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Let $p_r$ be the proportion of red balls out of $N.$ Filling in the details of the previous answer, the variance of a binomial is: $$Sp_r(1-p_r)$$ And for sampling with replacement, the variance of the hypergeometric is $$Sp_r(1-p_r)\frac{N-S}{N-1}$$ The last term (the finite population correction factor) will be close to $1$ if $S<<N.$

3

Let $X_1$ be the number of rolls until the first $5$, $X_2$ the number of rolls between the first $5$ and the second (but not including the first $5$), and so on. Then the number $Y$ of rolls until the fourth $5$ is given by $Y=X_1+X_2+X_3+X_4$. By the linearity of expectation $E(Y)=E(X_1)+\cdots+E(X_4)$. Each of the $X_i$ is a geometric random variable, ...

3

Stokes is off-topic here but elementary set theory and algebra are very much on-topic. To see why, consider, for every $i$ in $\{1,2,\ldots,n\}$, the events $A_i=[X_i\leqslant a_i]$ and $B_i=[X_i\leqslant b_i]$, then $$[X\in(a,b]]=C,\qquad C=\bigcap_{i=1}^n(B_i\setminus A_i).$$ In terms of indicator functions, this reads $$\mathbf 1_C=\prod_{i=1}^n\mathbf ... 2$$\begin{align*}f(\vec{x},θ)&=\prod_{i=1}^{n}(θ+1)x_i^{θ}=(θ+1)^n\prod_{i=1}^{n}x_i^{θ}\\\mathcal L(\vec{x},θ)&=\ln(f(\vec{x},θ))=n\ln(θ+1)+θ\sum_{i=1}^{n}\ln x_i\\\dfrac{d}{dθ}\mathcal L(\vec{x},θ)&=\dfrac{n}{θ+1}+\sum_{i=1}^{n}\ln x_i\overset{!}=0 \implies \hat{θ}=-\dfrac{n}{\sum_{i=1}^{n}\ln x_i}-1\end{align*}$$2 For 0\le x \le y \le 1 you have that$$f_{Y|X}(y|x)=\dfrac{f_{XY}(x,y)}{f_X(x)}=\frac{2}{\int_{x}^{1}f_{XY}(x,y)dy}=\dfrac{2}{\int_{x}^{1}2dy}=\dfrac{1}{1-x}$$for all x\le y \le 1. That is Y|X=x is uniformly distributed in [x,1]. Thus$$Var(Y|X=x)=\frac{(1-x)^2}{12}$$2 The distribution of the sample mean X_n is approximately normal with parameters μ=x_{mean} and σ/\sqrt{n}=x_{sd}/\sqrt{n}, in symbols$$X_n \sim N\left(μ, \dfrac{σ}{\sqrt{n}}\right)$$approximately (where the approximation is better as n grows to \infty). This follows from the Central Limit Theorem and you do not need any further particular ... 2 Ok, then I guess when you say E(\mathbf X_i \otimes \mathbf u_i)=0, you mean that 0 is a matrix ?... In this case, \mathrm{vec}(\mathbf X_i^\top \mathbf G \mathbf u_i) = (\mathbf u_i^\top \otimes \mathbf X_i^\top)\mathrm{vec}(\mathbf G). So if you have (\mathbf X_i \otimes \mathbf u_i)=0, then obviously \mathrm{vec}(\mathbf X_i^\top \mathbf G ... 2 I think it can be helpful to leave the matrix world temporarily to see what it means in terms of scalar variables. Let's see if you agree... Let \mathbf X_i consist of elements x_i^{(j, k)} and \mathbf u_i of elements u_i^{(l)}, where j=1, \dots G, k=1, \dots, K and l=1, \dots, G. The G^2\times K matrix E(\mathbf X_i \otimes \mathbf u_i) ... 2 We have$$\begin{align}\operatorname{Cov}(X+Y,X-Y)&=\operatorname{Cov}(X,X-Y)+\operatorname{Cov}(Y,X-Y)\\&=\operatorname{Cov}(X,X)-\operatorname{Cov}(X,Y)+\operatorname{Cov}(Y,X)-\operatorname{Cov}(Y,Y)\\&=\operatorname{Var}(X)-\operatorname{Var}(Y).\end{align}$$1 In conventional frequentist statistical inference, a quantity is considered "random", and therefore has a probability distribution assigned to it, only if it changes when a new sample is taken. "Sample" does not mean a single observation; it means a set of observations on which inferences are based. You randomly take 50 men out of a population. That's a ... 1 Not sure if it's really a math question, anyway here's what I think. It's a quite complex situation because you have a (nearly) 5-dimensional vector and you also want to represent for each vector the probability of the vector belonging to a certain class. I suppose that you have these probabilities and that you have graphing capabilities. One solution ... 1 I suppose that it should be X,Y are independent random variables. Then:$$Cov(X+Y,X-Y)=E((X+Y)(X-Y))-E(X+Y)E(X-Y)=\\=E(X^2-Y^2)-(E(X)+E(Y))(E(X)-E(Y))=\\=E(X^2)-E(Y^2)-E(X)^2+E(Y^2)=\\=E(X^2)-E(x)^2-(E(Y^2)-E(Y)^2)=Var(X)-Var(Y)=0$$1 Let's look at each of the possibilities: A: less than 84. Taking a look at our summary statistics, we see that 84 is actually the First Quartile or Q_{1}, which is by definition the 25th percentile. This means that 25% of our data points lie below 84. Since we have 100 data points, we expect 25 students to have IQ below 84. B: less than 110. Again, we ... 1 That is the right answer. The PDF = 1/4 on the range [4,8) and is zero elsewhere. 1 Substitute \frac{z^{2}}{2}=t, then zdz=dt and z^{3}=(2t)^{3/2}. \begin{eqnarray} \mathbb{E}[Z^{4}]&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^{4}e^{-z^{2}/2}\;dz\\ &=&\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}z^{4}e^{-z^{2}/2}\;dz\\ &=&\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}(2t)^{3/2}e^{-t}\;dt\\ ... 1 OK, so B^c is the complement event of B. In other words, it's the case where B hasn't happened. If B hasn't happened, then the probability that A has happened is 0.2 out of 0.4. Therefore, the answer to your question is \dfrac{1}{2}. Of course, you can also use conditional-probability, as explained in other answers given here... Please ... 1 For A) This is true (with a small questionmark). Since we restrict attention in the sample, I believe that the statement is true. Assume there was only the sample and no other unsampled units in the population. Then the statement would be true. But that is exactly what he means, since he restricts our attention only in the sample. For B) This is definitely ... 1 You want to show that$$P\left(X_1=x_1,X_2=x_2,...,X_n=x_n\Big|\frac{1}{n}\sum_{k=1}^{n}X_k=t\right)$$does not depend on p (in your expression this t is absent). Now the term in the condition is equivalent to \sum_{k=1}^nX_k=nt but \sum_{k=1}^{n} X_k is binomially distributed with parameters 2n and p, in symbols$$S_n:=\sum_{k=1}^nX_k \sim ...

