need to show that (⋃ 𝐺𝑖)𝑐 = ⋂ (𝐺𝑖
𝑐).𝑖∈𝐼𝑖∈𝐼
But ⋂ (𝐺𝑖𝑐) = 𝐺𝑖
𝑐.𝑖∈𝐼
Since 𝐺𝑖𝑐 is countable for each 𝑖 and since subset of a countable set is countable, it
follows that
(⋃ 𝐺𝑖)𝑐 = ⋂(𝐺𝑖
𝑐)𝑖 is countable. Thus, ⋃ 𝐺𝑖 ∈ 𝜏.𝑖
(ii) Let 𝐺1, 𝐺2, … , 𝐺𝑛 be finite collection of elements of 𝜏. Then it follows that
𝐺1𝑐, 𝐺2
𝑐 , … , 𝐺𝑛𝑐 are all countable subsets of X. Then by De-Morgan’s theorem:
(⋂ 𝐺𝑖)𝑐 = ⋃ 𝐺𝑖
𝑐.𝑛
𝑖=1
𝑛
𝑖=1
Since finite union of countable sets is countable, then
(⋂ 𝐺𝑖)𝑐 = ⋃ 𝐺𝑖𝑐𝑛
𝑖=1𝑛𝑖=1 is countable and so ⋂ 𝐺𝑖 ∈ 𝜏𝑛
𝑖=1 .
Hence, 𝜏 is a topology on X.
(c) (i) Since 𝜏𝑖 is a topology on X, then ∅ ∈ 𝜏𝑖 for each 𝑖 ∈ 𝐼. It then follows that
∅ ∈ ⋂ 𝜏𝑖𝑖 and 𝑋 ∈ ⋂ 𝜏𝑖𝑖 .
(ii) Suppose {𝐺𝑖: 𝑖 ∈ 𝐼} are collection of elements in ⋂ 𝜏𝑖𝑖 , then it follows that {𝐺𝑖}𝑖∈𝐼
is a collection of elements in 𝜏𝑖 for each 𝑖 ∈ 𝐼. Since for each 𝑖 ∈ 𝐼, 𝜏𝑖 is a topology,
then ⋃ 𝐺𝑖 ∈ 𝜏𝑖𝑖 for each 𝑖. Thus, it follows that ⋃ 𝐺𝑖 ∈ ⋂ 𝜏𝑖 .𝑖𝑖
(iii) Let 𝐺1, 𝐺2, … , 𝐺𝑛 ∈ ⋂ 𝜏𝑖𝑖 , this implies that 𝐺1, 𝐺2, … , 𝐺𝑛 ∈ 𝜏𝑖 for each 𝑖 ∈ 𝐼. Since
each 𝜏𝑖 is a topology on X, then ⋂ 𝐺𝑖 ∈ 𝜏𝑖 ,𝑛𝑖=1 hence
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⋂ 𝐺𝑖 ∈ ⋂ 𝜏𝑖.𝑖𝑛𝑖=1 Thus, ⋂ 𝜏𝑖𝑖 is a topology on X.
Question 2
(a) Let (𝑋, 𝜏) be a topological space. A class 𝛽 of open sets (i.e. 𝛽 ⊂ 𝜏) is called open base
for 𝜏 if and only if every open subset of X is a union of members of 𝛽. i.e. 𝛽 ⊂ 𝜏 is called
an open base for 𝜏 if given any 𝐺 ∈ 𝜏, then 𝐺 = ⋃ 𝐵𝑖𝑖 where 𝐵𝑖 ∈ 𝛽.
(b) 𝑅 = ⋃ (−𝑛, 𝑛],𝑛 where 𝑛 ∈ 𝑅. If (𝑎, 𝑏] and (𝑐, 𝑑] are elements in 𝛽 if (𝑎, 𝑏] ∩ (𝑐, 𝑑] =
∅, since ∅ is a subset of any set then ∅ ∈ 𝛽 and so 𝛽 is a base for a topology on R.
If (𝑎, 𝑏] ∩ (𝑐, 𝑑] ≠ ∅, then we have 𝑎 < 𝑐 < 𝑏, 𝑏 < 𝑑. We will have the shaded region
(𝑐, 𝑏]. The shaded are in the intersection of the two intervals (𝑎, 𝑏] ∩ (𝑐, 𝑑] = (𝑐, 𝑑] ∈
𝛽. Hence, 𝛽 is a base for a topology on R.
(c) Firstly, we need to show that 𝑋 ∈ 𝜏∗ and ∅ ∈ 𝜏∗. Since 𝑋 = 𝑓−1(𝑌) and ∅ = 𝑓−1(∅), it
follows that 𝑋, ∅ ∈ 𝜏∗. Secondly, we need to show that if {𝐴𝑖} is a class of sets in 𝜏∗, then
⋃ 𝐴𝑖 ∈ 𝜏∗.𝑖 By definition of 𝜏∗, there exists 𝐺𝑖 ∈ 𝜏 such that 𝑓−1(𝐺𝑖) = 𝐴. Thus
⋃ 𝐴𝑖 = ⋃ 𝑓−1(𝐺𝑖) = 𝑓−1(⋃ 𝐺𝑖).𝑖𝑖𝑖 Since 𝜏 is a topology, then it follows that ⋃ 𝐺𝑖 ∈ 𝜏𝑖
and so ⋃ 𝐴𝑖 ∈ 𝜏∗.𝑖
Thirdly, let 𝐴1, 𝐴2, … , 𝐴𝑛 ∈ 𝜏∗ then there exists 𝐺1, 𝐺2, … , 𝐺𝑛 such that
𝐴1 = 𝑓−1(𝐺1), 𝐴2 = 𝑓−1(𝐺2), … , 𝐴𝑛 = 𝑓−1(𝐺𝑛).
𝐴1 ∩ 𝐴2 ∩ … ∩ 𝐴𝑛 = 𝑓−1(𝐺1) ∩ 𝑓−1(𝐺2) … ∩ 𝑓−1(𝐺𝑛)
= 𝑓−1(𝐺1 ∩ 𝐺2 ∩ … ∩ 𝐺𝑛)
Hence, ⋂ 𝐴𝑖 = 𝑓−1(⋂ 𝐺𝑖).𝑛𝑖=1
𝑛𝑖=1 Since 𝜏 is a topology, then ⋂ 𝐺𝑖 ∈ 𝜏𝑛
𝑖=1 , hence ⋂ 𝐴𝑖 ∈ 𝜏∗.𝑛𝑖=1
Thus, 𝜏∗ is a topology on X.
Question 3
(a) Let X be a second countable space then the topology 𝜏 on X has a countable base 𝛽. Let
𝑝 ∈ 𝑋 and 𝛽𝑝 ⊂ 𝛽 where 𝛽𝑝 are members of 𝛽 which contains 𝑝 ∈ 𝑋. Since 𝛽 is
countable and 𝛽𝑝 is a subcollection of 𝛽, then 𝛽𝑝 is also countable. Thus 𝛽𝑝 form a
countable local base at p, hence X is first countable.
(b) Let 𝑝 ∈ 𝐴, since 𝐴 ⊂ 𝑋, then 𝑝 ∈ 𝑋. Since X is first countable then there exists a
countable local base 𝛽𝑝 = {𝐵𝑛|𝑛 ∈ ℕ}.
𝛽𝑝𝐴 = {𝐴 ∩ 𝐵𝑛: 𝑛 ∈ ℕ} is a 𝜏𝐴 −local base at p. Since 𝛽𝑝 is countable, then 𝛽𝑝
𝐴 is
countable, then (𝐴, 𝜏𝐴) is first countable.
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(c) Let S be the countable subbase of X. The base 𝛽 generated by S is the collection of finite
intersection of elements of S. Since S is countable, then 𝛽 is also countable. Hence, X is
second countable.
Question 4
(a) Let 𝑓: 𝑋 → 𝑌 be a continuous function and F a closed subset of Y. We have to show that
𝑓−1(𝐹) is a closed subset of X. Since F is closed, then 𝐹𝑐 = 𝑌 − 𝐹 is open and f is
continuous by the assumption, 𝑓−1(𝑌 − 𝐹) is open. But
𝑓−1(𝑌 − 𝐹) = 𝑓−1(𝑌) − 𝑓−1(𝐹) = 𝑋 − 𝑓−1(𝐹) = (𝑓−1(𝐹))𝑐.
Since (𝑓−1(𝐹))𝑐 is open in X then its compliment (𝑓′(𝐹)) is a closed subset of X.
Conversely, suppose the inverse image of each closed set in Y is closed in X. Let G be an
open subset of Y, then 𝑌 − 𝐺 = 𝐺𝑐 is closed in Y. By our assumption,
𝑓−1(𝑌 − 𝐺) = (𝑓−1(𝐺))𝑐 is closed subset of X. However,
𝑓−1(𝐺𝑐) = 𝑓−1(𝑌 − 𝐺) = 𝑓−1(𝑌) − 𝑓−1(𝐺)
= 𝑋 − 𝑓−1(𝐺) = (𝑓−1(𝐺))𝑐
which implies that the complement of 𝑓−1(𝐺) is a closed subset of X, hence 𝑓−1(𝐺) is open in
X. Thus f is a continuous function.
(b) We define a function 𝑓: (−1,1) → (𝑎, 𝑏) by 𝑓(𝑥) =𝑏+𝑎
2+ (
𝑏−𝑎
2) 𝑥 for all 𝑥 ∈ (−1,1).
Clearly, f is one-to-one because if 𝑓(𝑥1) = 𝑓(𝑥2), then
𝑏 + 𝑎
2+
𝑏 − 𝑎
2, 𝑥1 =
𝑏 + 𝑎
2+
𝑏 − 𝑎
2𝑥2
Hence, 𝑥1 = 𝑥2, and so f is one to one.
To show f is onto, let 𝑦 ∈ (𝑎, 𝑏), we then want to find 𝑥 ∈ (−1,1) such that
𝑓(𝑥) =𝑏 + 𝑎
2+ (
𝑏 − 𝑎
2) 𝑥 = 𝑦
Then (𝑏 + 𝑎) + (𝑏 − 𝑎)𝑥 = 2𝑦,
Hence 𝑥 =2𝑦−𝑏−𝑎
𝑏−𝑎.
