Pre-requisite self assessment

Pre-requisite self assessment quiz

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1. let X be a normal random variable with mean equal to one and variance equal to four. Using statistical software if needed, answer the following two questions:

Q1

Enter the probability that X X is no more than one, P(X≤1) :

(pnorm(1, mean=1, sd=2))

[1] 0.5

Q2

Enter the probability that X X is no less than minus two, P(X\ge-2) :

(1-pnorm(-2, mean=1, sd=2))

[1] 0.9331928

N(x)=P(Z≤x) denotes the probability that a standard normal random variable Z is no more than x.

Consider the probability

P(a e^{bZ+c} \ge d). ## Q3

Here, a,b,c,d are positive constants and e^x is the exponential function.

What is the expression for that probability in terms of function N(x)?

P(a e^{bZ+c} \ge d) = N\left( \frac{c- \log(d) + \log(a) }{b}\right)

Q4

Consider the function

f(x) = e^{-\frac{1}{2}(x-m)^2}.

Knowing that the integral of cf(x) from minus infinity to plus infinity is equal to one, what is the value of c

since the formula for the normal distribution is

f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} we substitute \mu = m and \sigma = 1 to get

f(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}(x-m)^2}

since this is a distribution, the integral of f(x) from minus infinity to plus infinity is equal to one.

Therefore, the constant c is equal to \frac{1}{\sqrt{2\pi}}.

what is

\int_{-\infty}^{\infty} cxf(x) dx equal to?

\int_{-\infty}^{\infty} cxf(x) dx = \int_{-\infty}^{\infty} \frac{x}{\sqrt{2\pi}} e^{-\frac{1}{2}(x-m)^2} dx but this is the expected value of the normal distribution, which is equal to the mean of the distribution. Therefore, the integral is equal to m.

Q5

f(t) be the solution to the ordinary differential equation

f'(t) = 2 f(t)

with the boundary condition

f(0) = 3

the a solution to this differential equation is of the form

f(t) = e^{2t} + c

substitute the boundary condition f(0) = 3 to get the value of the constant c: as

f(0) = e^{2*0} + c = 1 + c = 3 \implies c = 2 Therefore, the exact solution to the differential equation is

f(t) = e^{2t} + 2

Enter the numerical value value of f(0):

f <- function(t) {
  return(exp(2*t) + 2)
}
f(1)

[1] 9.389056

f(0)

[1] 3

6

Consider the function

f(x,y) = x^2 y^3.

Enter the value of its derivative with respect to y at the point (x=2, y=3)

the derivative of f(x,y) with respect to y is

\frac{\partial f}{\partial y} = 3x^2y^2 substitute x=2 and y=3 to get

\frac{\partial f}{\partial y} = 3(2)^2(3)^2 = 3*4*9 = 108

Q7

the second partial derivative of f(x,y) with respect to y and x is

\frac{\partial^2 f}{\partial y \partial x} = 6xy^2

substitute x=2 and y=3 to get

\frac{\partial^2 f}{\partial y \partial x} = 6(2)(3)^2 = 6*2*9 = 108

Q8

X be a random variable that takes values 0, 1 and 2 with probabilities 1/4, 1/4 and 1/2, respectively.

Enter the numerical value for the expected value of

E[X] E[X] = 0*\frac{1}{4} + 1*\frac{1}{4} + 2*\frac{1}{2} = 0 + 0.25 + 1 = 1.25 ## Q9

Enter the numerical value for the variance of X Var[X]:

\begin{align*} Var[X] & = E[X^2] - (E[X])^2 \\ & = 0^2*\frac{1}{4} + 1^2*\frac{1}{4} + 2^2*\frac{1}{2} - (1.25)^2 \\ & = 0 + 0.25 + 2 - 1.5625 \\ & = 0.6875 \end{align*}

Q10

Let g(x) = f(x, y(x)) where y(x) is a function of x and f(x,y) is a function of two variables. Denote the derivative of y(x) with respect to x by y_x, the derivative of f with respect to the first variable, and evaluated at the point (x, y(x)), by f_x, and the derivative of f with respect to the second variable, and evaluated at the point (x, y(x)), by f_y.

Write, in terms of only y_x, f_x, f_y, the derivative of the function g(x) = f(x, y(x)) with respect to x. Explicitly indicate multiplication with a “*” symbol, and write the derivatives as y_x, f_x, f_y.

\frac{d}{dx} g(x) = f_x + f_y*y_x where I used the chain rule for differentiation.