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1

a) 60% chance she knows the question 40 % chance she doesn't know but with a 20 % chance of getting it right giving 0.6+0.4*0.2 b)P(a given b) = p(a and b)/ p(b) =0.6/(0.6+0.4*0.2)


0

The formula for the number of ways to pick $m$ items out of $n$ possible is $$\binom{n}{m} = \frac{n!}{m!(n-m)!}$$ So you have to calculate $$\frac{4!}{3!1!}*\frac{5!}{2!3!}*\frac{6!}{4!2!} = 600$$


1

Standard deviation is very useful in metrology and when testing analog to digital converters (ADCs) input noise. Given that a ADC input has noise, to measure its noise one can plot an histogram of values x ocurrences and then calculate the standard deviation of this data. The standard deviation value representes de rms noise of the ADC input. Using only ...


2

The simplest counter example for the case of positive integers is probably $$ A=\{1,3\} \qquad B=\{1,2,3\}, $$ both with arithmetic mean $2$. The geometric mean of $A$ is $\sqrt{3}\approx 1.73$, while that of $B$ is $6^{1/3}\approx 1.81$. Hence the geometric mean of the larger set ($B$) is larger than that of the smaller set ($A$).


0

Let $U_{i}:=\sigma^{-1}\left(X_{i}-\mu\right)$ so that: $$\mathbb{E}\overline{X}^{2}=\mathbb{E}\left(\sigma\overline{U}+\mu\right)^{2}=\sigma^{2}\mathbb{E}\overline{U}^{2}+2\sigma\mu\mathbb{E}\overline{U}+\mu^{2}=\sigma^{2}\mathbb{E}\overline{U}^{2}+\mu^{2}$$ (It is easy to verify that $\mathbb{E}\overline{U}=0$) Next to that: ...


0

(independence assumed) \begin{align} E\bar{X}^2&=\frac1{n^2}E\Big(\sum_{i=1}^n X_i\Big)^2\\ &=\frac1{n^2}E\Big(\sum_{i=1}^n X_i^2+{\sum \sum}_{i\neq k} X_iX_k\Big)\\ &=\frac1{n^2}\Big(\sum_{i=1}^n EX_i^2+{\sum \sum}_{i\neq k}E X_iX_k\Big)\\ &=\frac1{n^2}\Big(\sum_{i=1}^n (\sigma^2+\mu^2)-{\sum \sum}_{i\neq k}(E X_i)(EX_k)\Big)\\ ...


1

Hint: $P(|Z|>2.4)= 1-P(|Z|\leq2.4)=1-P(-2.4\leq Z \leq 2.4)=1-2\times P(0\leq Z \leq2.4)=1-2 \times \big(\frac{1}{2}-P(Z>2.4)\big)=2\times P(Z>2.4)$ $$P(Z>x)= \int_{x}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} dz$$


0

Hint. Events $C$ and $D$ are independent if $$P(C\cap D)=P(C)P(D)\ .$$ You should be able to find $P(C)$ and $P(D)$ from the given information and then $P(C\cap D)$ from the above equation. Then, knowing the probability that $C\cap D$ occurs will give you the number of apartments in which $C\cap D$ occurs.


0

As $X_i$ follows $N(0,\sigma^2)$ $\Sigma_{i=0}^nX_i$ follows $N(0,n\sigma^2)$. So $\bar{X}=\frac{\Sigma_{i=0}^nX_i}{n}$ follows $N(0,\frac{\sigma^2}{\sqrt{n}}).$ Thus $\frac{\sqrt{n}\bar{X}}{\sigma^2}$ follows $N(0,1)$. and $\frac{(n-1)s}{\sigma^2}$ follows $\chi_{n-1}^2$ and is independent of $\bar{X}$. so ...


0

I've done the following with $n$ observations but if you set $n=1$ you get exactly what you have. Test of hypothesis are about parameters not data. \begin{equation} \begin{cases} H_0 : \mathbf{\mu} = \mathbf{\mu}_0 \\ H_a : \mathbf{\mu} \neq \mathbf{\mu}_0 \end{cases} \end{equation} Above we are testing the hypothesis that the data arises from a ...


1

Let $(Y,X_1,X_2,\ldots,X_p)^T\stackrel{d}{=}\mathcal{N}_{p+1}(0,\mathbf{C})$. Prove that $\mathbb{E}(Y|X_1,X_2,\ldots,X_p)$ maximizes Corr($Y,f(X_1,X_2,\ldots,X_p)$) in the space of $f$ linear functions. Notation: \ The matrix $\mathbf{C}$ be decomposed as \begin{equation*} \mathbf{C}=\begin{pmatrix} c_{yy}&\mathbf{c}^T_{yx}\\ ...