Thus f is onto and so f is bijective. We now show that f is continuous. We take any open set in
(𝑎, 𝑏) and show that the inverse image is open in (−1,1). For 𝜀 > 0 but small, then
(𝑎 + 𝜀, 𝑏 − 𝜀) is an open subset of (𝑎, 𝑏), since
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𝑓−1(𝑦) =2𝑦 − 𝑏 − 𝑎
𝑏 − 𝑎
Then it follows that
𝑓−1(𝑎 + 𝜀, 𝑏 − 𝜀) = (−1 +2𝜀
𝑏 − 𝑎, 1 −
2𝜀
𝑏 − 𝑎)
Which is an open subset of (−1,1), hence f is an open map. Let (−1,0) be an open subset of
(−1,1), then
𝑓(−1,0) = (𝑎,𝑏 − 𝑎
2)
Which is an open subset of (𝑎, 𝑏). This shows that f is an open map and so f is an
homeomorphism.
(c) f is bijective implies f is one-to-one and onto. Suppose f is an homeomorphism then by
definition of homeomorphism f is continuous. Suppose H is a closed subset of X, then 𝐻𝑐
is open subset of X. Since f is an homeomorphism, then f is an open map so 𝑓(𝐻𝑐) is
open in Y. This implies f(H) is closed in Y, hence f is a closed map. Conversely, suppose f
is continuous and closed. To show f is an homeomorphism, we only need to show that
𝑓−1 is continuous. Let G be open in X, then 𝐺𝑐 = 𝑋 − 𝐺 is closed. Since f is closed then
(𝑓−1)−1(𝐺𝑐) = 𝑓(𝐺𝑐) = (𝑓(𝐺))𝑐 is closed, hence f(G) is open in Y. This shows that 𝑓−1
is continuous, hence f is an homeomorphism.
Question 5
(a) Suppose X is a 𝑇1 − space and 𝑝 ∈ 𝑋. We want to show that {𝑝} is a closed subset of X.
This is equivalent to showing that {𝑝}𝑐 is open. Let 𝑦 ∈ {𝑝}𝑐 then 𝑦 ≠ 𝑝 and since X is a
𝑇1 −space, there exists open sets 𝐺𝑦 such that 𝑦 ∈ 𝐺𝑦𝑐and 𝑝 ∉ 𝐺𝑦. Since 𝑝 ∉ 𝐺𝑦, then
𝐺𝑦 ⊂ {𝑝}𝑐 =∪ {𝐺𝑦: 𝑦 ∈ {𝑝}𝑐}. Since {𝑝}𝑐 is a union of open sets 𝐺𝑦, then {𝑝}𝑐 is open
and this implies {𝑝} is closed. Conversely, suppose{𝑝} is a closed subset of X for each
𝑝 ∈ 𝑋. We need to show that X is a 𝑇1 −space. Let 𝑎, 𝑏 ∈ 𝑋 where 𝑎 ≠ 𝑏. Since {𝑎} and
{𝑏} are closed sets, by the hypothesis, then their complements {𝑎}𝑐 and {𝑏}𝑐 are open
and 𝑏 ∈ {𝑎}𝑐 and 𝑎 ∉ {𝑎}𝑐, 𝑎 ∈ {𝑏}𝑐 and 𝑏 ∉ {𝑏}𝑐. This shows that {𝑎}𝑐 and {𝑏}𝑐 are the
required open sets, hence X is a 𝑇1 −space.
(b) (i) A topological space X is called Hausdorff space if each pair of distinct points belong to
distinct open subsets of X. That is, X is a Hausdorff iff given 𝑎, 𝑏 ∈ 𝑋 such that 𝑎 ≠ 𝑏,
there exists open set G and H such that 𝑎 ∈ 𝐺, 𝑏 ∈ 𝐻 and 𝐺 ∩ 𝐻 = ∅.
(ii) Since X is a co-finite topological space, then finite subsets of X are closed. In
particular, the singletons {p} are closed, hence X is a 𝑇1 −space. To show that X is not
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Hausdorff, we need to show that two distinct points cannot be separated by disjoint
open sets. Let 𝑎, 𝑏 ∈ 𝑋, where 𝑎 ≠ 𝑏. If G and H are open subsets of X such that 𝑎 ∈ 𝐺
and 𝑏 ∈ 𝐻. We want to show 𝐺 ∩ 𝐻 ≠ ∅.
Suppose on the contrary 𝐺 ∩ 𝐻 = ∅, then 𝐺 ⊂ 𝐻𝑐. Since the topology is co-finite, the
compliment of any open set is finite, hence 𝐻𝑐 is infinite set. However, G is open and so
an infinite set. This is because X is infinite and 𝐺𝑐 = 𝑋 − 𝐺 is finite. It is impossible for a
finite set 𝐻𝑐 to contain an infinite set G, hence G and H cannot be disjoint. Thus 𝐺 ∩
𝐻 ≠ ∅ and so X is not Hausdorff.
(c) Let 𝐻1, 𝐻2 be closed subset of F where 𝐻1 ∩ 𝐻2 = ∅. It follows that there exist closed
subset 𝐹1 and 𝐹2and X such that 𝐻1 = 𝐹 ∩ 𝐹1 and 𝐻2 = 𝐹 ∩ 𝐹2. Since 𝐻1 ∩ 𝐻2 = ∅, then
𝐹1 ∩ 𝐹2 = ∅. Since X is normal then there exists open sets 𝐺1 and 𝐺2such that 𝐹1 ⊂ 𝐺1
and 𝐹2 ⊂ 𝐺2 and 𝐺1 ∩ 𝐺2 = ∅. Since 𝐹1 ⊂ 𝐺1 then 𝐹 ∩ 𝐹1 ⊂ 𝐺1and similarly 𝐹 ∩ 𝐹2 ⊂
𝐺2. Thus we have 𝐻1 ⊂ 𝐺1 and 𝐻2 ⊂ 𝐺2and since 𝐺1 ∩ 𝐺2 = ∅, then it follows that F is
normal.
Question 6
(a) (i) 𝜏1 is not a topology on X since {𝑎, 𝑏}, {𝑎, 𝑐} ∈ 𝜏1 but {𝑎, 𝑏} ∪ {𝑎, 𝑐} = {𝑎, 𝑏, 𝑐} ∉ 𝜏1.
(ii) 𝜏2 is not a topology on X since {𝑎, 𝑏. 𝑐}, {𝑎, 𝑏, 𝑑} ∈ 𝜏2, but {𝑎, 𝑏, 𝑐} ∩ {𝑎, 𝑏, 𝑑} =
{𝑎, 𝑏} ∉ 𝜏2.
(iii) 𝜏3 is a topology on X.
(b) Suppose 𝛽∗is a base for X. Since any singleton set {p} is open in a discrete space, {p}
must be a union of members of 𝛽∗. But a singleton set can only be the union of itself or
itself with the empty set. Hence, {p} must be a member of 𝛽∗ so 𝛽 ⊂ 𝛽∗. On the other
hand, since 𝛽 is a base for the discrete space X, any superset of 𝛽 is also a base for X.
(c) Note first that any open interval is the union of open-closed intervals. For example
(𝑎, 𝑏) =∪ {(𝑎, 𝑏 −1
𝑛): 𝑛 ∈ ℕ}
Since the class of open intervals is a base for U, 𝑈 ∈ 𝜏, i.e. any U-open set is also
𝜏 −open.
11
COVENANT UNIVERSITY
B. Sc. Degree Examination
COLLEGE: SCIENCE AND TECHNOLOGY DEPARTMENT: MATHEMATICS
SESSION: 2014/2015 SEMESTER: Alpha
COURSE CODE: MAT412 (3C) COURSE TITLE: Differential Equations II
QUESTION #1
(i) The integral transform of f(t) is defined as [ ( )] ( ) ( ). ( , )b
aI f t s f t K s t dt where K(s,t) is
called the kernel of the transform .
(ii)Ordinary differential equation is the one involving only one independent variable and derivatives
of the dependent variable with respect to the independent variable.
(iii) Initial Value Problems An initial value problem is a differential equation along with subsidiary
conditions on the unknown and its derivatives in which all the conditions are specified at one point
of the independent variable.
(iv) Fourier transforms the Fourier transforms of function f(t) is defined by
(v) Convolution theorem the convolution of two functions f(x) and g(x) is defined as a function of x
as
(b)(i)
2 2
1 1 1
2 2 2
1 1 1 1 1 1
2 2 2 2 2 2
7 13( )
( 4 13) 4 13
1 3 1 3
4 13 ( 2) 3
1 ( 2) 5 1 ( 2) 1 5
( 2) 3 ( 2) 3 ( 2) 3
s A Bs cF s
s s s s s s
s sL L L
s s s s s
s sL L L L L L
s s s s s s
2 251 cos3 sin3
3
t te t e t
(ii)
1
( ) ( ) ( )2
i xF F f x f x e dx
1( )* ( ) ( ) ( )
2f x g x f t g x t dt
12
1 1 1
2 222 2
3 1 3 13
7 77 7
3sinh 7
7
L L Ls s s
t
(C)
13
14
Using partial fraction
15
16
2(a)
2(b)
17
3(a) 22 3 1y y y x
2 3 0y y y
22 3 1y y y x
2
3
1 2
2 3 ( 1)( 3)
1, 3
x x
c
r r r r
r r
y C e C e
Is a polynomial of degree 2
hence the PS will be a polynomial of degree 2,then
2( ) 1G x x
18
SOLUTIONS TO Question #4
(a) 2 21
( )2
ax ax ikxe F k e e dx
F
21
2
ax ikxe dx
Note that
2 22 .