1

It suffice to compute $$f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_{X}(x)}=\frac{2*\chi{0\le x\le y\le 1}}{\int^{1}_{x}2dy *0\le x\le y\le 1}=\frac{1}{1-x}*\chi{0\le x\le y\le 1}$$ Therefore as a function of $y$, $f_{Y|X}(y|x)$ is well defined on $0\le x\le y\le 1$ except at $y=1,x=1$. Excluding this case we may compute that $$... 1 After N trials, you can either have 9 failures in a row or not, and the sequence of trials can end in anywhere between 0 and 8 failures in a row if 9 failures have not occurred in a row so far. Write down recurrence relations that give you the probabilities for these cases for N+1 trials in terms of the probabilities for these cases for N trials. ... 1 Hint: For n=1:$$P_X(a,b] = P(\{X\leq b\}\setminus \{X\leq a\}) = P(\{X\leq b\}) - P(\{X\leq a\}) = F_X(b)-F_X(a).$$For n=2:$$P_X(a,b] = P(\{X\leq (b_1,b_2)\}\setminus \left( \left(\{X\leq (b_1,a_2)\} \setminus \{X\leq (a_1,a_2)\}\right) \cup\left(\{X\leq (a_1,b_2)\} \setminus \{X\leq (a_1,a_2)\}\right) \cup \{X\leq (a_1,a_2)\} ...

1

Tables of functions in books are usually printed in a format like this. To look up a function value, you find a number at the head of a row and a number at the head of a column that add up to your desired value of $x,$ then find the value of $f(x)$ at the intersection of that row and column. For example, to look up the cumulative distribution $\Phi(x)$ for ...

1

I have given the hand-written solution of the problem. Let me know if it is legible and clear

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$\mathbb{E}X^2=\mathbb{E}X(X-Y+Y)=\mu_X(\mu_X-\mu_Y)+\mu_X\mu_Y=\mu_X^2\Rightarrow P(X=c)=1$ which follows from CS inequality.

1

Here's a sketch of a proof that does not require existence of moments. The characteristic function of $(X,X-Y)$ is $$E[e^{iXt+i(X-Y)s}]=E[e^{iXt}]E[e^{i(X-Y)s}]=E[e^{iXt}]E[e^{iXs}]E[e^{-iYs}]$$ where the equalities follow by independence of $X$ and $X−Y$ and the independence of $X$ and $Y$. We also have by independence of $X$ and $Y$ that, $$... 1 You are correct. As a check, given X=1, you could say Y is a Bernoulli random variable with parameter p=\frac13. The mean of a Bernoulli random variable is p and its variance is p(1-p), as you have found in this particular case. 1 A joint density function will have an always positive measure over the joint support, and the (double) integral over the joint support will equal 1. The function (x^2+\frac{xy}{2}) is always positive over (x,y)\in[0,1]\times[0,2]. So perform the integration. If the result is equal to 1, you have a probability mass function over that support. ... 1 Hint: You need to perform a test for equality of sample proportions from two independent populations. The test statistic for testing the difference in two population proportions, that is, for testing the null hypothesis$$H_0:p_1−p_2=0$$is:$$Z=\dfrac{(\hat{p}_1−\hat{p}_2)−0}{\sqrt{\hat{p}(1−\hat{p})\left(\dfrac{1}{n_1}+\dfrac{1}{n_2}\right)}} \sim ...

1

Your question is a little confusing because at first you state that all $5$ girls are chosen then you ask what is the probability that $2$ girls are chosen so what was the point of mentioning $5$ girls chosen? If the $7$ student choices are truly random then we have the following: $5 \choose 2$ * $6 \choose 5$ / $11 \choose 7$ = $2/11$ = $.1818$... Your ...

1

You should use that if $X$ is binomially distributed with parameters $p=0.1$ and $n=100$ then $X$ can be approximated by the following distributions $X \sim$ Poisson($\lambda=10$), where $10=\lambda=np=0.1\cdot100$, $X \sim$ Normal($\mu=10, \sigma^2=9$), where $10=\mu=np=0.1\cdot100$ and $9=\sigma^2=np(1-p)$. Of course, since $0.1$ is not close to ...

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