# Tag Info

-1

To get the pdf of $X$, let us first get the cdf of $X$: $$F_X(x) = P(X \le x) = P(2Z + 1 \le x) = P(Z \le \frac{x-1}{2})$$ Now $P(Z \le \frac{x-1}{2}) = F_Z(\frac{x-1}{2})$ so: $$F_X(x) = F_Z(\frac{x-1}{2})$$ $$\to f_X(x) = f_Z(\frac{x-1}{2}) (\frac{x-1}{2})'$$ $$\to f_X(x) = f_Z(\frac{x-1}{2}) (\frac{1}{2})$$

1

The easier way to proceed is to note that $2Z+1 \leq x$ if and only if $Z \leq \frac{x-1}{2}$. This gives you the CDF, which you can differentiate as necessary.

3

Mean: You integrate $y=e^{-x^2}$ times a Gaussian pdf. Hence, you get an exponential with exponent $$-x^2-\frac{(x-\mu)^2}{2\sigma^2}=-\frac{2\sigma^2x^2+x^2-2\mu x+\mu^2}{2\sigma^2}.$$ By completing the square, $$2\sigma^2x^2+x^2-2\mu x+\mu^2=(2\sigma^2+1)(x-\frac\mu{2\sigma^2+1})^2-\frac{\mu^2}{2\sigma^2+1}.$$ This yields the pdf of a Gaussian of ...

1

The mean of $e^{-X^2}$ is the integral of $e^{-x^2}$ weighted by the density function of $X$ (which is another exponential function). For an analytic expression, notice that the integral now contains a product of two exponentials. Combine the product of exponentials into an exponential of a sum and re-write the resulting exponent as a single negative square ...

0

In the stattrek example, you are not told the true distribution of the population of students. You only know the average weight and the sd. I take a sample of 50 students and then take the average of these 50 students. I call that $\bar X$. I need to know what is the probability that the average weight $\bar X$ of a sampled student will be less than 75 ...

0

Since you were given information about all the students taking the test, you know the true population mean and standard deviation. In other words, the problem is trying to get you to relate a score of 70 to the area under the curve. The area under the curve to the left of 70 represents the percentage $p$ of students who got a 70 or worse. In other words, if ...

0

It means the percentage of students getting a score of 142 or less (either that or the percentage getting 142 or more, depending on the definition of percentile rank in your course notes).

2

$$\mathbb{E}[X^2+X]=\mathbb{E}[X^2]+\mathbb{E}[X]=\mathbb{E}[(X-\mathbb{E}[X])^2]+\mathbb{E}[X]=\sigma^2 +0=1$$

0

$$\int x^2e^{\frac{-x^2}{2}}=\int x\cdot xe^{\frac{-x^2}{2}}=-x\cdot e^{\frac{-x^2}{2}}+\int e^{\frac{-x^2}{2}}$$ $$\int xe^{\frac{-x^2}{2}}=-e^{\frac{-x^2}{2}}$$ $$\therefore \int (x^2-x)e^{\frac{-x^2}{2}}=-(x-1)\cdot e^{\frac{-x^2}{2}}+\int e^{\frac{-x^2}{2}}$$

1

Since $p>\frac{1}{2}$, we can write $p= \frac{1}{2}+q$ for some $q>0$. Now $$\frac{(n-1)^p}{\sqrt{n}} = \sqrt{\frac{n-1}{n}} (n-1)^q \geq \frac{1}{2} (n-1)^q, \qquad n \in \mathbb{N},$$ implies $$\mathbb{P} \left(B_1 > C \frac{(n-1)^p}{\sqrt{n}} \right) \leq \mathbb{P} \left(B_1 > \frac{C}{2} (n-1)^q\right).$$ Choose $k \in \mathbb{N}$ ...

1

No, in general, if $X$ follows a normal distribution, then in shorthand it is written as $$X\sim N(\mu,\sigma^2).$$ Thus, for example, the variance of $X_1$ is 3 not $3^2$. Or as you have written it $$\sigma_1^2 = 3$$ not $\sigma_1 = 3$.