2 4
ik kax ikx a x
a a
So, we write
2
2
2
p
p
p
y Ax Bx c
y Ax B
y A
2 22 2(2 3 ) 3( ) 1A Ax B Ax Bx c x
Substituting the results in the equation
Collecting like terms
2 23 ( 4 3 ) (2 2 3 ) 1Ax A B x A B C x
2
2
: 3 1,
: 4 3 0,
: 2 2 3 1
1 4 5, ,
3 9 27
1 4 5
3 9 27p
x A
x A B
ct A B C
A B C
y x x
Equating coefficients
19
2
2 4 21( )
2
ika x
k a aF k e e dx
Introducing a change of variable ,2
iku x
a we obtain
2 22
4
2
1( )
2
ik
ak a au
ik
a
F k e e du
The value of the integral is simply
22
0
12
2.uu e de u ud
Thus,
2 24 .1
( )2
k aF k ea
That is,
2 2 41;
2
ax k ae k ea
F [SUB-TOTAL MARKS = 4]
(b) SOLUTION
1
'( ) '( )2
ikxf x f x e dx
F
1
( ) ( )2 2
ikx ikxikf x e f x e dx
(After integrating by parts)
( )2
ikxikf x e dx
(since ( ) 0f x as x )
( )ik f x F
Thus,
'( ) ( ) ( )f x ik f x ikF ké ù é ù=ë û ë ûF = F [SUB-TOTAL MARKS = 4]
(c) The convolution theorems of the Fourier transform and its inverse variant. SOLUTION
If ( )F k and ( )G k are the Fourier transforms of ( )f x and ( )g x respectively, then the
Fourier transform of the convolution ( * )f g is the product ( ) ( )F k G k . That is,
( ) * ( ) ( ) ( )f x g x F k G ké ùë ûF =
For inverse Fourier transform, we have
1 ( ) ( ) ( ) * ( ) ( ) * ( )F k G k f x g x f x g x- é ùê úë û
F = . [SUB-TOTAL MARKS = 2]
(d) Applying the FT wrt to the PDE givesx
2
2
1
4
u u
t x
F F
20
2
4
dU kU
dt
2
4
dU kdt
U
Integrating yields, 2ln ( 2) ln ( )U k t A k
The general solution is
2( 2)( , ) ( ) .k tU k t A k e
Taking FT wrt of the initial condition :x
2
( ,0) xu x eF F2( 2)1
( ,0) .2
kU k e
Applying the to the solution :IC
20 ( 2)1
( ,0) ( )2
kU k A k e e
2( 2)1
( )2
kA k e
The solution now becomes
2 2 2( 2) ( 2) ( 2)1
( , ) ( )2
k t k k tU k t A k e e e
This is a case of convolution : i.e.,
2 41
( , )2
kF k t e 2
( , ) xf x t e
2
4( , )k
t
G k t e
22
( , ) x tg x t et
The inverse FT gives the solution as
2
21 4 41
( , )2
kt
ku x t e e
F
Using the convolution relation : 1 ( , ) ( , ) ( , ) ( , )F k t G k t f x t g x t F
We have
2 22
( , ) x x tu x t e et
[SUB-TOTAL MARKS = 7½]
[TOTAL MARKS FOR QUESTION ONE = 1 7½]
SOLUTIONS TO Question #5 (a) Proof: We have, by the definition
0
2cos( ) ,ax ax
c e e kx dxp
¥- -é ù=ê úë û òF
( ) ( )
0
1 2,
2
a ik x a ik xe e dxp
¥- - - +é ù= +ê úë ûò
21
1 2 1 1
,2 a ik a ikp
é ùê ú= +ê ú- +ë û
2 2
2.
( )
a
a kp=
+ [SUB-TOTAL MARKS = 4½]
(b) (Proof)
For 0,c ¹
{ }1
( ) ( )2
ikxf cx f cx e dxp
¥
-
- ¥
= òF
If we put u cx= , then
{ }( )1 1
( ) ( )2
k c uf cx f u e du
c p
¥-
- ¥
= òF
( )1
F k cc
= [SUB-TOTAL MARKS = 4]
(c) If ( )F k and ( )G k are the Fourier transforms of ( )f x and ( )g x respectively, then the Fourier
transform of the convolution ( * )f g is the product ( ) ( ).F k G k That is,
( ) * ( ) ( ) ( ).f x g x F k G ké ùë ûF =
Proof:
By definition, we have
1
( * ) ( * )2
ikxf g x f g x e dxp
¥
-
- ¥
é ù é ù=ë û ë ûòF
1 1
( ) ( ) ,2 2
ikxe dx f x g dx x xp p
¥ ¥
-
- ¥ - ¥
= -ò ò
( )1( ) ( ) ,
2
ikx ik xg e d f x e dxxx x xp
¥ ¥
- - -
- ¥ - ¥
= -ò ò
With the change of variable, xh x= - , we have
1 1
( * ) ( ) ( ) ,2 2
ikx ikf g x g e d f e dhx x h hp p
¥ ¥
- -
- ¥ - ¥
é ù=ë û ò òF
( ) ( )F k G k=
[SUB-TOTAL MARKS = 6]
(d) (i) a linear integral operator: A linear integral operator is an operator given by
2
2
2 1 2 2 2ˆ ( ) ( , ) ( )d ,
b
a
Hf x c x x f x x
where the kernel c(x1, x2) is a continuous complex function defined for
22
1 1 1 2 2 2[ , ], [ , ],x a b x a b
and f(x2) is a continuous function. [SUB-TOTAL MARKS = 1½]
(ii) An operator of the form
0
dˆ ( ) ,d
nN
n nn
fHf c x
x
where the coefficients cn(x) are complex functions of a real variable x, is called a linear
differential operator of order n. [SUB-TOTAL MARKS = 1½]
[TOTAL MARKS FOR QUESTION FIVE = 17½]
3 2
1 2
( )
1 4 5
3 9 27
c p
x x
y x y y
C e C e x x
23
MARKING GUIDE MAT413 QUESTIONS 1-3. BY DR. E.A OWOLOKO
Question One
a. i. A probability space is a triple , , .f The first object is an arbitrary set,
representing the set of outcomes, sometimes called sample space. The second object f is
a collection of events, that is, a set of subsets of . The object f is a - algebra,
meaning it satisfy the following conditions
i. , f
ii. cA f A f
iii. 1 2, ,... ,A A f then
1
.n
i
i
A f
Finally, the third object is a number attached to every event A and it satisfies the
following three axioms
i. 0,A for all A
ii. 0, 1
iii. If 1 2, ,...A A is a sequence of pair-wise disjoint events, then i ii
i
A A
(ii) The random variables 1 2, ,..., nX X X on , ,f are said to be independent, if for any
choice of subsets 1 2, ,..., ,n the events 1 1 2 2, ,..., n nX X X B are
independent.
(b). 2 2, 1
,0,
vx y if x yf x y
otherwise
24
2
2
2
1 1 2
1
1 1
2
1
1 1
2
1
11 22
1
1
2 6
1
1
1
1
12
12
1 1 1 11
2 3 7 3 7
4 41
2 21 21
8. 1
2 21
41
21
21.
4
x
x
x
x
vx dxdy
v dx x ydy
v dx x y dy
yv dx x
vdx x x dx
v
v
v
v
v
b.(ii). Marginal density of
2
2 2 2
1
11 1 22 2 2
42 2 6
( ) ( , )
21 21 21
4 4 4 2
21 1 21
4 2 2 8
x
x
x x x
X f x f x y dy
yx ydy x ydy x
xx x x
Marginal density of 1
1
( , )Y f x y dx
25
1
2
1
1
2
1
13
1
21
4
21
4
21 21 2 7. .
4 3 4 3 2
x ydx
yx dx
y xy y
The joint density function is
2
2
2
1 1
2
1
1 1
2
1
11 22
1
1 42
1
12 62 6
1
13 7
1
21
4
21
4
21
4 2
21 1.
4 2 2
21 21
4 2 2 8
21 21 4 1
8 3 7 8 21 2
x
x
x
x ydxdy
dx x dy
ydx x
xdx x
x xdx x x dx
x x
Since 2 61 21 7,
2 8 2x x y then X and Y are not independent.
QUESTION 2:
(a). Let X be non-negative integer-valued random variable such that
, 0,1,2,...,kX k p k is a probability mass function. Then the probability
generating function is given by
0
( ) .k k
k k
k
G s E s s p
Y have a Poisson distribution. Then
, 0,1,2...!
keP Y k k
k
Then the probability generating function of Y is given by
26
0 0
0 1 20 1 2
0 1 2
2
( )!
....0! 1! 2!
...0! 1! 2!
1 ... .1! 2!
kk k
Y k
k k
eG s s p s
k
e s e s e s
s s se
sse
The items in the brackets are easily identified as an exponential series, the expansion ofse , so the generating function is given as
1( ) .
ss
XG s e e e
(b).
( ) , (sin ).
( ) ( ), ( )
( ) ( ).
Z X Y
Z
X Y
X Y
G s E s E s ce Z X Y
E s E s independence
G s G s
QUESTION 3
(a). We define a stochastic process as a collection of random variables ( ), ,X t t T
defined on a state space (set) ,S where each random variable is indexed by a parameter
(the index parameter) t which varies in an index set .T
The properties of an exponential random variable X , are as stated below:
i. Density: , 0
( )0 , 0
xe if xf x
if x
ii. 1
( )E X
iii. 2
1( )Var X
iv. xX x e
v. Memory less property: xX x y X y e .
(b)
i. Let T be the arrival time of the bus. It is exponential with parameter 1
.2
Then its
density function is given as
21
( ) , 0.2
t
Tf t e for t
27
Since,
323 ,
13, . 0.75
2
xT e X x e
x
(ii) Let X be Bimbo’s arrival time. It is uniform on 0, 2 .
1
2 .3
X
The number of late days among 10 days is Binomial 1
10,3
and therefore,
2 10or more late working days among working days
10 9
10
1 0 1
2 1 21 10
3 3 3
1 0.6667 3.3333 0.0260
1 0.0174 0.0867
0.8959
late late
(iii)
We have that 2,
1, , 0,3 0.
