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29

$P(\text{hit by bus in 1000 crossings}) = 1-P(\text{not hit by bus in 1000 crossing}) = 1-(999/1000)^{1000} \approx 0.63$


13

You can approximate this very well using a Poisson distribution: If over a large number $N$ of trials an event occurs an average of $\lambda \ll N$ times, the probability that it occurs $k$ times in a set of $N$ trials is $$P(k) \approx \frac{\lambda^k e^{-\lambda}}{k!}.$$ In our case, over $1000$ trials we expect our event with per-trial probability of ...


3

It is not totally obvious to me what you are asking for, but I would have thought the argument is something like $$\Pr(R \le \mu + \sigma \Phi^{-1}(u)) = \Pr(\epsilon \le \Phi^{-1}(u)) = \Phi(\Phi^{-1}(u)) =u$$


3

The natural approach uses transition matrices. For ease of typesetting we write up the solution another way. Let $e$ be the expected number. Let $a$ be the expected number of additional letters, given that the last letter was an A. Let $b$ be the expected number of additional letters, given the last two letters were AB. And let $c$ be the analogous thing, ...


2

Standardize so that you can use the standard normal distribution to find the numerical value for the answer. So, we have $$\begin{align} \mathbb{P}(X \geq 203) &= \mathbb{P}\left(\frac{X - 200}{10} \geq \frac{203 -200}{10}\right) \\ &= \mathbb{P}\left(Z \geq \frac{3}{10}\right) \\ &= 1 - \mathbb{P}\left(Z < \frac{3}{10}\right) \\ &= ...


2

We have $$\mathbb{E}(\|Y_n\|^4) = \frac{1}{n^4} \sum_{j,k,l,m=1}^n \mathbb{E}(\langle X_j, X_k \rangle \langle X_l, X_m \rangle).$$ It follows from the independence of the random variables and the fact that $X_n$ has mean zero (i.e. $\mathbb{E}(\langle X_n,x \rangle)=0$ for all $x \in H$) that $$\mathbb{E}(\langle X_j, X_k \rangle \langle X_l, X_m ...


2

If your confidence level is $1-\alpha$, then the absolute earliest you could stop would be if you got a string of heads. The minimum length ($N_0$) of such a run is calculated as: $N_0 = \lceil -\log_{20} \alpha \rceil \approx 1$ (if $\alpha = 0.05$) so if you get a heads on your first toss, you're done - which makes sense because there is only a 5% chance ...


2

This sounds like a Beta-Binoimial purchase rate model would be useful. The link has a lot of details on this, but I'll give the salient highlights: We are assuming that when a product $i$ is introduced, we are uncertain as to its actual bought rate $BR_i$. We model this uncertainty with a beta distribution on [0,1]. For example, the uniform distribution on ...


2

Economics is probably a good choice if you want those (diminishing returns of capital and labour in production functions)


2

A "favourable" hand can be selected as follows: choose two different values from $13$, order not important. . . . . $C(13,2)$ ways; choose three of the four cards from each of these values. . . . . $C(4,3)^2=4^2$ ways. Since the total number of hands is $C(52,6)$, the required probability is $$\frac{C(13,2)4^2}{C(52,6)}\ .$$ The problem with your ...


2

When $n$ is large $$ W=\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\sim N(0,1) $$ approximately, or equivalently, $$ \bar{X}= \mu+\frac{\sigma}{\sqrt{n}}W\sim N(\mu,\sigma^2/n) $$ approximately. That is, the sample mean $\bar{X}$ is approximately normal distributed with mean $\mu$ and variance $\sigma^2/n$ when $n$ is large.


2

I suppose the sample is iid. You can find the distribution of $F_{\hat\theta}(t)$ by noticing $$F_{\hat\theta}(t)=P(\hat\theta\leq t)=P(Y\leq t)^n$$ Thus you can find $f_{\hat\theta}$, and then its expected value $$E(\hat\theta)=\int_{[0,\Theta]} t f_{\hat\theta}(t)\;\mathrm{d}t$$ Here you have $$P(Y\leq t)=\int_0^t f_Y(u)\;\mathrm{d}u=\int_0^t ...


1

I would say $\Phi(\frac{c-\mu}\sigma)=0.08\Rightarrow \frac{c-\mu}\sigma =\Phi^{-1}(0.08)=-1.406$ $\Phi(u)=\,\,$ distribution function of the standard normal distribution $N(0,1)$ You'll find in the tables N(0,1), in Ecxel (function "=NORMSINV(0,08)) etc.


1

Here notice that $X\cap Y=X$ because $X\subset Y$ since $2$ is even. So generally when events $A$ and $B$ and $A\subseteq B$ then we have $P(A\cap B)=P(A)$. Also since $P(A\cap B)=P(A|B)P(B)$ so if $P(A|B)=1$ you have $P(A\cap B)=P(B)$ but also $P(A|B)=1$ will generally be true when $A\subseteq B$


1

For $H(X)$, it seems there's a known closed form: $$ \frac{1}{2}\log(2\pi e np(1-p)) + O(1/n) $$ where $p=0.5$ in your case and $n$ is the number of trials By the Weak Asymptotic Equipartition Property (see, e.g. Yeung - Information Theory and Network Coding, Theorem 5.1, or Wikipedia), the "empirical" entropy converges in probability to the true entropy. ...


