MAT 2371 - Fall 2014
Assignment 2 - Solutions
1.3-2 (3 points)
1041
a) P (A1 ) = 1456 = 0.715
1 S
b) P (A1 |S1 ) = N (A(S1 ) 1 ) =
N
N (A1 S2 )
N (S2 )
392
633
649
823
= 0.619
=
= 0.789
c) P (A1 |S2 ) =
d) In parts b) and c), we have calculated the propor
MAT 2379
Introduction to biostatistics
Summary of R commands (Part 1)
1. Binomial distribution with parameters n and p:
To compute f (x) = P (X = x) :
dbinom(x,n,p)
To compute F (x) = P (X x) :
pbinom(x,n,p)
To compute the binomial coefficient
n
k
:
choos
1
MAT2379, Introduction to biostatistics
Assignment 3
Due date: Friday November 18, 2016 at 3:00 p.m.
Total = 100 marks
Solve the following problems from the textbook using a Faculty-standard calculator 1 and a Standard
Normal Table:
10.4 (20 marks),
10.7
MAT 2379, Introduction to Biostatistics, Lecture Notes for Chapter 5
1
MAT 2379, Introduction to Biostatistics
Chapter 5. Independence
In Chapter 4, we discussed the concept of mutually exclusive events. These are events which
can not occur in the same ti
A special type of deductive argument
is a categorical syllogism.
Categorical Statements
In
categorical
reasoning
the
statements, or claims, of interest are
categorical statements.
Categorical statements make
simple
assertions
about
categories, or classes
1. Some athletes are not baseball players, and some
baseball players are not basketball players. Therefore,
some athletes are not basketball players.
2. At least one lawyer is not a golfer. All lawyers are
persons who have attended law school. So, at leas
Deductive Argument
An argument intended to provide logically
conclusive support for its conclusion.
The defining characteristic of a deductive argument is that
it is valid or invalid.
The support that a deductive argument provides its
conclusive is:
F
MAT 2379, Introduction to Biostatistics, Section 9.1 (Part 1)
1
MAT 2379, Introduction to Biostatistics
Chapter 9. Introduction to statistics
9.1 Random Sampling and Data Description
We will start by examining the question:
What is a statistical problem?
Argument by Analogy
An analogy is a comparison of two or more things
that are alike in specific ways.
Analogies can also be used to argue inductively
for a conclusion.
Such arguments are known as analogical
induction or argument by analogy.
It works li
MAT 2379, Introduction to Biostatistics
Midterm Formula Sheet
Addition Rule: P (A B) = P (A) + P (B) P (A B)
Conditional probability of A given B:
P (A B)
P (B)
P (A|B) =
Let E1 , E2 , . . . , Ek be a partition of the sample space (i.e. they are
mutual
MAT 2371 - Fall 2014
Assignment 3 - Solutions
2.1-6 (3 points)
(a) The space of possible outcomes of X is cfw_2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. In
each case, fX (x) = P (X = x) = P cfw_(i, j) : i + j = x, 1 i, j 6, and
1
P (i, j) = 36 . Therefore,
1
fX
MAT 2371 - Fall 2014
Assignment 1 - Solutions
1.1-4 (4 points)
(a) Let H = head and T = tail. The sample space is:
S = cfw_HHHH, HHHT, HHT H.HT HH, T HHH, HHT T, HT HT, T HHT,
HT T H, T HT H, T T HH, HT T T, T HT T, T T HT, T T T H, T T T T .
(b) The even
MAT 2371 - Fall 2014
Assignment 5 - Solutions
3.1-8 (6 points)
Graphs are not required for this problem.
4
c
c4
(a) (i) 1 = f (x)dx = 0 x3 /4dx = x |c = 16 c = 2.
16 0
x<0
0
x
x 3
x4
(ii) F (x) = f (y)dy =
y /4dy = 16 0 x 2
0
1
x>2
2
x3
x5 2
32
(iv) = x
MAT 2371 - Fall 2014
Assignment 6 - Solutions
3.3-2 (4 points)
Using Table Va and noting that (z) = 1 (z),
(a) P (0 Z 0.87) = (0.87) (0) = .8078 .5000 = .3078
(b) P (2.64 Z 0) = (0) (2.64) = .5000 (1 .9959) = .4959
(c) P (2.13 Z 0.56) = (2.13) (0.56) = .9
MAT 2371 - Fall 2014
Assignment 8 - Solutions
5.6-4 (2 points)
Using the central limit theorem with n = 32,
P (39.75 X 32 < 41.25)
= P
39.75 40
8/32
X 32 40
8/32
<
41.25 40
8/32
P (0.50 Z < 2.5) = 0.6853
5.6-8 (3 points)
(a) E[X] = = 24.43
2
(b) Var(X) =
MAT 2371 - Fall 2014
Assignment 7 - Solutions
5.3-18 (2 points)
Let Xi be the lifetime of the ith part, i = 1, 2, 3. Xi gamma with = = 2.
2
Therefore, E[Xi ] = = 4 and Xi = 2 = 8. The total lifetime is
T = X1 + X2 + X3 . By independence of the Xi s,
E[T ]
MAT 2379A
Midterm Examination (with Solutions)
October 22, 2014
Time: 80 minutes
Professor Raluca Balan
Student Number:
Family Name:
First Name:
This is a closed book examination. A formula sheet and some statistical
tables are included with the exam. Onl