6
t
T Xf t x e for x and t
Therefore,
3
2
0
3
2
0
32
1
3
1
3
2 21 1 0.75 0.6667 0.25
3 3
0.1667
t
x
x
X T dx e dt
e dx
e
28
COVENANT UNIVERSITY
CANAANLAND, KM 10, IDIROKO ROAD
P.M.B 1023, OTA, OGUN STATE, NIGERIA. TITLE OF EXAMINATION: B.Sc EXAMINATION
COLLEGE: Science & Technology
SCHOOL: Natural & Applied Sciences
DEPARTMENT: Mathematics
SESSION: 2014/2015 SEMESTER: Alpha
COURSE CODE: MAT 415 CREDIT UNIT: 3
COURSE TITLE: Design and Analysis of Experiment
COURSE COORDINATOR: Mr. Odetunmibi O.A.
COURSE LECTURERS: Mr. Odetunmibi O.A. & Mr. Oguntunde P.E.
MARKING GUIDES
QUESTION 1
a) For a Latin Square (LS) design, increase in the size of the squares results in the increase in
the error degree of freedom. For instance; 1 mark
Source of Variation Degree of Freedom 2x2 3x3 8x8
Rows r-1 1 2 7
Columns r-1 1 2 7
Treatments r-1 1 2 7
Error (r-1)(r-2) 0 2 42
Total r2
-1 3 8 63
2½ marks
b) The model for Latin Square design is;
𝑦𝑖𝑗𝑘 = 𝜇 + 𝛼𝑖 + 𝛽𝑗 + 𝛾𝑘 + 𝑒𝑖𝑗𝑘
𝑖 = 1,2, … , 𝑟 ; 𝑗 = 1,2, … , 𝑟 ; 𝑘 = 1,2, … , 𝑟
Where;
𝑦𝑖𝑗𝑘 is the observation of the 𝑗𝑡ℎ row and the 𝑘𝑡ℎ column for the 𝑖𝑡ℎ treatment.
𝜇 is the overall mean.
𝛼𝑖 is the 𝑖𝑡ℎ treatment effect.
29
𝛽𝑗 is the 𝑗𝑡ℎ row effect.
𝛾𝑘 is the 𝑘𝑡ℎ column effect.
𝑒𝑖𝑗𝑘 is the random error. 1 mark
To get the Sum of Square Treatment (SSTrt);
Treatments Total
A 36
B 28
C 43
D 10
22 ..... 2
1i
i
ySSTrt y
r r
22 2 2 2
2
1 11736 28 43 10
4 4
4029 13689
4 16
1007.25 855.5625
151.6875SSTrt 1 mark
22 ...
2ijk
ySSTotal y
r
22 2 2 2
2
1177 7 3 .... 9
4
136891035
16
1035 855.5625
179.4375SSTotal 1 mark
22 ...
2
1 ySSRow Row
r r
22 2 2 2
2
1 11728 30 31 28
4 4
30
3429 13689
4 16
857.25 855.5625
1.6875SSRow 1 mark
22 ...
2
1 ySSCol Col
r r
22 2 2 2
2
1 11723 27 35 32
4 4
3507 13689
4 16
876.75 855.5625
21.1875SSCol 1 mark
SSE SSTotal SSTrt SSRow SSCol
179.4375 151.6875 1.6875 21.1875SSE
4.875SSE 1 mark
ANOVA TABLE
Source of Variation Degree of Freedom Sum of Squares Mean Square F-Value
Treatment 3 151.6875 50.5625 62.2308
Row 3 1.6875 0.5625 0.6923
Column 3 21.1875 7.0625 8.6923
Error 6 4.8750 0.8125
Total 15 179.4375
1 mark
For Treatment;
0 : 0iH i
Versus
1 : 0iH for at least one i
62.2308calF
0.05,3,6 4.34tabF F
31
Decision Rule: Reject 0H if
cal tabF F 1 mark
Decision: Reject 0H since 62.2308 4.34cal tabF F
Conclusion: The treatment effects are significantly different from zero.
For Row;
0 : 0jH j
Versus
1 : 0jH for at least one j
0.6923calF
0.05,3,6 4.34tabF F
Decision Rule: Reject 0H if
cal tabF F 1 mark
Decision: Do not reject 0H since 0.6923 4.34cal tabF F
Conclusion: The row effects are not significantly different from zero.
For Columns;
0 : 0kH k
Versus
1 : 0kH for at least one k
8.6923calF
0.05,3,6 4.34tabF F
Decision Rule: Reject 0H if cal tabF F 1 mark
Decision: Reject 0H since 8.6923 4.34cal tabF F
Conclusion: The column effects are significantly different from zero.
32
c) Disadvantages of Latin Square Designs
1. The number of treatments must be equal to the number of replicates.
2. The experimental error is likely to increase with the size of the squares.
3. Small squares have very few degrees of freedom for experimental error
4. You cannot evaluate interactions between (a) rows and columns (b) rows and
treatments (c) columns and treatments.
4 marks
QUESTION 2
a) 𝑦𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝛽𝑗 + 𝑒𝑖𝑗
𝑒𝑖𝑗 = 𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗 1 mark
𝑆𝑆𝐸 = ∑ ∑ 𝑒𝑖𝑗2
𝑗𝑖
𝑆𝑆𝐸 = ∑ ∑(𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)2
𝑗𝑖
Differentiating with respect to 𝜇;
𝜕𝑆𝑆𝐸
𝜕𝜇= −2 ∑ ∑ (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)𝑗𝑖 1 mark
Setting 𝜕𝑆𝑆𝐸
𝜕𝜇 to zero
∑ ∑(𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)
𝑗𝑖
= 0
∑ ∑ 𝑦𝑖𝑗
𝑗𝑖
− 𝑎𝑏�̂� − 0 − 0 = 0
∑ ∑ 𝑦𝑖𝑗
𝑗𝑖
− 𝑎𝑏�̂� = 0
�̂� =∑ ∑ 𝑦𝑖𝑗
𝑏𝑗
𝑎𝑖
𝑎𝑏
�̂� =𝑦. .
𝑎𝑏= �̅�. . 1 mark
Differentiating SSE with respect to 𝛼𝑖 and equating the result to zero;
𝜕𝑆𝑆𝐸
𝜕𝛼𝑖= −2 ∑ (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)𝑗 1 mark
33
∑(𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)
𝑗
= 0
Since ∑ 𝛽𝑗 = 0𝑏𝑗
∑ 𝑦𝑖𝑗𝑏𝑗 − 𝑏�̂� − 𝑏𝛼�̂� − 0 = 0 1 mark
∑ 𝑦𝑖𝑗
𝑏
𝑗
− 𝑏�̂� − 𝑏𝛼�̂� = 0
𝛼�̂� =∑ 𝑦𝑖𝑗
𝑏𝑗
𝑏− �̂�
𝛼�̂� = �̅�𝑖 . − �̅�. . 1 mark
Differentiating SSE with respect to 𝛽𝑗 and equating the result to zero gives;
𝜕𝑆𝑆𝐸
𝜕𝛽𝑗= −2 ∑ (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)𝑖 1 mark
∑(𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)
𝑖
= 0
Since ∑ 𝛼𝑖𝑎𝑖=1 = 0;
∑ 𝑦𝑖𝑗𝑎𝑖 − 𝑎�̂� − 0 − 𝑎𝛽�̂� = 0 1 mark
𝛽�̂� =∑ 𝑦𝑖𝑗
𝑎𝑖
𝑎− �̂�
𝛽�̂� = �̅�. 𝑗 − �̅�. . 1 mark
b) 𝑒𝑖𝑗 = 𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗
𝑆𝑆𝐸 = ∑ ∑ (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)2
𝑗𝑖 1 mark
= ∑ ∑ (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)𝑗𝑖 (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)
= ∑ ∑ 𝑦𝑖𝑗(𝑦𝑖𝑗 − �̂� − 𝛼�̂� − 𝛽�̂�) −𝑗𝑖 �̂� ∑ ∑ 𝑦𝑖𝑗𝑗𝑖 (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗) − ∑ 𝛼𝑖𝑖 ∑ (𝑦𝑖𝑗 −𝑗
𝜇 − 𝛼𝑖 − 𝛽𝑗) − ∑ 𝛽𝑗𝑗 ∑ (𝑦𝑖𝑗 − 𝜇 − 𝛼𝑖 − 𝛽𝑗)𝑖
= ∑ ∑ 𝑦𝑖𝑗(𝑦𝑖𝑗 − �̂� − 𝛼�̂� − 𝛽�̂�)𝑗𝑖 1 mark
= ∑ ∑ 𝑦𝑖𝑗2
𝑗𝑖 − �̂� ∑ ∑ 𝑦𝑖𝑗𝑗𝑖 − ∑ 𝛼�̂�𝑖 ∑ 𝑦𝑖𝑗𝑗 − ∑ 𝛽�̂�𝑗 ∑ 𝑦𝑖𝑗𝑖
= ∑ ∑ 𝑦𝑖𝑗2
𝑗𝑖 − [∑ ∑ 𝑦𝑖𝑗𝑗𝑖
𝑎𝑏. ∑ ∑ 𝑦𝑖𝑗𝑗𝑖 ] − ∑ [
𝑦𝑖 .
𝑏−
𝑦. .
𝑎𝑏]𝑖 ∑ 𝑦𝑖𝑗𝑗 − ∑ [
𝑦.𝑗
𝑎−
𝑦. .
𝑎𝑏]𝑗 ∑ 𝑦𝑖𝑗𝑗
1 mark
𝑆𝑆𝐸 = [∑ ∑ 𝑦𝑖𝑗2
𝑗𝑖 −𝑦. .
2
𝑎𝑏] − [
∑ 𝑦𝑖 .2
𝑏−
𝑦. .2
𝑎𝑏] − [
∑ 𝑦.𝑗2
𝑎−
𝑦. .2
𝑎𝑏]
34
𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝐴 − 𝑆𝑆𝐵 3 marks
c) The missing value of ‘x’ is given by;
𝑥 =𝑎𝑦𝑖 .
′ +𝑏𝑦.𝑗′ −𝑦. .
′
(𝑎−1)(𝑏−1) ½ mark
3, 4a b
'
. 2 3 1 6iy
'
. 31 16 47jy
'
.. 157y ½ mark
Therefore;
3 6 4 47 157
3 1 4 1x
½ mark
18 188 157
6x
8.17 1 mark
QUESTION 3
a) Technique for Controlling Experimental Error
1. The use of appropriate experimental design. 1 mark
2. Choice of size and shape of experimental units in field experiments
1 mark
b) Distinction Between Replication and Randomization
Replication is the repetition of the basic experiment or appearance of a treatment
more than once in an experiment while Randomization is a technique used to ensure
that no particular treatment is consistently favored or handicapped.