1

a) For $x$ in the interval from $0$ to $1$, we want $\int_0^x 3(1-t)^2\,dt$. One way to evaluate is to make the substitution $1-t=u$. Then $dt=-du$ and after not much manipulation we end up at $$\int_{u=1-x}^{1} 3u^2\,du.$$ The rest of the calculation is straightforward. Alternately, we can expand $3(1-t)^2$, getting $3-6t+3t^2$ and integrate from $0$ to ...


1

Calling $p$ the probability of being hit on one crossing and $q=1-p$ the probability to be safe in one crossing you can compute $p_{1000}$ (the probability of being hit at least once in $1000$ crossings) like this: $$p_{1000} = p + qp + qqp +...+q^{999}p$$ This mean: you get hit at the first attempt, or you escape the first and get hit at the second, or ...


1

It seems that I need to give a more detailed answer in the post but not only refer to the references for the main technical ideas. Let me start with a general picture. Estimating the entropy, from a statistical perspective, is by no means a unique problem among the problems of estimating functionals of parameters. The reason why it has attracted so much ...


1

The two kinds that we can have $3$ of can be chosen in $\binom{13}{2}$ ways. Given the kinds, the actual $3$ cards of each kind can be chosen in $\binom{4}{3}\binom{4}{3}$ ways, for a total of $\binom{13}{2}\binom{4}{3}^2$. For the probability, divide by $\binom{52}{6}$. Remark: In the attempted solution, there is multiple counting going on. You are ...


1

Preliminary comment: The result follows immediately from the meaning of binomial distribution. But the problem is asking for a computation. We deal with the induction step. Assume that the sum $Y_n=X_1+\cdots+X_n$ has binomial distribution. This means that for all suitable $i$, we have $$\Pr(Y_n=i)=\binom{n}{i}q^i(1-q)^{n-i}.$$ We want to prove that ...


1

Here How exactly are the beta and gamma distributions related? you can find in one of the solutions posted. X1/X1+X2 has betta distribution


1

We have that $X_1$ and $X_2$ are independent $\Gamma(\alpha_i,1)$ random variables. Note that $X_1$ and $X_2$ are non-negative with values in $[0,\infty).$ Given the transformation, the joint density of $Y_1$ and $Y_2$ is: $$ f_{Y_1Y_2}(y_1,y_2)=f_{X_1X_2}[X_1(y_1,y_2),X_2(y_1,y_2)]|J|= f_{X_1}(y_1y_2)f_{X_2}(y_2-y_1y_2)y_2= ...


1

The probability that they have a fourth child is the probability of having three boys, namely $(\frac{1}{2})^3 = \frac{1}{8}$. The probability that they have a fifth is the probability of having $4$ boys, or $(\frac{1}{2})^4 = \frac{1}{16}$, and so on. So, the probability they stop after three is $\frac{7}{8}$, the probability that they stop after exactly ...


1

Hint: $$P\left(A\right)=P\left(A\cap E\right)+P\left(A\cap E^{c}\right)=P\left(A|E\right)P\left(E\right)+P\left(A|E^{c}\right)P\left(E^{c}\right)$$


1

I understand your confusion. What happens is that there is all this talk about the sample space in the first part of a stats class and then they introduce the idea that a random variable "maps" the sample space to a new sample space. However, either the people, as people, or their associated heights, can be validly called the sample space. However, it ...


1

Here's some steps and rationale: Make a box plot. You can do a rough check of normality by verifying that the upper and lower quartiles are approximately 0.7 standard deviations from the mean and that extreme outliers are less than 4 standard deviations from the mean. The box plot should look symmetric. If the data appear normal, then calculate the ...


1

For $i=0,1,2,3,\dots$, you have found that $$\Pr(Y=i+1)=e^{-i}(1-e^{-1}).$$ Let $k=i+1$. You have found that for $k=1,2,3,\dots$, we have $$\Pr(Y=k)=(1-e^{-1})(e^{-1})^{k-1}.\tag{1}$$ That's because $e^{-i}=(e^{-1})^i=(e^{-1})^{k-1}$. Formula (1) is exactly the formula for the probability that $Y=k$, where $Y$ has geometric distribution with parameter ...


1

Consider the winners of the triples. $S = \{\underbrace{HHH}_A, \underbrace{HHT}_A, \underbrace{HTH}_A, \underbrace{THH}_B, \underbrace{HTT}_A, \underbrace{THT}_B, \underbrace{TTH}_C, \underbrace{TTT}_{\text{repeat}}\}$ Now can you tell the probabilities of each winning?


1

$A$ wins if the sequence of tosses is any one of the following: $$H, TTTH, TTTTTTH, \cdots, T^{3i}H, \cdots $$ that is, Tails $3i$ times in succession, $(i = 0, 1, 2, \ldots)$ followed by a Head. and so $A$ wins with probability $$\left.\left.\frac 12 \right[1 + \left(\frac 12\right)^3 + \left(\frac 12\right)^6 + \cdots\right] = \left.\left.\frac 12 ...


1

The more the number of variables we include in the model, the seemingly better the fit and higher the $R^2$. Yes it is true. Every time you add a predictor to a model, the R-squared increases, even if due to chance alone. It never decreases. Consequently, a model with more terms may appear to have a better fit simply because it has more terms.But for this ...



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