1

Suppose $X$ is normal with mean $\mu$ and standard deviation $\sigma$. Then $Z=\frac{X-\mu}{\sigma}$ is normal with mean $0$ and standard deviation $1$, and $X=\sigma Z + \mu$. Then $$E[X \mid X \in [a,b]]=E[\sigma Z + \mu \mid X \in [a,b]] \\ = \mu + \sigma E[Z \mid X \in [a,b]] \\ = \mu + \sigma E \left [Z \left | Z \in \left [ ... 0 Suppose for simplicity that you have a standard normal X with pdf f. One of the main properties of f is that it satisfies f'=-xf which implies \int_a^b xf\,dx=-\int_a^b df=f(a)-f(b). It follows that$$E(X|X\in[a,b])=\frac {f(a)-f(b)}{\Phi(b)-\Phi(a)}$$Set b=\infty (\Phi(\infty)=1, f(\infty)=0) if you want a one-sided bound. 0 The "mean" of a continuous probability distribution, P(x), is, by definition, the integral of xP(x). To restrict a normal distribution, y= Ae^{\frac{(x- \mu)^2}{\sigma^2}}, between x= a and x= b, with a< b, we have to divide by the probability x is between a and b, the integral of P(x) between a and b. Here, mean is \frac{\int_a^b xe^{\frac{(x- ... 0 I think you just need to take the values/points (which are above 10MW), sum them up and divide the sum by their count. That's all, no? Whether the distribution is normal or not, that's not relevant here, I think. 0 If T = X_1 + 2X_2 were sufficient, then the conditional distribution of (X_1, X_2) given T would be independent of \theta. In particular, the conditional density of X_1 given T = t should be independent of \theta. So let's compute this conditional density. Since X_1 and X_2 are independent, it follows that$$\begin{pmatrix}X_1 \\ ...

1

I don't know an answer in the general case but concerning the specific case $b=2$ we may start by defining : \begin{align} \tag{1}f(a,c)&:=\int_0^{\infty} \frac{x^{-1}}{a+x^{-2}} e^{-c x^2} dx\\ &=\int_0^{\infty} \frac{x}{1+ax^2} e^{-c x^2} dx\\ &=e^{\,\large{\frac ca}}\int_0^{\infty} \frac{c\;x}{c\;(1+ax^2)} e^{\,\large{-\frac ca (1+ax^2)}} dx\\ ...

0

The last inequality is untrue for general distributions. Suppose $X$ and $Y$ are +1 or -1 independently with probability $\frac12$. Then $E[X]+E[Y]=0+0=0$. But $E[\max(x,y)]=\frac14\cdot-1+\frac34\cdot1=\frac12$. For normally distributed random variables both with mean 0, I would say it is also untrue (In that case it conflicts with the first two ...

1

This question seems to belong to StackOverflow's R-tag In R we do not generally care about whether a variable is integer or floating if they are used for calculations as an operation such as 2L/3.0 will return a floating variable anyways. For example an integer is also a floating variable: is.numeric(2L) returns TRUE while is.integer(2) returns FALSE. For ...

1

$$\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}x\mathrm{e}^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$$ let $z=\frac{x-\mu}{\sigma}$ then we get $$\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\sigma z+\mu)\mathrm{e}^{-z^2/2}dz$$ where you should use $$\dfrac{d}{dz}\mathrm{e}^{-z^2/2}=-z\mathrm{e}^{-z^2/2}$$ to help. But, you should really know how to ...

1

We will transform $|Y−μ|<1.96\sigma$ into the form $a(Y)<\mu<b(Y)$: $$|Y−μ|<1.96\sigma\Leftrightarrow1.96\sigma<Y−μ<1.96\sigma\Leftrightarrow1.96\sigma-Y<−μ<1.96\sigma-Y$$$$\Leftrightarrow Y-1.96\sigma<μ<Y+1.96\sigma.$$ From $P(\left | Y-\mu \right | < 1.96\sigma) \approx0.95$ and transforming done above we obtain ...