2 marks
c) i) The ANOVA Table
Source of
Variation
Degree of
Freedom
Sum of
squares
Mean
Square
F-
value
Treatment 3 102 34.0000 3.8491
Error 12 106 8.8333
Total 15 208
3 marks
35
ii) The mathematical model is;
𝑦𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝑒𝑖𝑗 ½ mark
𝑖 = 1, … . , 𝑎 and 𝑗 = 1, … . , 𝑛
Where
𝑦𝑖𝑗 is the ijth observation in the ith treatment
𝜇 is the overall mean effect
𝛼𝑖 is the ith treatment effect
𝑒𝑖𝑗 is the random error ½ mark
Assumptions
1. The population is approximately normally and identically distributed with mean 𝜇
and variance 𝜎2. That is, 𝑦𝑖𝑗~𝑁(𝜇, 𝜎2). ½ mark
2. The error term 𝑒𝑖𝑗 are normally and identically distributed with mean zero and
variance 𝜎2. That is, 𝑒𝑖𝑗~𝑁(0, 𝜎2). ½ mark
iii) Hypothesis
0 :H The four types of subgrade soil have the same effect on moisture content in the top
soil. 1 mark
Versus
1 :H The four types of subgrade soil do not have the same effect on moisture content in
the top soil. 1 mark
iv) 3.8491calF
0.05,3,12 3.77tabF F ½ mark
Decision Rule: Reject 0H if
cal tabF F ½ mark
36
Decision: Reject 0H since 3.8491 3.77cal tabF F ½ mark
Conclusion: The four types of subgrade soil do not have the same effect on moisture content
in the top soil. 1 mark
d) Reasons for Experiments
1. To compare responses achieved at different settings of controllable variables.
2. To determine the principal causes of variation in a measured response.
3. To find the conditions that gives rise to a maximum and minimum response.
4. To obtain the mathematical model in order to predict future response.
5. To minimize any bias in the treatments comparison.
5 marks
QUESTION 4
a) 𝑦𝑖𝑗𝑘 = 𝜇 + 𝛼𝑖 + 𝛽𝑗 + (𝛼𝛽)𝑖𝑗 + 𝑒𝑖𝑗𝑘 𝑓𝑜𝑟 𝑖 = 1,2, 𝑗 = 1,2, 𝑎𝑛𝑑 𝑘 = 1,2, … , 𝑟
1 2
m
a
r
k
Where:
yijk = denote the observation k arising from level i of factor A and level j of factor B.
µ = denote the overall mean or grand mean.
αi = denote the effect of level I of factor A
βj = denote the effect of level j of factor B
(αβ)ij = denotes the (ij)th interaction term.
eijk = denote the (ijk)th error term.
1 2 mark
37
23 4 22 ...
1 1 1
2(7102)2,110,768
24
2,110,768 2,101,600.167
9,167.833
ijk
i j k
ySST y
N
1 mark 23
2 .....
1
22 2 2
1
4 2
1 (7102)2440 2374 2288
8 24
1,452.33
i
i
ySSA y
N
1 mark
24
2 .... j.
1
22 2 2
1
3 2
1 (7102)1896 1791 1172
6 24
4,008.16
i
ySSB y
N
1 mark
SSAB SSsubtotal SSA SSB 23 4
2 ...ij.
1 1
22 2
1
2
1 (7102)667 ... 579
6 24
7,676.83
i j
ySSsubtotal y
N
1 mark
7676.83 1452.33 4008,16
2216.34
SSAB
SSAB
1 2 mark
9,167.83 7,676.83
1,491
SSE SSTotal SSsubtotal
SSE
SSE
1 2 mark
ANOVA TABLE
38
Source of
Variation
Degree of
Freedom
Sum of Squares Mean of Squares f-Calculated
Factor A 2 1452.33 726.165 5.96
Factor B 3 4008.16 1336.05 10.97
AB 6 2216.34 369.39 3.034
Error 12 1461 121.75
Total 23 9167.83
3 marks
4Ai) HYPOTHESIS
FOR MAIN EFFECT:
A) FACTOR A:
NULL HYPOTHESIS {H0}: α1 = α2 = α3 = 0
VS
1 2 mark
Alternative Hypothesis {H1}: at least one αi ≠ 0
B) FACTOR B:
NULL HYPOTHESIS {H0}: β1 = β2 = β3 = β4 = 0
VS
1 2 mark
Alternative Hypothesis {H1}: at least one βi ≠ 0
FOR INTERACTION EFFECT:
NULL HYPOTHESIS {H0}: (αβ)ij = 0 for all i,j
VS
1 2 mark
39
Alternative Hypothesis {H1}: at least one (αβ)ij ≠ 0
4Aiv) FOR FACTOR A:
0 1, 2,
0 2,12,0.05
0 3.89
v vF F
F F
F
DECISION: Since (5.96)calF > (3.89)talF , we therefore reject H0
1 mark
Conclusion: The results of the missile system are not the same.
FOR FACTOR B:
0 1, 2,
0 3,12,0.05
0 3.49
v vF F
F F
F
DECISION: Since (10.97)calF > (3.49)talF , we therefore reject H0
1 mark
Conclusion: The effects of each propellant type are not the same.
FOR INTERACTION EFFECT (AB):
0 1, 2,
0 6,12,0.05
0 3.00
v vF F
F F
F
1 mark
DECISION: Since (3.034)calF > (3.00)talF , we therefore reject H0
1 mark
Conclusion: At least one ( ) 0ij
4B) FACTOR A
Factor A is worth carrying out LSD for since null hypothesis is rejected.
FACTOR B
Factor B is also worth carrying out LSD for since null hypothesis of equal propellant is rejected
2 marks
40
QUESTION 5
5A). FACTORIAL DESIGNS: Factorial designs are those designs in which there are two or
more cross classification of treatments or in which the main effects and interaction effect of two or
more variables are simultaneously studied.
11 2 mark
5B). Factorial design is an improvement over all other designs in the following ways:
i) In factorial design, the main effect of two or more variables is studied simultaneously by
conducting an experiment.
ii) In this design, joint effect or interaction effect of two are more variables is studied
whereas the question of interaction in earlier designs does not arise.
4 marks
5C).
𝑦𝑖𝑗𝑘𝑙 = 𝜇 + 𝛼𝑖 + 𝛽𝑗 + 𝜏𝑘 + (𝛼𝛽)𝑖𝑗 + (𝛼𝜏)𝑖𝑘 + (𝛽𝜏)𝑗𝑘 + (𝛼𝛽𝜏)𝑖𝑗𝑘 + 𝑒𝑖𝑗𝑘𝑙 𝑓𝑜𝑟 𝑖 = 1,2,
𝑗 = 1,2 , 𝑘 = 1,2 , 𝑙 = 1,2, … , 𝑟
1 mark
Where:
yijkl = denote the observation k arising from level i of factor A and level j of factor B.
µ = denote the overall mean or grand mean.
αi = denote the effect of level I of factor A
βj = denote the effect of level j of factor B
τk = denote the effect of level k of factor C
(αβ)ij = denotes the (ij)th interaction term
(ατ)ik = denotes the (ik)th interaction term
(βτ)jk = denotes the (jk)th interaction term
(αβτ)ijk = denotes the (ijk)th interaction term
eijk = denote the (ijk)th error term. 1 mark
41
5D). FACTORS EFFECTS
Treatments
Combination
I A B C AB AC BC ABC
000 (I) + - - - + + + -
100 a + + - - - - + +
010 b + - + - - + - +
110 ab + + + - + - - -
001 c + - - + + - - +
101 ac + + - + - + - -
011 bc + - + + - - + -
111 abc + + + + + + + +
4 marks
5E.) FOR SSA:
2
3
2
2
1
1
8
CASSA
r
CA a b ab c ac bc abc
a b ab c ac bc abcSSA
r
1 mark
2
1
8
a b ab c ac bc abcSSB
r
1 mark
2
1
8
a b ab c ac bc abcSSAB
r
1 mark
2
1
8
a b ab c ac bc abcSSAC
r
1 mark
42
2
1
8
a b ab c ac bc abcSSBC
r
1 mark
2
1
8
a b ab c ac bc abcSSABC
r
1 mark
QUESTION 6
6A.) Hypothesis:
0 1 2 3
1 1 2 3
: 0
: 0
H
VS
H
1 2 mark
Model:
𝑦𝑖𝑗 = 𝜇 + 𝛼𝑖 + 𝑒𝑖𝑗
For i = 1, 2, 3 and j = 1, 2, …,5
1 2 mark
SUM OF SQUARES:
2
2 ..
1 1
22 2 2 480
25 31 ... 2815
450
a b
ij
i j
ySST y
ab
SST
SST
1 mark
43
22 ...
2 2
1
1 450160 ... 135
5 15
250
a
i
i
ySSTr y
b ab
SSTr
SSTr
1 mark
450 250
200
SSE SST SSTr
SSE
SSE
1 2 mark
ANOVA TABLE
Source of Variation
Degree of Freedom Sum of Square Mean of Square F –Calculated
Treatment 2 250 125 7.5
Error 12 200 16.67
Total 14 450
2
marks
0 1, 2,
0 2,12,0.05
0 3.89
v vF F
F F
F
1 2
mark
DECISION: Since (7.5)calF > (3.89)talF , we therefore reject H0
1 mark
Conclusion: The results above shows that at least one of the machine is faster than the
other.
6B.) Treatment Means
44
1
1
1
1
160
5
32
y
n
2
2
2
2
185
5
37
y
n
3
3
3
3
135
5
27
y
n
1 2
mark
Arrange treatments means in ascending order
3 1 2, ,
27,32,37
y y y
Arrange in step apart:
Two step apart: 3 1
1 2
&
&
y y
y y
Three step apart: 3 2&y y 1 2
mark
Hypothesis for Two Step Apart.