0

Note that $\bar{X}$ may not be a standard normal. Your $X$ r.v. representing $t$-distribution $$X = \frac{Z}{\sqrt{U/n}}$$ where U = variance. IF you have a copy of Mood Graybill and Boes (or download from Colorado University), refer to page 249 and 250. Let Z be a standard normal and U be a chi-square distribution with n degrees of freedom, then we can ...


1

Lets start with the univariate case: $$X_1 \sim \mathcal{N}(\mu,\sigma)$$ with pdf $\phi(x)$ standing in for the currently specified gaussian (I'm using it as shorthand for "the density for the variable" its not necessarily the standard density). Now, lets find a linear predictor that minimizes the mean squared error: $$f(x):=ax+b \implies MSE_f(a,b) := ...


0

Well, assuming the two distributions are for continuous random variables, and that they have well behaved probability density functions $f_1, f_2$, then: $$\Pr(X_{(1)}>Y_{(1)}) = \displaystyle\int_\Bbb R \binom{k}{k}\, (1-F_1(y))^k\, \binom{k}{k-1}\,(1-F_2(y))^{k-1}\,f_2(y)\operatorname d y$$ By reason that we want the probability that all of the $k$ ...


1

Let $X$ and $Y$ be independent random variables each with normal distribution (the means and variances need not be the same). Let $W=XY$. Then $$\text{Var}(W)=E((XY)^2)-(E(XY))^2.$$ We need to compute the two expectations on the right. By independence, $E(XY)=E(X)E(Y)$. Also, $E(X^2Y^2)=E(X^2)E(Y^2)$. Since $E(X^2)=\text{Var}(X) +(E(X))^2$, with a similar ...


0

There are at least two versions of the Wilcoxon signed-rank test, used to test whether the population median of paired differences is $0.$ Suppose you have $N$ differences $d_i = x_i - y_i.$ Delete any pairs with $d_i = 0$ because they contain no useful information for our purposes. Call the reduced sample size after any deletions $N_r.$ Then rank the ...


-2

First question I can't help with graphing the CDF, but I can help with proving that this is a valid distribution. To prove that a distribution is not valid you need to prove that: 1) some range of values has P>1 2) some range of values has P<0 (remember that in a continuous distribution, probabilities can be nonzero only if they refer to intervals. ...


0

Historically, across time and geographical regions, votes by machine or paper ballot (on election day) and votes by absentee ballot have shown significant differences in percentages for candidates, propositions, parties, etc. I think there is less data about the voting patterns of affidavit voters. However different demographic groups vote by different ...


0

Is this what you are looking for? $$\displaystyle \frac{\bar{X}\,-\,\mu}{S} \sqrt{n}= \displaystyle \frac{\frac{\bar{X}\,-\,\mu}{\sigma}}{\frac{S}{\sigma}} \sqrt{n}= \displaystyle \frac{\frac{\bar{X}\,-\,\mu}{\sigma}\sqrt{n}}{\sqrt{\frac{S^2}{\sigma^2}}} = \displaystyle ...


3

Take a corn of rice and try to measure its weight with your standard scale in your bathroom. Your result will not be very accurate as its weight is out of the bounds of your scale. What can you do? Take 50.000 corns of rice (given all corn of rice have exactly the same weight), measure the result and divide it by 50.000. Now you have a value you can work ...


1

Errors tend to be random, so they will tend to be of opposite signs and cancel when $n$ measurements are added, and then you divide by $n$.


0

Sum up the number 14+15+14+17+16+18+14+13+13 = 134 So median is at 134/2 = 67, this corresponds to interval 25-29.


0

(C) P(X < 65) = P(Z < (65 - 70)/5) = 0.1586552539 Given sample of 250 adults, the no of adults expected to have resting heart rate below 65 bpm is 250 * 0.1586552539 = 40 (round up from 39.66381348)


0

(B) P(75 < X < 80) = P(X < 80) - P(X < 75) = P(Z < (80 - 70)/5) - P(Z < (75 -70)/5) = 0.9772498681 - 0.8413447461 = 0.135905122


0

(A) Let Z be standard normal and X be the resting heart rate, then P(Z > (x-70)/5) = 1 - (68%/2) = 0.16 Note that standard normal is symmetric. So if you have a z table that z value at upper tail (or use Excel or calculator), you will find that this z value is 0.9944578832 Thus, (x-70)/5 = 0.9944578832 => x = 74.97228942 is the higher value By symmetry ...


