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The $1-\alpha$ confidence interval for $\hat p$ is $$\large{\left[ \hat p-z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} , \hat p+z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} \right]}$$ $z_{(1-\frac{\alpha}{2})}$ is the z-value of the standard normal distribution. $1-\alpha$ is the confidence level. In your case ...

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Hint:$$\Phi^{-1}(U)\leq x\iff U\leq \Phi(x)$$ Here $\Phi$ denotes the CDF corresponding with standard normal distribution.

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Use the inverse transform method: With $X \sim N(0,1)$ and CDF $F$. $1.$ Generate a random number $u$ from $U$ in the interval $[0,1]$ $2.$ Compute the value $x$ such that $F(x) = u$ $3.$ Take $x$ to be the random number drawn from the distribution described by $F$

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If $$Y = X_{(n)} = \max_i X_i,$$ then $$\Pr[Y \le y] = \Pr\left[\max_i X_i \le y\right] = \Pr\left[\bigcap_{i=1}^n (X_i \le y) \right] \overset{\text{ind}}{=} \prod_{i=1}^n \Pr[X_i \le y].$$ This is because the largest observation is at most $y$ if and only if each observation is at most $y$. Since $$\Pr[X_i \le y] = F_X(y) = \frac{y}{\theta}$$ for $0 \le ... 0 The only (somehow) meaningful origin that I can think of, is an estimator of$EX = \frac{a+b}{2} = \frac{\theta}{2}$, that was modified appropriately. I.e., clearly $$\hat{\theta}_n = \frac{1}{2}(X_{(1)} + X_{(n)})$$ is consistent estimator of$EX$(which for some cases can be unbiased as well). I don't know if it has any special name, but I would call it ... 1 Let's start from the beginning. The joint probability of the sample$\boldsymbol x = (x_1, \ldots, x_N)$where each $$x_i \sim \operatorname{DiscreteUniform}(1,n)$$ is IID, is given by $$f(\boldsymbol x \mid n) = \prod_{i=1}^N f(x_i \mid n) = n^{-N} \mathbb 1(1 \le x_{(1)} \le x_{(N)} \le n).$$ Consequently, the joint likelihood is proportional to ... 1 Guessing that$G^*$denotes the set of non-zero elements of$G$, and hoping that the question has finally stabilized: If$X$and$Y$are independent then$X+Y$is uniform on$G$. Without assuming independence I don't see how you can expect to say anything about$X+Y$. 1 Let$U$be uniform continuous on$[0,1]$and$V$be uniform discrete on$\{0,1,\cdots,m\}$. Then$U+V$is uniform continuous on$[0,m+1]$. More generally, assume$U$to be uniform continuous on$[0,1]$and$V$to be another random variable such that$U+V$is uniform continuous on some other interval. By translation, we can assume that this interval is ... 0 Hint. The probability density function of$X$is $$f(x)=\begin{cases} \frac{1}{2\pi} & \mathrm{for}\ -\pi \le x \le \pi, \\[8pt] 0 & \mathrm{for}\ x<-\pi\ \mathrm{or}\ x>\pi \end{cases}$$ then you have $$P(X\leq0)=\int_{-\infty}^0f(x)\:dx$$ $$P\left(X\leq \frac{\pi}{2}\right)=\int_{-\infty}^{\pi/2}f(x)\:dx.$$ Can you take it from ... 2$W$, the conditional expectation of$Y$given$X$will be a piecewise function partitioned on$X's enumeration. \begin{align}W~=~& \mathsf E(Y\mid X) \\[2ex]~=~& \sum\limits_{\omega\in X^{-1}(X)} Y(\omega)\cdot\mathsf P^{Y\mid X}(\omega) & :~ \mathsf P^{Y\mid X}(\omega) \mathop{:=} \Pr(Y{=}Y(\omega)\mid X{=}X(\omega)) \\[2ex] =~& \tfrac ... 0 This can be determined analytically. P(Bomb hits the target)=P(correct X) P(correct Y) P(correct Y)=30/55 easily, since Y follows a uniform distribution P(correct X) is a difference of the cumulative distribution, given byP_x=1-e^{150/75}=1-e^2$$So the expected number of bombs hitting the target is given by$$E=20 \cdot 30/55 \cdot (1-e^2)$$0 I am trying to use simulation methods to estimate the expected number of bombs hit by the twenty aircraft, given 20 pairs of random numbers from [0,1]. Details will depend on the programming language, but the idea is to have a procedure that simulates twenty random points (x,y) and determines how many of those twenty points, say N_{20}, fall in ... 