Hypothesis I
0 3 1
1 3 1
:
:
H M M
VS
H M M
1 2
mark
45
𝑊𝑝 = 𝐷𝑝,𝑣,𝛼√𝑀𝑆𝐸
�̅� 1 2
mark
When P=2, V=12, and α=0.05
2 2,12,0.05
2
200
5
19.49
W D
W
1 2
mark
Absolute Difference Value of M3 and M1 is 5
Decision: Since 2 3 1(19.49) (5)W y y we do not reject H0
Conclusion: 3 1M M 1 mark
Hypothesis 2
0 1 2
1 1 2
:
:
H M M
VS
H M M
1 2
mark
𝑊𝑝 = 𝐷𝑝,𝑣,𝛼√𝑀𝑆𝐸
�̅�
When P=2, V=12, and α=0.05
2 2,12,0.05
2
200
5
19.49
W D
W
Absolute Difference Value of M1 and M2 is 5
Decision: Since 2 1 2(19.49) (5)W y y we do not reject H0 1 mark
Conclusion: 1 2M M
Hypothesis 3
46
0 3 2
1 3 2
:
:
H M M
VS
H M M
1 2
mark
𝑊𝑝 = 𝐷𝑝,𝑣,𝛼√𝑀𝑆𝐸
�̅�
When P=3, V=12, and α=0.05
3 3,12,0.05
3
200
5
20.39
W D
W
1 2
mark
Absolute Difference Value of M3 and M2 is 10
Decision: Since 3 3 2(20.39) (10)W y y we do not reject H0 1 mark
Conclusion: 3 2M M
6C.) Blocking Efficiency is the method that is used in maximizing the difference among blocks
leaving the difference among plots of the same blocks as well small as possible. It is expected that
the result of every randomized complete block (Latin square) design experiment should be
examined to see how this objective has been achieved.
21 2 mark
47
COVENANT UNIVERSITY
CANAANLAND, KM.10, IDIROKO ROAD
P.M.B.1023, OTA, OGUN STATE. NIGERIA
College: Science and Technology Department: Mathematics
Session: 2014/2015 Semester: Alpha
Course Title: Functional Analysis Course Code: MAT 417
Credit Units: 3 Time: 2.5 Hours
Instruction: Attempt ANY THREE (4) questions .The mark for each questions is 17.5
Question 1(a) Let V be a linear space. Define a map : by N V V V V
1 2 1 2 1 2 1 2 ( , ) ( 2 , 2 ), ,N h h h h h h h h V V
i) Is N linear? (10 marks)
ii) Find the inverse map of N if it exists.
(b) Show that N is bounded if 1 2 1 2( , )V
h h h h (7.5 marks)
Question 2 (a) Every linear operator if bounded. Prove or disprove this claim. (12 marks)
(b) Define a metric space related to the set X . (5.5 marks)
Question 3: (a) Prove that ( , , ),p
pL is a normed linear space, where ( , , )pL is the
set of all equivalent class of -measurable function on K valued satisfying
( ) ( ) , , , 1p pf d f L p
such that p
is defined by ( ) ( )p p
pf f d
. (12 marks)
(b) Let and X Y be two linear spaces, and T , a map defined on and X Y .
When is
T said to be continuous? (5.5 marks)
Question 4: (a) Suppose g a linear functional on [ 1,1]C is defined by
0 1
1 0
( )( ) ( ) ( )gx t x t dt x t dt
Then, show that g is bounded, and hence, find the norm of .g (7.5 marks)
48
(b) Let X be the normed space of all polynomial on [0,1] with the norm given as:
[0,1]
sup ( ) , X
t
x x t x X
Suppose Y X with the same norm and :T X X is a map defined by:
( ) , [0,1], Tx t x t x is the derivative of x X .
Then prove or disprove the claim that T is linear but unbounded. (10 marks)
Question 5: (a) Define the followings:
(i) Cauchy sequence in a metric space ,X d (4 marks)
(ii) Banach spaces
(b) Show that the Euclidean space n is complete (7 marks)
(c) Let X be the space of all ordered 1 2n-tuples , , nx x x x of real
numbers and
, max .j j jd x y x y Show that ,X d is complete. (6.5marks)
Question 6: (a) Define the followings:
(i) an inner product space
(ii) a Hilbert space (5 marks)
(b) State and prove the Schwarz inequality. (8.5 marks)
(c) State the Hahn-Banach theorem (state without proof). (4 marks)
49
MAT417_MARKING GUIDE_OMEGA_2014/15
1a Solution/Hint: 1(ai) 10 marks:
By definition, :N V V V V is given by:
1 2 1 2 1 2, 2 , 2 , N h h h h h h V a linear space. (1*)
ai) For the linearity of N , let , , ,u v V V K , then to show that:
* , , , ,N x y u v N x y N u v
** , , ,N x y N x y x y V V
So, award marks if (*) and (**) are shown correctly.
aii) 1N exists iff ( ) : 0sp sp
N x D N x , so:
1 2 1 2 1 2, 2 , 2 0,0N h h h h h h
1 2 1 22 0 and 2 0h h h h
Showing that 1 2, 0,0h h .
For the only if, 1 2, (0,0) (0,0) (0,0)h h N , hence, 1N exists.
aiii) To obtain 1N , we pre-multiply both sides of (1*) by 1N , thus,
1 1
1 2 1 2 1 2, 2 , 2N N h h N h h h h
Showing that:
1 1
1 2 1 2 1 2
1 1, , ,
2 4N x y N h h h h h h
1b Solution/Hint: 1(b) 7.5 marks:
For the boundedness of N , let ,m K then to see if
1 2 1 2, ,N h h m h h
50
1 2 1 2 1 2
1 2 1 2
, 2 , 2
,V
N h h h h h h
h h h h
given
1 2 1 2 1 2, 2 , 2N h h h h h h
1 2 1 22 2h h h h
1 2 1 22 2h h h h
1 24 h h
So, N is bounded with a norm 4M
2a Solution/Hint: 2(a) 12 marks:
Every linear operator is bounded. Prove or disprove.
This statement is not tue, as such to be disproved as follows:
Suppose H is a normed space of all polynomials defined on the interval [0,1]I with a norm
given by:
[0,1]
sup ( ) , H
t
h h t h H
And claiming H P with the same norm, such that a map M is defined by:
, : , , Mh t h M H H t I h denotes the derivative of in h H .
Showing that H is linear is straight forward, so for boundedness of H , let nh H be a monomial
such that:
( ) , n
nh t t t I
1n n
nMh t nt
1sup ( ) = sup n
nHt I t I
h Mh t nt n
Since n is arbitrary, it thus follows that M is unbounded though linear. Therefore, every linear
operator is not bounded. Q.E.D.
2b Solution/Hint 1 (ai) 5.5 marks:
51
Let X be a non-empty set and d a function defined on X such that: then
: [0, ),d X X
(i) ( , ) 0, , , with equality iff ( property).d x y x y z X x y positivity
( ) ( , ) ( , ) , (symmetry property).ii d x y d y x
(iii) ( , ) ( , ) ( , ) (Triangle inequality).d x y d x z d z y
Then the pair ( , )X d is called a metric space.
3a Solution/Hint 3(a) 12 marks:
The non-trivial part to show is the in the aboveclaim is the Minkowski’s inequality:
, , ( , , ),p
p p p pf g f g f g L
For p=1, the proof is trivial.Thus, we consider a case when 1 p as follows:
By definition,
( ) ( ), pp
pf f u du u
(1)
But
1( ) ( ) ( ) ( ) ( ) ( )
p pf u g u f u g u f u g u
1( ( ) ( ) ) ( ) ( )
pf u g u f u g u
1 1( ) ( ) ( ) ( ) ( ) ( )
p pf u f u g u g u f u g u
1 1( ) ( ) ( ) ( )
p p
p pf f u g u g f u g u
1p p
p p p pf g f g f g
Thus , , ( , , ), Q.E.D.p
p p p pf g f g f g L
3b Solution/Hint 3(a) 5.5 marks:
1
( 1)1 ( ) ( ) ( ) ( ) ( ) ( ) ( )
p
ppp p
pp p
f u g u du f g f u g u du
52
Let : ( )T D T X Y be a map, then T is said to be continuous at a point 0 ( )x D T if for any
0 0 >0, ( ) s.t. , ( ) , whenever ( )Tx Tx x D T x x .
4a Solution/Hint 4a 7.5 marks:
For h a linear functional on [1,3]C defined by:
3
1
( )hx t x t dt
Recall that: [ , ]
( ) ( ( )( )t a b
f x Sup f x t
[1,3]
( ) ( ( )( )t
h x Sup h x t
3
[1,3]1
( )t
Sup x t dt
Showing that: ( ) 2X
h x x
Whence, h is bounded, and the norm is 2h .
4b Solution/Hint 4b 10 marks:
To show that :T X X defined by ( ) , :[0,1]Tx t x t is linear but not bounded. Let
1 2 1 2, , ,x x X K , Therefore,
1 1 2 2 1 1 2 2( )d
T x x t x x tdt
1 1 2 2
d dx t x t
dt dt
Thus, :T X X is indeed linear. For the unboundedness, see Qstn 2a.
5a Solution/Hint 5a :4 marks:
Let ( , )X X d be a metric space, then a sequence nx in X is said to be Cauchy if for every
0, there is ( )N N such that:
53
( , ) , for every ,m nd x x m n N .
5aii) X is a complete space if every Cauchy sequence in X converges to a point in .X
Thus, a complete metric space is a Banach space.
5b Solution/Hint 5b :7 marks:
To show that n is complete , we define:
1
22
1
( , )n
i i
i
d x y
Where and with in n
i i nx y x as a Cauchy sequence.
54
MAT 418
Question 1
a. 7 6y x
Therefore, 2
2
1
67 119(z)dz 7 6 7
3
if x ix x dx i dx
b. Consider the function f z
z a which is analytic at all points within a circle C except at point z a .
With the point a as centre and radius r , draw a circle 1C lying within C .