3

The first thing you must do is find $a,b$ from the given mean ($7$) and variance ($4$).   Only then can you try to find $\mathsf P(a\leq X\leq 6\mid 4\leq X\leq b)$ The mean and variance of a uniform discrete distribution, $X\sim\mathcal U\{a..b\}$ are: $$\mathsf E(X) = \frac{a+b}{2} = 7\\\mathsf{Var}(X)= \frac{(a-b+1)^2-1}{12} = 4$$


1

Yes, your approach is correct. This manipulation is justified by Slutsky's theorem. See the article If $a_n \rightarrow 1$, does $X_n \overset{p}{\rightarrow} X$ implies $a_n X_n \overset{p}{\rightarrow} X$? for a reference to, and discussion about, Slutsky's theorem (and also a proof).


0

To answer the first question, it seems this procedure is called the Serial Approximate Entropy Test, according to section 2.12 of the NIST Statistical Test Suite for the Validation of Random Number Generators and Pseudo Random Number Generators.


1

You have been given that $$f_{X,Y}(x,y) = \begin{cases}4xy & : 0 \leq x \leq 1,\; 0 \leq y \leq 1\\ 0 & : \text{ elsewhere}\end{cases}$$ You wish to know where $f_{X,Y}(x, z-x)$ is supported (ie: not zero) with respect to $x$, for values of $z$ where $0\leq z\leq 2$.   (Since $z=x+y$, then $0+0\leq z\leq 1+1$.) $$f_{X,Y}(x,z-x) = ...


0

If the lower 3-sigma limits are really computed as $\bar X \pm 3s$, where s is the sample SD, or as $\bar X \pm 3\sigma$, where $\sigma$ is some 'known' population SD, then you know $\bar X$ is the midpoint of the 3-sigma interval. If you know the sample size $n$ and it is large, then a 95% CI for the population mean is $\bar X \pm 1.96s/\sqrt{n}$. (For $s$ ...


0

$$f_{X_1,\ldots,X_n}(x_1,\ldots,x_n \mid \theta) = (2\pi)^{-n/2} \prod_{i=1}^n \exp\left( -\frac{1}{2} (x_i - i\theta)^2 \right) $$ $$= (2\pi)^{-n/2} \exp\left( -\frac{1}{2} \sum_{i=1}^n(x_i - i\theta)^2 \right) $$ $$= (2\pi)^{-n/2} \exp\left( -\frac{1}{2} \left(\sum_{i=1}^n x_i^2 -2\theta\sum_{i=1}^n ix_i + \theta^2\sum_{i=1}^n i^2 \right)\right) $$ ...


1

Whenever you see $$\bar{X}\mid \mathbf{x}$$ or some other form of it, it means to fix the value of $\mathbf{x}$. So, in your case, we seek $$\text{Var}\left[\bar{X}\mid X_1, \cdots, X_n\right]\text{.}$$ But here's the thing. By definition, $$\bar{X} = \dfrac{1}{n}\sum\limits_{i=1}^{n}X_i$$ so thus $\bar{X}$ is constant because $X_1, \cdots, X_n$ are fixed ...


1

When you know all your $X_i$ then you know your $\bar{X}$, hence $E(\bar{X}|X_1,X_2,...,X_n)=\bar{X}$ as your sample mean is no longer a random variable but a constant now. Thus, $Var(\bar{X}|X_1,X_2,...,X_n)=E(\bar{X}^2|X_1,...,X_n)-(E(\bar{X}|X_1,...,X_n))^2=\bar{X}^2-\bar{X}^2=0$


1

It's essentially due to Runge's Phenomenon for interpolating polynomials. Even without uncertainty, the behavior of higher order polynomials at their endpoints is very sensitive to the parameter values. Here is a link to a technical explanation that I will regurgitate here. It relates Runge's Phenomenon to regression. Below is a intuitive, but admittedly ...


0

As has already been pointed out, the precise and careful choice of notation is paramount. The event $C$ of choosing the same color is the disjoint union of two separate events, so it is better to choose the following notation: Let $(b_1, b_2)$ be the random outcome of the two ball draws in order, where $b_i \in \{W, B\}$ for $i = 1, 2$. Thus, we want: ...


1

Split it into disjoint events, and add up their probabilities: The probability of choosing white from the first box and then from the second box is: $$\frac{4}{9}\cdot\frac{6}{10}=\frac{24}{90}$$ The probability of choosing black from the first box and then from the second box is: $$\frac{5}{9}\cdot\frac{5}{10}=\frac{25}{90}$$ So the probability ...