1

Outline: Let $a$ be the probability that a randomly chosen female Smurf is between $1$ and $1.3$, and let $b$ be the corresponding probability for male Smurfs (Smurves?). Then our required probability is $(0.6)a+(0.4)b$.

0

$$P\{ Y \le y\} = P\left\{\frac{1}{Z}\le y\right\} = P\left\{Z \ge \frac{1}{y}\right\} = \int_{1/y}^\infty f_Z(z)\mathrm{d}z$$ $$\implies f_Y(y) = \frac{\partial}{\partial y} P\{Y \le y\} = \frac{1}{y^2} f_Z(1/y)$$ Alternatively, recall that densities of transformed random variables are in the ratio of the Jacobian of the transformation. $$z = ... 0 The change of variables formula is (from the chain rule):$$\begin{align} f_{g(Z)}(y) & = f_Z(g^{-1}(y)) \cdot \lvert \mathcal D_y\; g^{-1}(y)\rvert & \textrm{where $g(z)$ is an invertable function} \\[2ex] f_Y(y) & = f_Z(1/y)\cdot\left\lvert\dfrac{\operatorname d y^{-1}}{\operatorname d y\quad}\right\rvert & \textrm{since $g(z)=1/z$} ...

1

By the properties of a continuous density, $$\mathbb P(0 < Y \le b) = \int_0^b f_Y(y) \, dy.$$ Therefore, for $Y = 1/Z$, $$\mathbb P(0 < Y \le y) = \mathbb P(Z \ge 1/y) = \int_{1/y}^{\infty} f_Z(z) \, dz = 1/2 - \int_0^{1/y} f_Z(z) \, dz.$$ Differentiating with respect to $y$ to recover $Y$'s density, $$f_Y(y) = \frac d{dy} \left( 1/2 - ... 1 Hint. You are looking for a number a such that$$ P(L>a)=0.8 $$or equivalently, by the change of variable L \to Z,$$ P(Z>\frac{a-50000}{5000})=0.8 $$or$$ P(Z\leq\frac{50000-a}{5000})=0.8 $$Using a table your Z-table you can find that$$ P(Z\leq \color{red}{0.84}) \approx 0.8. $$Then$$ \frac{50000-a}{5000}=\color{red}{0.84} $$and ... 0 Here's a hint. Using your z-table (or whatever you are using), find the value z such that the cumulative probability is .2. We do this understanding that this is a right-tail. So, maybe you could go backwards with that z-value? 1 The univariate normal p.d.f. is$$ x \mapsto \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{1}{2}\left( \frac{x-\mu} \sigma \right)^2 \right). \tag 1 $$The expression \left(\dfrac{x-\mu} \sigma\right)^2 is the same as (x-\mu) (\sigma^2)^{-1} (x-\mu). The 1\times1 matrix x-\mu is its own transpose, and the inverse of the 1\times1 matrix whose ... 0 In part b) you should divide \sigma by \sqrt n, dont subtract it from \sigma.so, you need to recompute z-statistics and your conclusion on part c will depend on the new z-statistics. 1 You need to complete the top-right of the covariance matrix by reflecting the bottom-left to give$$ \Sigma = \left( \begin{array}{cccc} 1037.21 & -80.02 & 1430.7 & 271.44 \\ -80.02 & 219.84 & 92.1 & -91.58 \\ 1430.7 & 92.1 & 2624 & 210.3 \\ 271.44 & -91.58 & 210.3 & 177.36\end{array} \right)$$... 0 Let X \sim Poisson(n). You are correct that$$ X = \sum_{i=1}^n X_i $$where each X_i \sim Poisson(1) and independent. Thus X is indeed a sum of i.i.d. random variables with finite mean and variance, so we may apply the CLT. First note E(X) = Var(X) = n. Then$$ P\left(\left| \frac{X}{n} - 1\right| > 0.01\right) = P\left(\left| ...