Now f z
z a is analytic within the region between
1C and C . Hence by Cauchy’s integral theorem, we
have,
1
2
0
i
i
i
C C
f a re f af z f zdz dz ire d
z a z a re
2
0
if a re f a id
Since r can be made as small as possible i.e., r tends to zero, we have
2
0
2if a re f a id i
Hence, we have
0 2C
f zdz if a
z a
Therefore,
55
1
2C
f zf a dz
i z a
Question 2
Take any point z inside C . Draw a circle 1C with centre a , enclosing the point z . Let w be a point on
circle 1C .
1 1 1
w z w a a z w a z a
11
1z a
w a w a
By applying Binomial theorem
2
2 3 1
1 1... ...
n
n
z a z az a
w z w a w a w a w a
Since z a w a or 1z a
w a
, the series converges uniformly. Hence, the series is integrable.
Multiplying the series by f w , we have
2
2 3 1
( )... ...
n
n
f w f w f w z a f w z a f w z a
w z w a w a w a w a
Integrating with respect to w , we get
1 1 1 1
2 1... ...
( ) ( )n
C C C C
f w f w f w f wdz dw z a dw dw
w z w a w a w a
By substitution, we have
2''
' ... ...2! 2!
nnf a f a
f z f a f a z a z a z a
56
b. Given 2
1
4f z
z
singularities are 2z i and 2z i . The distances of singularities 2z i and 2z i from the centre
are 1 and 3. Within the circle 1z i , f z is analytic, hence can be expanded in a Taylor’s series
within the circle 1z i which is the circle/region of convergence.
With 2
1
4f z
z
,
2
1 1
4 3f z i
i
, 2 2
2 2'
( 4) 9
z if z
z
,
42''
81f z
By Taylor’s series expansion of 2
1
4f z
z
, the solution gives
2
2
2 201 1...
4 3 9 162
i z i z if z
z
Question 3
a.
2
0
1 /
15 4sin 1 125 4
22c c
dz iz dzd
z i z izi z
the poles are 2z i and 1
2z i
2 2
1 2Re Re [ ]
1 32
2
z i z i
is f z s
z i z i
and
1 1
2 2
1 2Re Re [ ]
1 52
2
z i z i
s f z s i
z i z i
57
b. 2 2
2 32
0 0
3 111 1 2
4 2 12 2
i ii i
z dz x i dx i
Question 4
a.
2 3
2 2
1 1 11 ...
1f z z z z
z z z z
b. By L’Hospital rule
0
0
2cos sinRe arg 2
coszz
z z zs f z
z
For sin 0z
0, 1, 2,...z m m
Outside 0, 2D
Question 5
a. i. A point 0z is called a singular point of a function f if f fails to be analytic at
0z but is analytic at
some point in every neighbourhood of 0z
ii. A singular point 0z is said to be isolated if there is a deleted neighborhood 00 z z of
0z
throughout which f is analytic.
b. i. 20
1Re 1
zs
z z
ii. 0
Re zcot(1/ z) 1/ 2z
s
iii. 0
sinRe 0
z
z zs
z
58
Covenant University, Canaanland,
Km 10 Idiroko Road, P.M.B. 1023, Ota,
Ogun State, Nigeria
B.Eng Degree Examination
College: Science and Technology Department: Mathematics
Session: 2014/2015 Semester: Alpha
Course Code: GEC 410 Course Title: Probability &
Statistics
Credit Units: 2 Time: 2 Hrs
Instruction: Attempt any THREE questions.
QUESTION 1
(a). Find the probability of getting exactly 2 heads in 6 tosses of a fair coin.
5 marks.
(b). Consider an experiment of drawing two cards at random from a bag containing
four cards marked with integer 1 through 4. (i) Find the sample space S1, of
the experiment if the first card is replaced before the second is drawn. (ii) Find
the sample space S2, of the experiment if the first card is not replaced.
8 marks
(c). If the probability that an individual will suffer a bad reaction from injection
of a given serum is 0.001, determine the probability that out of 2000
individuals, (i) exactly 3, (ii) more than 2, individuals will suffer a bad
reaction.
10 marks
QUESTION 2.
(a) A lot of 100 semiconductor chips contains 20 that are defective. Two chips
are selected at random, without replacement, from the lot.
(i). What is the probability that the first one selected is defective?
3 marks
(ii). What is the probability that the second one selected is defective given that the
first one was defective?
4 marks
59
(iii) What is the probability that both are defective?.
3 marks
(b). Suppose that a game is to be played with a single die assumed fair. In this
game a player wins ₦20 if a 2 turns up; ₦40 if a 4 turns up; loses ₦30 if a 6
turns up; while the player neither wins nor loses if any other face turns up.
Find the expected sum of money to be won, variance and standard deviation
for the game.
13 marks
QUESTION 3.
(a). Find the probability (i). p(Z < -1.32) (ii). p(Z > - 1.32) (iii). p( - 2.1 < Z < -
1.3 ) (iv). p (- 1.1 < Z < 2.3)
8 marks
(b). Lifetimes of batteries in a certain application are normally distributed with
mean 50 hours and standard deviation 5 hours. Find the probability that a
randomly selected and chosen battery lasts between 42 and 52 hours.
6
marks
(c). Explain the term “Central Limit Theorem’’
4
marks
(d). Bottles filled by a certain machine are supposed to contain 12oz of liquid. In
fact the fill volume is random with mean 12.01oz and standard deviation of
0.2oz. What is the probability that the mean volume of a
random sample of 144 bottles is less than 12oz? Use the theorem in (c).
5
marks QUESTION 4.
(a). If X1, X2 , X3 and X4 are independent random variables having the
Bernoulli distribution with parameter θ. Show that the following estimators
of θ are unbiased.
(i). 𝑋1+𝑋2+𝑋3+𝑋4
4 (ii).
3𝑋1+2𝑋2+3𝑋3
8
4 marks
60
(b). A sample of 12 measurements of the breaking strength of cotton threads gave
a mean of 7.38cm and a standard deviation of 1.24cm. Find the 95%
confidence limits for the actual mean breaking strength.
5 marks
(c) State four properties of estimators. 4 marks
(d). It has been found from experience that the mean lifetime of a sample of 50
bulbs produced by a company is computed to be 1570 hours with a standard
deviation of 100 hours. If µ is the mean lifetime of all bulbs produced by the
company. Test the hypothesis: µ= 1000 hours against the alternative µ ≠1000
hours.α =0.01. 10 marks
QUESTION 5.
(a). Stainless steels can be susceptible to stress corrosion cracking under certain
conditions. A materials engineer is interested in determining the proportion of
steel alloy failures that are due to stress corrosion cracking. In a sample of 200
failures, 30 of them were caused by steel corrosion cracking. Find the 95%
confidence interval for the proportion of failures caused by stress corrosion
cracking.
5 marks
(b). Plates are evaluated according to their finish, and placed into four categories:
premium, conforming, downgraded and unacceptable.
A quality engineer claims that the proportions of plates in the four categories
are 10%, 70%, 15% and 5% respectively. In a sample of 200 plates; 19 were
classified as premium, 133 were classified as conforming, 35 were classified
as downgraded and 13 were classified as unacceptable. Can you conclude at
0.05 level of significance that the engineer\s claim is incorrect?
8 marks
(c). The following table presents shear strengths (in kNmm) and weld diameter
(in mm) for a sample of spot welds.
Diameter (x) 4.2 4.4 4.6 4.8 5.0 5.2 5.4 5.6
Strength (y) 51 54 69 81 75 79 89 101
Obtain a regression line of y on x. What is the value of the strength when the
diameter is 5.5?
10 marks
61
Covenant University,
Canaanland,
Km 10 Idiroko Road, P.M.B. 1023, Ota,
Ogun State, Nigeria
B.Eng Degree Examination MARKING GUIDE
College: Science and Technology Department: Mathematics
Session: 2014/2015 Semester: Alpha
Course Code: GEC 410 Course Title: Probability &
Statistics
Credit Units: 2 Time: 2 Hrs
Question 1 a 𝑃(𝑋 = 𝑥) =
(𝑛𝑥
) 𝑝𝑥𝑞𝑛−𝑥 =𝑛!
𝑥!(𝑛−𝑥)! 𝑝𝑥𝑞𝑛−𝑥
Given 𝑛 = 6, 𝑋 = 𝑥 = 2, 𝑝 =1
2, 𝑞 = 1 −
1
2=
1
2.
𝑃(𝑋 = 2) = (62
) (12
)2
(12
)6−2
=6!
2!4!(
12
)2
(12
)4
=15
64 5 marks
(b)
(i) The sample space 𝑆1, contain 16 ordered pairs (𝑖, 𝑗), 1 ≤ 𝑖 ≤ 4, 1 ≤ 𝑗 ≤ 4, where
the first number indicates the first number drawn. Thus,
𝑆1 = {
(1,1), (1,2), (1,3), (1,4)(2,1), (2,2), (2,3), (2,4)(3,1), (3,2), (3,3), (3,4)(4,1), (4,2), (4,3), (4,4)
}
(ii) The sample space 𝑆2, contain 12 ordered pairs (𝑖, 𝑗), 𝑖 ≠ 𝑗, 1 ≤ 𝑖 ≤ 4, 1 ≤ 𝑗 ≤4, where the first number indicates the first number drawn. Thus
𝑆2 = {
(1,2), (1,3), (1,4)(2,1), (2,3), (2,4)(3,1), (3,2), (3,4)(4,1), (4,2), (4,3)
} 8 marks
62
c Let X denote the number of individuals suffering a bad reaction X is Bernoulli distributed,
but since bad reactions are assumed to be rare events, we can suppose that X is Poisson
distributed i.e.
𝑃(𝑋 = 𝑥) =𝜆𝑥𝑒−𝜆
𝑥!, where 𝛌= (2000)(0.001) = 2
(i) 𝑃(𝑋 = 3) =23𝑒−2
3!= 0.18,
(ii) 𝑃(𝑋 > 2) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2)]
= 1 − [20𝑒−2
0!+
21𝑒−2
1!+
22𝑒−2
2!]
