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MATH 1081 UO Mathematical Methods for Data Analytics 2 Continuous Assessment 1.2: Problem Solving Exercise Instructions: • Structure of the assessment: This assessment is worth 25% of your...

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MATH 1081 UO Mathematical Methods
for Data Analytics 2
Continuous Assessment 1.2: Problem Solving Exercise
Instructions:
• Structure of the assessment: This assessment is worth 25% of your final
grade and is due no later than 12 pm on Monday, Week 9. This assessment
consists of 9 questions which ca
y 100 marks.
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1
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2
1. Let
dy1
dt
= y′1 = 3y2
dy2
dt
= y′2 = −y1 + 4y2,
where y1(0) = 1 and y2(0) = 1. Find the solution of the system and comment on
its stability. [10 marks]
2. Three sky-jumpers attempt independently of each to land on a straight line on the
ground marked ABC, where B is the half-way point of the line AC. Each jumpe
is able to land somewhere on the line with probability 0.8. Otherwise the jumpe
ends up landing some where away from the line. If any one of the jumper happens
to land on the line, then he/she will land in the AB section with probability 0.5,
or in the BC section also with probability 0.5.
(a) What is the probability that any 2 out of the 3 jumpers manage to land
somewhere on the line? [2 marks]
(b) What is the probability that any 2 out of the 3 jumpers manage to land
somewhere on the AB section of the line? The other one could have eithe
landed in the BC section or not on the line at all. [2 marks]
(c) Jenny is one of the jumpers. What is the probability that Jenny is one of
only two jumpers who manage to land somewhere on the line? [3 marks]
(d) What is the probability that two jumpers manage to land somewhere on the
line, given that Jenny is one of them? [3 marks]
[10 marks]
3. For a particular COVID test, there is a probability of 0.2 that the test result
comes out positive if it is randomly applied to a randomly chosen person off the
street in Gotham City. Also in the same city, the probability that the person
actually has COVID if tested positive is 0.90 and the probability that the person
actually does not have COVID if tested negative is 0.99. Assume that everyone
has the same probabilities of having COVID and testing positive for COVID.
(a) If I randomly pick a person in Gotham City, what is the probability that the
person actually does not have COVID? [6 marks]
(b) What is the probability that the test result is positive given that the person
actually has COVID? [3 marks]
(c) What is the probability that the test result is negative given that the person
actually does not have COVID? [3 marks]
[12 marks]
3
4. A fair coin is tossed 3 times. Let X = (X1, X2) where X1 counts the number of
heads in the 3 tosses, and X2 counts the number of tails in the 3 tosses.
(a) Write down the sample space S. [1 mark]
(b) Show the mapping of each sample point in S to the plane R2 by the random
variable X. [2 marks]
(c) Tabulate the joint probability mass function of X1 and X2. [2 marks]
(d) Calculate the marginal probability mass function of X1 and X2. What can
you deduce about them? Will each co-ordinate have the same values fo
their mean and variance? [2 marks]
(e) Calculate ρX1,X2 , the co
elation coefficient. Are they independent?
- [5 marks]
[12 marks]
5. Suppose X1 ∼ Bin(m, p) and X2 ∼ Bin(n, p) and are independent. E.g., X1
counts the number of heads for the 1st m independent coin tosses and X2 counts
the number of heads for the next n independent coin tosses. The probability of
getting heads for any toss is p where 0 ≤ p ≤ 1. Let S = X1 +X2.
(a) Using
pS(s) = P [S = s] =
s∑
x1=0
P [X1 = x1]P [X2 = s− x1],
obtain the probability mass function of S. Here Cba ≡ 0 if a > b.
Note: You will need the identity
∑k
y=0C
m
y C
n
k−y = C
m+n
k for y a non-negative
integer. [3 marks]
(b) What do you observe? [1 mark]
(c) Obtain an expression fo
P [X1 = x1|S = s],
where x1 ≤ s and x1 is a non-negative integer. [4 marks]
(d) Obtain an expression fo
P [S = s|X1 = x1],
where x1 ≤ s and x1 is a non-negative integer. [4 marks]
[12 marks]
4
6. Suppose X has the probability density function
fX(x) =

k(x+ 1)2, for − 1 ≤ x ≤ 0,
k(1− x)2, for 0 ≤ x ≤ 1,
0 otherwise.
(a) Find k so that fX(x) is a probability density function. [3 marks]
(b) Derive the cumulative distribution function FX(x). [3 marks]
(c) Calculate P [−0.5 ≤ X ≤ 0.5] by integration. [3 marks]
(d) Calculate P [−0.5 ≤ X ≤ 0.5] by using the cumulative distribution function.
- [1 mark]
(e) Calculate E[X] and Var[X]. [6 marks]
[16 marks]
7. The number of visitors clicking an online shopping website is distributed as Pois-
son with a mean visit rate of 7 per hour. Any individual visitor who goes onto the
website has a 65% chance of making a purchase. Assume each individual visito
acts independently of each other, and assume that the visitor only makes 1 click
to the website within that hour.
(a) What is the probability that within a particular hour, there are 5 visits to
the site? [2 marks]
(b) What is the probability that given n visits to the site, there are x purchases
where x = 0, 1, 2, · · · , n? [2 marks]
(c) What is the probability that there are 5 visits to the site with only 3
purchases? [2 marks]
(d) Simulate using R the number of actual purchases over 10 hour period.
- [5 marks]
[11 marks]
8. Refer to the previous question. The number of visitors clicking an online shopping
website is distributed as Poisson with a mean visit rate of 7 per hour. Any indi-
vidual visitor who goes onto the website has a 65% chance of making a purchase.
Assume each individual visitor acts independently of each other, and assume that
the visitor only makes 1 click to the website within that hour. If the ith visito
makes a purchase, the log of the amount spent ln(Yi) is normally distributed as
N(3, 2). Simulate the number of visitors in a one hour period, the number from
these visitors who make a purchase, and if they make a purchase, their purchase
amounts. [5 marks]
5
9. Look at the treering data in R . This give the standardised values of annual
treering outward growth of a particular tree from 6000 BC to 1979 AD. Calculate
a 99% confidence interval of the mean standardised value of the annual treering
growth. [6 marks]
10. Look at the sleep data in R . The extra column gives the hours of extra sleep of
that particular participant in this study. Test the hypothesis at the 1% significance
level that the extra hours of sleep is 0 hours. [6 marks]
6
Answered 10 days After Feb 28, 2023