1

What "nullUser" said is correct. Looking at the table you find P(z<-1) = .16 and p(z<1) = .84 roughly. You can solve then by just subtracting .84-.16 to get the probability in-between. If you just calculate p(z<1) like it seems you did, it overlaps with z<-1 so there is no way to discern the correct answer unless you "delete" the overlapping ...

1

You don't want $P(X-\mu < \sigma)$ if $X>\mu$ and $P(\mu-X < \sigma)$ if $X < \sigma$, you want $P(|X-\mu|<\sigma) = P(|z|<1) = P(z<1) - P(z<-1)$.

1

I will assume upvotes validate the accuracy of the following answer: The first value is not a probability, but a density value, $D_p$ given by $$D_p=\frac1{\sigma\sqrt{2\pi}}e^\frac{-(x-\mu)^2}{2\sigma^2}.$$ The density for a value $x=a$ should be used to compare whether a value $x_i$ is more likely to be chosen from the population if it is near (within a ...

0

It is unlikely but could it be the probability of being within 3 standard deviations of the mean ? If you computed statistics you generated using few samples, outliers may have more weight than they should ie 1-96.8% instead of 1-99.7%. How did you find 96.788% ?

0

First, notice that the second-to-last line is of the form $$k_1e^{a}\times k_2e^{b}$$ for appropriate replacements of $k_1,k_2,a,b$. Therefore since $p(\theta|x)$ is proportional to $k_1e^{a}\times k_2e^{b}$, it is also proportional to just $e^ae^b$ (dropping constants, as you said). And this is just $e^{a+b}$. Since $a$ and $b$ are quadratics in ...

0

Not quite sure if the one on Wiki is right or if I understand it, but anyway I'm basing my answer on CLT in Larsen and Marx (an elementary probability book): Anyway, this answer might be weird or very wrong or stupid but... Consider iid $X_i$ ~ $Poi(\lambda/n)$. Then $X := \sum_{i=1}^{n} X_i$ ~ $Poi(\lambda)$ Then $$(\frac{1}{n} \sum_{i=1}^{n} X_i - ... 1 General forecasting rule Suppose X_1 has mean \mu_1, X_2 has mean \mu_2, and that they are Gaussian with covariance matrix$$ \left[\begin{matrix} \Omega_{11},\Omega_{12} \\ \Omega_{21},\Omega_{22} \end{matrix} \right]$$then$$ X_2∣X_1 \sim N(\mu_2+(\Omega_{21}\Omega_{11}^{-1}(X_1-\mu_1),\Omega_{22}-\Omega_{21}\Omega_{11}^{-1}\Omega_{12}) $$... 1 If X_1 and X_2 are iid random variables such that X_1\sim\mathcal N(0,\sigma^2) and X_2\sim\mathcal N(0,\sigma^2), then$$ X_1-X_2\sim\mathcal N(0,2\sigma^2). $$If X\sim\mathcal N(0,\sigma^2), then Y=|X| has the half-normal distribution and$$ \operatorname EY=\frac{\sqrt2\sigma}{\sqrt\pi}. $$Hence, we have that$$ \frac12\operatorname ...

0

If I understand your setup (maybe I don't), then you are after: $\mathrm P_\textsf{Answer} =$ $\displaystyle \int_{0}^\infty\int_{0}^\infty f_{X,Y}(x,y)\operatorname d x\operatorname d y \\ \displaystyle +\int_{-\infty}^0\int_{0.7x}^\infty f_{X,Y}(x,y)\operatorname d y\operatorname d x \\ \displaystyle + \int_{-\infty}^0\int_{0.8y}^\infty ... 0 Things are prety clear here, your notations are the ones that confuse everything. Let$I_1 (r) = \int \limits _{-r} ^r$and$I_1 = \lim \limits _{r \to \infty} I_1 (r)$. Let$D(r)$be the disk of centerpoint$(0,0)$and radius$r$. Let$I_2 (r) = \iint \limits _{D(r)}$(in polar coordinates). Now$I_1 ^2 = \left( \lim \limits _{r \to \infty} I_1 (r) ...