= 1 − 5𝑒−2
= 0.323 10 marks
QUESTION 2 a
(i). Let A denote the event that the first one selected is defective;
Then
𝑃(𝐴) =20
100= 0.2 3 marks
(ii). Let B denote the event that the second one selected is defective. After the first one selected
is defective, there are 99 chips left in the lot with 19 chips that are defective. Thus, the
probability that the second one selected is defective given that the first one was defective is
𝑃(𝐵/𝐴) =19
99= 0.192 3 marks
(iii). The probability that both are defective is
𝑃(𝐴 ∩ 𝐵) =19
99= 𝑝 (
𝐵
𝐴) 𝑝(𝐴) = (0.2)(0.192) = 0.038 4
marks Total
10 marks
b
Let 𝑋 be the random variable giving the amount of money won on any toss. The possible
amounts won when the die turns up 1, 2, … ,6 are 𝑥1, 𝑥2, . . , 𝑥6, respectively, while the
probabilities of these are 𝑓(𝑥1), 𝑓(𝑥2), … , 𝑓(𝑥6). The probabilities function for 𝑋 is given
by
X 0 +20 0 +40 0 -30
63
f(x) 1
6
1
6
1
6
1
6
1
6
1
6
Therefore, the expected value is defined as 𝐸(𝑋) = ∑ 𝑥𝑓(𝑥)𝑥
𝐸(𝑋) = (0) (1
6) + (20) (
1
6) + (0) (
1
6) + (40) (
1
6) + (0) (
1
6) + (−30) (
1
6) = 5
It follows that the player can expect to win $5. In a fair game, therefore, the player should
expect to pay $5 in order to play the game.
Since 𝜇 = 5, thus, the variance is given by 𝜎𝑥2 = 𝐸[(𝑥 − 𝜇)2] = ∑ (𝑥𝑗 − 𝜇)2
𝑓(𝑥𝑗)𝑛𝑗 .
𝜎𝑥2 = (0 − 5)2 (1
6) + (20 − 5)2 (
1
6) + (0 − 5)2 (
1
6) + (0 − 5)2 (
1
6) + (40 −
5)2 (1
6) + (−30 − 5)2 (
1
6) =
2750
6= 458.33 .
The standard deviation is √𝜎𝑥2= 𝜎𝑥.
𝜎𝑥 = √458.333 =21.40872096.
Total 13
marks
Question 3 a
(i). 𝑝(𝑍 < −1.32) = 0.0934
(ii). 𝑝(𝑍 > −1.32) = 1 − 𝑝(𝑍 < −1.32) = 1 − 0.0934 = 0.9066
(iii). 𝑝(−2.1 < 𝑍 < −1.30) = 𝑝(𝑍 < −1.30) − 𝑝(𝑍 < −2.1) = 0.0968 − 0.0179 =
0.0789
(iv). 𝑝(−1.1 < 𝑍 < 2.3) = 𝑝(𝑍 < 2.3) − 𝑝(𝑍 < −1.1) = 0.9893 − 0.1357 = 0.8536
2 marks each Total 8 marks
b
Let X be the lifetimes of battery. Required probability is p(42 < X < 52).
𝑍 =𝑋−𝜇
𝜎 , 𝜇 = 50, 𝜎 = 5.
𝑊ℎ𝑒𝑛 𝑋 = 42, 𝑍 =42−50
5= −1.60 ,
𝑊ℎ𝑒𝑛 𝑋 = 52, 𝑍 =52−50
5= 0.40,
𝑝(42 < 𝑋 < 52) = 𝑝(−1.60 < 𝑍 < 0.40)
64
𝑝(−1.60 < 𝑍 < 0.40) = 𝑝(𝑍 < 0.40) − 𝑝(𝑍 < −1.60) = 0.6554 − 0.0548 = 0.6006
6 marks
c
The central limit theorem states that for a population with a mean µ and standard deviation σ, the
sampling distribution of the means of all possible samples of size n generated from the
population will be approximately normally distributed – with the mean of the sampling
distribution equal to µ and the standard deviation equal to 𝜎
√𝑛 assuming that the sample size n is
sufficiently large.
Let X1, X2, …..Xn be a random sample from a distribution with mean µ and finite variance σ2.
𝐿𝑒𝑡 �̅�𝑛 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑎 𝑠𝑎𝑚𝑝𝑙𝑒 𝑜𝑓 𝑠𝑖𝑧𝑒 𝑛.
𝑇ℎ𝑒𝑛 �̅�𝑛−𝜇
√𝜎2
𝑛
~𝑁(0, 1) 𝑎𝑠 𝑛 → ∞ 4 marks
d
Let X1, X2, . . . .X144 be the volume of sample of bottles.
𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 𝑝(�̅� < 12) 𝜇 = 12.01, 𝜎2 = 0.04, 𝑛 = 144
𝐼𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑓𝑟𝑜𝑚 𝐶𝐿𝑇, 𝑍 = �̅�𝑛−𝜇
√𝜎2
𝑛
=12−12.01
√0.04
144
= −0.60
𝑝(�̅� < 12) = 𝑝(𝑍 < −0.60) = 0.2743 5 marks
Question 4 a
(i). �̂�𝟏 =𝑿𝟏+𝑿𝟐+𝑿𝟑+𝑿𝟒
𝟒
𝑬[�̂�𝟏] = 𝑬 [𝑿𝟏+𝑿𝟐+𝑿𝟑+𝑿𝟒
𝟒] =
𝟏
𝟒(𝑬[𝑿𝟏] + 𝑬[𝑿𝟐] + 𝑬[𝑿𝟑] + 𝑬[𝑿𝟒])
𝑬[�̂�𝟏] =𝟏
𝟒(𝜽 + 𝜽 + 𝜽 + 𝜽) = 𝜽 𝒖𝒏𝒃𝒊𝒂𝒔𝒆𝒅.
(ii). �̂�𝟐 =𝟑𝑿𝟏+𝟐𝑿𝟐+𝟑𝑿𝟑
𝟖
𝑬[�̂�𝟐] = 𝑬 [𝟑𝑿𝟏+𝟐𝑿𝟐+𝟑𝑿𝟑
𝟖] =
𝟏
𝟖(𝟑𝑬[𝑿𝟏] + 𝟐𝑬[𝑿𝟐] + 𝟑𝑬[𝑿𝟑])
𝑬[�̂�𝟐] =𝟏
𝟖(𝟑𝜽 + 𝟐𝜽 + 𝟑𝜽) = 𝜽 𝒖𝒏𝒃𝒊𝒂𝒔𝒆𝒅
2 marks each Total 4
marks
b
𝑛 = 12 , �̅� = 7.38, 𝑠 = 1.24 𝑡0.025 ,11 = 2.2
�̅� ± 𝑡𝛼
2
𝑠
√𝑛= 7.38 ± (2.2)
1.24
√12= 7.38 ± 0.79 = (6.59 , 8.17) 5 marks
c
unbiasedness efficiency sufficiency consistency 4 marks
65
d
H0 : µ = 1000 hours
H1 : µ ≠ 1000 hours
It is a two tailed test.
𝑇𝑜 𝑡𝑒𝑠𝑡 𝛼 = 0.01 ,𝛼 ,
2= 0.005 1 −
𝛼
2= 1 − 0.005 = 0.995
𝐹𝑟𝑜𝑚 𝑡ℎ𝑒 𝑡𝑎𝑏𝑙𝑒 𝑍𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙 = ±2.58
𝑇𝑒𝑠𝑡 𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 𝑍𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =�̅�−𝜇
𝜎 √𝑛⁄ �̅� = 1570 , 𝜎 = 100 , 𝑛 = 50 , 𝜇 = 1000
𝑍𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 =1570−1000
100 √50⁄= 40.31 𝑍𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 > 𝑍𝑐𝑟𝑖𝑡𝑖𝑐𝑎𝑙
We REJECT H0 since it lies on the critical part. 10 marks
Question 5 a
�̂� =𝑥
𝑛=
30
200= 0.15 1 − 𝑃 = 0.85 𝑍𝑐 = 1.96
�̂� ± 𝑍𝛼
2 √
𝑝(1−𝑝)
𝑛= 0.15 ± (1.96)√
(0.15)(0.85)
200= 0.15 ± 0.05 = (0.10 , 0.20) 5 marks
b
The total number of samples is 200.
Category Expected frequency
Premium 10% of 200 = 20
Conforming 70% of 200 = 140
Downgraded 15% of 200 = 30
Unacceptable 5% of 200 = 10
200
Bearing O E 𝑂𝑖 − 𝐸𝑖 (𝑂𝑖 − 𝐸𝑖)2 (𝑂𝑖 − 𝐸𝑖)2
𝐸𝑖⁄
Premium 19 20 - 1 1 0.05
Conforming 133 140 - 7 49 0.35
Downgraded 35 30 5 25 0.83
Unacceptable 13 10 3 9 0.90
2.13
𝜒2𝑐𝑎𝑙
= 2.13
The degrees of freedom k – 1 = 4 – 1 = 3.
𝜒2𝑡𝑎𝑏
= 𝜒20.05 ,3
= 7.81
𝜒2𝑐𝑎𝑙
< 𝜒2𝑡𝑎𝑏
The engineer claim is not incorrect 8 marks
c
66
x y xy x2
4.2 51 214.2 17.64
4.4 54 237.6 19.36
4.6 69 317.4 21.16
4.8 81 388.8 23.04
5 75 375 25
5.2 79 410.8 27.04
5.4 89 480.6 29.16
5.6 101 565.6 31.36
39.2 599 2990 193.76
𝑏 =𝑛 ∑ 𝑥𝑦−∑ 𝑥 ∑ 𝑦
𝑛 ∑ 𝑥2−(∑ 𝑥)2=
8(2990)−(39.2)(599)
8(193.76)−(39.2)2=
439.2
13.44= 32.6
�̅� =∑ 𝑦
𝑛=
599
8= 74.875 �̅� =
∑ 𝑥
𝑛=
39.2
8= 4.9
𝑎 = �̅� − 𝑏�̅� = 74.875 − (32.6)(4.9) = −84.865
�̂� = 𝑎 + 𝑏𝑥 = −84.865 + 32.6𝑥 This is the regression line
𝑊ℎ𝑒𝑛 𝑥 = 5.5 𝑦 = −84.865 + 32.6(5.5) = 94.44 ≈ 94. 10 marks