ST107 Class 2 Additional exercises
1. Calculate the following probabilities. In each case you can assume that all outcomes are
equally likely.
(a) A die is thrown five times in a row. The score of each throw is one of the numbers
1, 2, 3, 4, 5 or 6. What
ST107 Exercise 1
You should make arrangements with your class teacher for the dates when
solutions to exercises should be handed in, and when they will be marked and
returned to you. You should spend a reasonable amount of time on each set
of exercises. D
ST107 Class 2 Solutions to Additional exercises
1. (a) The size of the sample space is the total number of ordered sequences with
replacement, which is 65 = 7776. To count the sequences which satisfy the
requirement, it is easiest to simply list them:
1
1
Sharing Econ Articles
Inc - Resistance Is Futile
Uber, the car service that is not a car service, knows that any time it moves into a new city, it will face
a fight. That's fine. Fighting is good, winning is better, and explosive growth is best of all.
"T
Wealth of networks
1 Enabled by technological change, we are beginning to see a series of economic, social, and
cultural adaptations that make possible a radical transformation of how we make the information
environment we occupy as autonomous individuals
MY451
Introduction to Quantitative Analysis
2014-15
Jouni Kuha
Department of Methodology
London School of Economics and Political Science
Course information
MY451: Introduction to Quantitative Analysis
Course Description
This course is intended for those
Chapter 2: Three Approaches to Qualitative Data
Analysis
Introduction
In this chapter, you will learn about the fundamental approach
to qualitative data analysis: thematic analysis. This is a
generic approach to data analysis that enables data sources to
The Net Delusion
Evgeny Morozov
p.13 This was globalization at its worst: A simple email based on the premise that Twitter
mattered in Iran, sent by an American diplomat in Washington to an American company in
San Francisco, triggered a worldwide internet
Textual Analysis Database Companies
Uber.com/about
Uber is evolving the way the world moves. By seamlessly connecting riders to drivers through our
apps, we make cities more accessible, opening up more possibilities for riders and more business for
driver
Zero Marginal Cost Society
Jeremy Rifkin
p.1-2 The capitalist era is passing. Not quickly, but inevitably. A new economic paradigm the Collaborative Commons - is rising in its wake that will transform our way of life. We are
already witnessing the emergen
ST102 Outline solutions to Exercise 7
1. We have:
MX (t) = E(etX ) =
k
X
etx
x=1
=
=
1
k
et e2t
ekt
+
+ +
k
k
k
1 t
(e + e2t + + ekt ).
k
The bracketed part of this expression is a geometric series with a first term of et and a
common ratio also of et . H
ST102 Exercise 8
In this exercise you will practise working with some common probability distributions.
Questions 1 and 2 involve the normal distribution. Questions 3 and 4 concern the exponential
distribution. A very important part of this weeks exercise
ST102 Outline solutions to Exercise 8
1. SinceX N (8, 4), we use the transformation Z = (X )/ with the values = 8 and
= 4 = 2.
(a)
i. We have P (X > 10.4) = P (Z > (10.4 8)/2) = P (Z > 1.2) = 0.1151.
ii. We have:
P (6 < X < 7) = P
78
68
<Z<
2
2
= P (1 <
ST102 Exercise 7
In this exercise you will practise working with some common discrete distributions, such as the
discrete uniform distribution (Question 1) and the binomial distribution (Questions 2 and 4).
The last two questions are theoretical. In Quest
ST102 Class 3 Additional exercises
1. Calculate the following probabilities. In each case you can assume that all outcomes are
equally likely.
(a) A die is thrown five times in a row. The score of each throw is one of the numbers
1, 2, 3, 4, 5 or 6. What
ST102 Class 2 Solutions to Additional exercises
1. [1] Just 1 possibility: No-one is connected with anyone else.
[2] 6: As given in the question.
[3] 3: This looks similar to [2], but is not quite the same. Here it is sufficient to pick
just one person, a
ST102 Class 6 Additional exercises
1. Show that:
(
1/x2
f (x) =
0
for x 1
otherwise
is a valid pdf, but that X does not have a finite expected value.