Solution

Vedant answered on Mar 01 2023
34 Votes
Answer1
To solve this system of differential equations, we can use the method of matrix exponentials. First, we can write the system in matrix form:
y' = Ay
where
A =
The matrix exponential of A is given by:
where n! denotes the factorial of n, and An denotes the matrix product of A with itself n times.
To compute eAt, we need to first find An. Since A is a 2x2 matrix, we can use the formula:
where P is the matrix of eigenvectors of A, and D is the diagonal matrix of eigenvalues of A. The eigenvalues of A are λ1 = 2+i and λ2 = 2-i, and the co
esponding eigenvectors are v1 = [1+i, -1] and v2 = [1-i, -1], respectively. Therefore, we have:
P = [1+i, 1-i; -1, -1]
D = [2+i, 0; 0, 2-i]
Using these matrices, we can compute A^n for any positive integer n.
Now, we can compute e^At using the matrix exponential formula:
We can write out the matrix exponential explicitly as:
= (1/2) [(1+i)e^(2+i)t + (1-i)e^(2-i)t, (1+i)e^(2+i)t - (1-i)e^(2-i)t; -e^(2+i)t, e^(2-i)t]
To find the solution of the system, we can use the initial conditions y(0) = [y1(0), y2(0)] = [1, 1]. Therefore, we have:
y(t) = e^At y(0) = (1/2) [(1+i)e^(2+i)t + (1-i)e^(2-i)t, (1+i)e^(2+i)t - (1-i)e^(2-i)t; -e^(2+i)t, e^(2-i)t] [1; 1]
Simplifying this expression using the properties of complex exponentials, we get:
y(t) = [e^(2t) cos(t) + (1/2) e^(2t) sin(t), (1/2) e^(2t) sin(t) - e^(2t) cos(t)]
This is the solution of the system of differential equations.
To comment on its stability, we can examine the behavior of the eigenvalues of the matrix A. Since both eigenvalues have positive real parts, the system is unstable and the solutions will grow without bound as t increases. This can also be seen from the explicit expression of the solution, which involves exponential functions that grow without bound as t increases.
Answer2
A)
p(land somewhere on line) = p = 0.8 {given}
1-p = 0.2
n=3
Required
P(2) = 3C2 * 0.8^2 * 0.2^1 = 0.384
B)
p(land somewhere on AB) = p = 0.5*0.8 = 0.4
1-p = 0.6
n=3
p(2) = 3C2 * 0.4^2 * 0.6^1 = 0.288
C)
P (jenny is one of the two | 2 land on line) = no. of ways to select jenny and on other jumper / no. of ways to select 2 jumpers
=
= = 0.667
P(jenny is one of the two | 2 land on line)  = 0.667
D)
P(jenny land on line) = P(land on line) = 0.8
P(2 land on line | jenny one of them) = P(jenny is one of the two | 2 land on line)*P(2 land on line) / P(jenny land on line) = = 0.32016
Answer 3
Let us denote COVID diseased 'D', not diseased 'ND', positive test result '+' and negative test result '-'.
Then,
P(+) = 0.2 , P(-) = 1 - P(+) = 1 - 0.2 = 0.8
P(D|+) = 0.90 , P(ND|+) = 1 - P(D|+) = 1 - 0.90 = 0.10
P(ND|-) = 0.99 , P(D|-) = 1 - P(ND|-) = 1 - 0.99 = 0.01
(A)
P(person does not have COVID) =  
(B)
P(D) = 1 - P(ND) = 1 - 0.812 = 0.188
P(positive test result given that person has COVID) = P(+| D)
Using Bayes' theorem,
 
(C)
P(negative test result given that person does not have COVID) = P(-| ND)
Using Bayes' theorem,
 
Answer4
Answer 5
A) We are given that S = X1 + X2 ~ Bin(m + n, p), where X1 ~ Bin(m, p) and X2 ~ Bin(n, p) are independent. We need to find the probability mass function of S.
We can use the formula for the probability mass function of the binomial distribution, which is given by:
Where is the binomial coefficient.
Using this formula for X1 and X2, we can write:
where s = X1 + X2.
Then, we can use the identity:
to express the probability mass function of S as:
where the last step follows from combining the terms with p and (1-p).
Using the identity for the binomial coefficient, we can simplify this expression to:
for s = 0, 1, 2, ..., m+n.
Therefore, the probability mass function of S is given by:
for s = 0, 1, 2, ..., m+n.
B) The probability mass function of S that we derived is consistent with the fact that S is a binomial random variable with parameters m+n and p, as we showed earlier. Specifically, the formula we obtained matches the standard formula for the probability mass function of the binomial distribution, with the parameters m+n and p substituted in.
We also observe that the distribution is symmetric around the mean, which...
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