2 \operatorname{E}(X\mid Y)=Y, so by the law of total expectation, \operatorname{E}(X) = \operatorname{E}(\operatorname{E}(X\mid Y)) = \operatorname{E}(Y) = 3. For the Poisson distribution, the variance is the same as the expected value, so \operatorname{var}(X\mid Y) = Y. Then the law of total variance tells us that$$ \operatorname{var}(X) = ... 0 Let's rescale to[0,1]$and a sum of$\frac43$to make it a bit easier. Fleshing out Brian's suggestion: At least one pair sums to at least$\frac43$if the maximum pair sums to at least$\frac43$. The maximum voltage has cumulative distribution function$x^3$and thus probability density function$3x^2$. Conditional on$x$, the second-highest voltage ... 0 Conditioned on$Y_{i-1}$and$Y_{i+1}$,$Y_i$has uniform distribution on$[Y_{i-1}, Y_{i+1}]$. Thus conditioned on$Y_{i-1}$and$Y_{i+1}$, both$Y_i - Y_{i-1}$and$Y_{i+1} - Y_i$are uniform on$[0, Y_{i+1} - Y_{i-1}]$. Since the conditional distributions given$Y_{i-1}$and$Y_{i+1}$are the same, the unconditional distributions are the same. 0 Because the joint distribution is so simple: $$f_{X,Y}(x,y) = 1, \quad 0 \le x, y \le 1,$$ it is easier to integrate directly: $$\operatorname{E}[|X-Y|] = \int_{x=0}^1 \int_{y=0}^1 |x-y| f_{X,Y}(x,y) \, dy \, dx = \int_{x=0}^1 \int_{y=0}^x x-y \, dy \, dx + \int_{x=0}^1 \int_{y=x}^1 y-x \, dy \, dx.$$ 0 Let us write the Marsaglia formulas under the form : $$X = \sqrt{-2\log{U}}\frac{V_1}{\sqrt{R}},Y = \sqrt{-2\log{U}}\frac{V_2}{\sqrt{R}}$$ By comparison with Box-Muller formulas, you see that the cosine and the sine are resp. replaced by $$C=\frac{V_1}{\sqrt{R}},S=\frac{V_2}{\sqrt{R}} \ \ \text{with} \ \ C^2+S^2=1$$ which is exactly the same because the ... 2 We can derive the pdf by calculating the cdf and taking the derivative. The probability that$X_{max}\le x$for$x$between$0$and$\theta$is the probability that all$n$measurements fall between$0$and$x$. But for a single measurement, this is$\frac{x}{\theta}$. The probability$n$independent measurements fall between$0$and$xis then ... 1 If so, should I add them together or should I separate the situation and say if x<=z, then the answer is xz and if x>z, then the answer is z^2? The latter. You have a piecewise function. Indeed there are a few other cases to consider. \begin{align}\mathsf P(U\leq x, T\leq z) ~=~& \begin{cases}\mathsf P(U\leq x, V\leq z) & : ... 1 Your mistake is using Bayes' Rule. Don't.\begin{align} \Pr(\lvert X-Y\rvert\leq 2) ~=~& \Pr(Y-2\leq X\leq Y+2) \\[1ex] = ~& \dfrac{1}{5} \end{align}$$Because whatever the value of Y is in [6;14], the probability that X will lie within the four second interval near that value is: 4/20. 1 You should be integrating over the support of the PDF. The support is the set of all numbers such that the PDF is not zero. Therefore, your integral should be \displaystyle\int_a^x \frac{1}{b-a} \, dx. This will give CDF=\dfrac{x}{b-a}-\dfrac{a}{b-a}=\dfrac{x-a}{b-a} just like wikipedia has. The reason we integrate over the support is simple. We'd ... 1 For company B, let \{X_n\} be the interarrival times, supposing that they are independent and identically distributed with mean \mathbb E[X_1]=30. Let S_n=\sum_{k=1}^n X_k, then \{S_n\} is a renewal process. Let N(t)=\sum_{n=1}^\infty\mathsf 1_{(0,t]}S_n be the number of renewals up to time t, then at a given time t, the time until the next ... 1 The conditional density of Y given X=x is f_{Y|X}(y|x)=\frac{1}{x+1}1_{0<y<x+1}, hence the joint density of X and Y is$$ f(x,y)=f_{Y|X}(y|x)f_X(x)=\frac{1}{x+1}1_{0<x<1,0<y<x+1}$$The marginal density can then be obtained by "integrating out" the x-variable:$$ f_Y(y)=\int f(x,y)\;dx\$

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