2. The waiting time X (in minutes) of a customer at a taxi rank has the following probability
density fun
ST102 Class 3 Solutions to Additional exercises
1. (a) The size of the sample space is the total number of ordered sequences with
replacement, which is 65 = 7776. To count the sequences which satisfy the
requirement, it is easiest to simply list them:
1
1
ST102 Class 2 Additional exercises
1. Consider the patterns of four-person networks below (introduced on page 40 of the course
pack).
[1]
[2]
[3]
[4]
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
s
@
s
s
s
s
s
@
@
@s
s
s
s
s
s
[5]
s
s
[6]
s
[9]
s
s
[7]
[10]
s
s
s
@
[
ST102 Outline solutions to Exercise 8 (201415)
1. These should be simple exercises in using statistical tables of the standard normal distribution. For (a), you can
use the table directly. For (b), you use it for the the standardised
variable Z = (X 2)/
ST102 Outline solutions to Exercise 2 (201415)
1. There is more than one way to answer this question, because the sets can be expressed
in different, but logically equivalent, forms. One way to do so is the following:
(a) A B c C c .
(b) Ac B c C c .
(c)
ST102 Outline solutions to Exercise 5 (201415)
1. We have:
p(x) = P (X = x) = P (X x) P (X x 1).
However, the event cfw_X x occurs when all three dice are x and that can occur in x3
ways. Therefore:
P (X x) = x3 /216.
Similarly:
P (X x 1) = (x 1)3 /216.
T
ST102 Outline solutions to Exercise 3 (201415)
1. (a) The size of the sample space is the total number of ordered sequences with
replacement, which is 65 = 7776. To count the sequences which satisfy the
requirement, it is easiest to simply list them:
1
1
ST102 Outline solutions to Exercise 13 (201415)
as the estimator of the population mean, which is the
1. (a) We use the sample mean, X,
method of moments estimator (MME), the least sqaures estimator (LSE) and also
the maximum likelihood estimator (MLE).
ST102 Outline solutions to Exercise 12 (201415)
1. (a) We have E
k
P
k
P
=
ai Xi
E(ai Xi ) =
i=1
i=1
k
P
ai E(Xi ).
i=1
(b) We have:
Var
k
X
!
ai Xi
= E
i=1
=
k
X
k
X
i=1
a2i E (Xi E(Xi )2 +
i=1
=
k
X
i=1
ai Xi
k
X
!2
ai E(Xi )
i=1
X
= E
k
X
!2
ai (Xi
ST102 Exercise 1, Outline solutions (201415)
1. (a) Insert your guess here. Chances are that you got the average correct, without
even thinking that it could possibly be anything else. The standard deviation was
probably less obvious.
(b)
i. Using the rul
ST102 Outline solutions to Exercise 11 (201415)
1. Each possible sample has an equal probability of occurrence of 1/6.
is:
distribution of X
(1, 2)
(2, 1)
(1, 3)
(3, 1)
(2, 3)
(3, 2)
=x
Sample mean, X
1.5
2
2.5
=x
Relative frequency, P (X
)
1/3
1/3
1/3
Examiners commentaries 2012
Examiners commentary 2012
ST102 Elementary Statistical Theory
General remarks
Learning outcomes
By the end of this module you should:
be able to summarise the ideas of randomness and variability, and the way in which these lin
ST102 Outline solutions to Exercise 12 (201314)
1. (a) E
(b)
k
P
ai Xi
i=1
Var
k
X
i=1
=
k
X
=
k
P
E(ai Xi ) =
i=1
a i Xi
!
a2i E (Xi
k
P
0
= E@
k
X
=
i=1
k
X
ai Xi
i=1
E(Xi )2 +
i=1
k
X
ai E(Xi ).
i=1
ai E(Xi )
i=1
X
!2 1
ai aj E (Xi
0
A = E@
k
X
ai (Xi