Answer To: MATH 1081 UO Mathematical Methodsfor Data Analytics 2Continuous Assessment 1.2: Problem Solving...
Vedant answered on Mar 01 2023
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 corresponding 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...