0

$X$ - numbers of "shows". So, $X \in \mathrm{Bin}(n,p)$ where $n=208$ and $p = 0.91$. Then, if $np(1-p) \geq 10$ we can make the following approximation, where the approximation was made in the last step  $$P_{\mathrm{Bin}}(X \geq 191) = 1- P_{\mathrm{Bin}}(X \leq 190) = 1- P_{\mathrm{Bin}}(X < 191) \approx 1- \mathrm{N}(\frac{190.5-np}{\sqrt{np(1-p)}} ... 1 Let X \sim N(\mu,\sigma^2). Take g := 1. Then Stein's lemma gives$$\mathbb{E}(1 \cdot (X-\mu)) = 0, \tag{1}$$i.e. \mathbb{E}(X) = \mu. For g(x):= (x-\mu)^{n-1}, n \geq 2, we have, by Stein's lemma,$$\mathbb{E}((X-\mu)^n) = \sigma^2 (n-1) \mathbb{E}((X-\mu)^{n-2}).$$If n is odd, i.e. n=2k+1, then we get by iteration$$\begin{align*} ...

1

Your question posits the existence of a normally distributed random variable $X$ with standard deviation $\sigma = 0.8$, for which $\Pr[X > 40] = 0.96$; that is, to say, the chance that $X$ is above $40$ is $96\%$. That is what those two conditions mean. From this, we are supposed to determine the mean $\mu$ of $X$. To do this, we can standardize $X$: ...

1

The standard normal distribution is symmetric. And therefore $P(X\geq 40)=1-\Phi \left( \frac{40-\mu}{0.8} \right)=\Phi \left( \frac{\mu-40}{0.8} \right)=0.96$ $\frac{\mu-40}{0.8}=\Phi^{-1}(0.96)$ $\frac{\mu-40}{0.8}=1.75$ To get your solution you multiply both sides by $(-1)$: $\frac{40-\mu}{0.8}=-1.75$

0

Yes, I was wrong. The pdf of $Y$ is indeed integrable. A function with a singularity might indeed be integrable, as in the example: $$\int_0^a{y^{-\frac{1}{2}}dy}=2a^{\frac{1}{2}}$$

2

I decided to move from a comment to an answer because some times I ended up using some facts without digging too much in the whys. So, I decided to dig a little more and hopefully answer the questions of the OP. This answer is mainly about why the RMS of the noise is equal to its standard deviation. As a side note, I will also do some comments about the ...

0

There shouldn't be a $0.11$ in that denom. In fact it should be your given SD because you are selecting only one cylinder. This gives a z-score of $-1.5$ for the left boundary. Looking in the cumulative z-table I read $0.0668$ Due to symmetry the region that falls outside the required measurement is then $2*0.0668$ and when you subtract that from $1$ (total ...

0

Let the random variable $X=$The price of the pound in euros on a randomly selected day within this period. Then $$X\sim N\left(1.459,0.033^2\right)$$ $$P(X \ge 1.492)=P\left(\cfrac{1.459 - 99}{0.033} \le X \le \cfrac{1.459-1.492}{0.033}\right)=P(-2955 \le z \le -1)=\color{red}{\fbox{0.158655253}}$$ as required. If you are wondering where the $99$ came ...

1

You appear to be considering something like: $$P\left ( x-\frac{\Delta x}{2} \leq X \leq x+\frac{\Delta x}{2} \right )$$ where $X$ is Gaussian with mean $\mu$ and standard deviation $\sigma$. This is exactly equal to $$\int_{x-\Delta x/2}^{x+\Delta x/2} \frac{1}{\sqrt{2 \pi} \sigma} e^{-(y-\mu)^2/(2\sigma^2)} dy$$ which can be straightforwardly ...

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