4

The calculation on the right-hand side is correct; just the notation is bad, because as you say the right-hand factors in both terms are conditional probabilities. A better way to write this would be $$ P(\text{C})=P(\text{W})+P(\text{Bl})= ...


3

Sub $u=x+y$, $v=x-y$ with Jacobian $J=1/2$: $$\int_0^{\infty} dx \, \int_0^{\infty} dy \frac{f(x+y)}{x+y} = \frac12 \int_0^{\infty} du \int_{-u}^u dv \frac{f(u)}{u} = \int_0^{\infty} du \, f(u) = 1$$


0

The outcomes, in tree form, are as follows, when $\rm Xy$ represents "$X$ wins against $Y$", a stop ($\cdot$) reprensent a termination (someone won) and an ellipsis ($\cdots$) represents a repeating pattern. Thus the probability that $C$ wins the game is obtained by solving the recursion, and we obtain the probability that $A$ wins by noting symmetry ...


0

Here's an example that may convey some intuition. Suppose we consider how much it rains in a given year in Los Angeles. This random variable, which we shall call $R$, has a distribution, which is a PDF that we may denote $f_R(r)$. Suppose, for the sake of argument, that $f_R(15) = 0.1$. Now what does that mean? Obviously, it doesn't mean that there's a ...


2

The velocity of a moving particle is the rate of change of the displacement at that time. The probability density of a continuous real valued random variable is the rate of change of the cumulative probability at that point.


4

Strictly speaking, at any given point the pdf doesn't mean anything. You can change the value of the pdf at that particular point to be whatever you want, and the distribution of the random variable is the same. Still, "generically", $f(x)$ is essentially the probability to find the random variable in the interval $(x-\epsilon,x+\epsilon)$, divided by $2 ...


0

To begin with, $Y = -\ln(X)$ only works when $[a,b] \subset \mathbb{R}_{>0}$. So, let $$ X \sim \mathcal{U}[a,b] \quad\text{where}\quad [a,b] \subset \mathbb{R}_{>0}. $$ then $Y$ is log-uniformly distributed, in symbols $$ Y \sim \mathcal{LU}[\alpha,\beta] \quad\text{with}\quad \alpha = -\exp(a) \ \text{and} \ \beta = -\exp(b). $$ This can ...


0

Think in simple terms as you don't really need to apply probability rules to intuitively understand how $(a)$ works. The first part $(i)$ $P(B \mid D) + P(B^c \mid D)$ can be stated in words as: the probability that you pick a component from factory B, given that it's defective, plus the probability that you pick a component that's not from factory B, given ...


1

For part a), $$p(B|D)+p(B^c|D)=\frac{p(B\cap D)}{p(D)}+\frac{p(B^c\cap D)}{p(D)}=\frac{p(B\cap D)+p(B^c\cap D)}{p(D)}$$ $$=\frac{p((B\cup B^c)\cap D)}{p(D)}=\frac{p(D)}{p(D)}=1$$


1

Let the result of a game be denoted by $XdY$, meaning that Player $X$ defeats Player $Y$. Then the space of possible results is $$ \{(AdB, AdC), (BdA, BdC), (AdB, CdA, CdB), (BdA, CdB, CdA), (AdB, CdA, BdC, BdA), \ldots\} $$ Partial solution follows. To determine the probability of winning, observe that $C$ wins if and only if the entire series ends ...


0

An example of a random process yielding multiple iid random variables could be one of the dimensions of some widget coming off an automated assembly line. We want this dimension of the widget to be within a certain tolerance of a certain value, but the machines that make it produce widgets with a variety of sizes of this dimension. Let $X_k$ be the size of ...


2

$${ n \choose r+1} = \frac{n-r}{n + 1} {n \choose r}$$


2

$\sum_iX_i=n\bar X$, so $$\begin{align*} \sum_i(X_i-\bar X)^2&=\sum_iX_i^2-2\bar X\sum_iX_i+\sum_i\bar X^2\\ &=\sum_iX_i^2-2\bar X(n\bar X)+n\bar X^2\\ &=\sum_iX_i^2-n\bar X^2\;. \end{align*}$$ Moreover, $$\begin{align*} \sum_i(X_i-\bar X)(Y_i-\bar Y)&=\sum_iX_iY_i-\bar X\sum_iY_i-\bar Y\sum_iX_i+\sum_i\bar X\bar Y\\ ...


1

Imagine the selected elements as black beads and the others as white. Add one more black bead (so there are $b+1$ black beads and $a+1$ beads in all) and close the row of beads into a circle. The circle is divided into $b+1$ segments, each consisting of some (possibly zero) white beads followed by one black bead. By symmetry, each of these segments has ...



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