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Fact 1: If $X\sim N(0,a)$ then $\mathbb{P}(|X|>\epsilon)\leq \frac{1}{\epsilon}\sqrt{\frac{2a}{\pi}}\exp[-\epsilon/2a]\leq \exp[-\epsilon/2a]$, whenever $a\leq \epsilon^{2}\pi/2$. Fact 2: If $t\to 0$ then $t^{-b}e^{-c/t}\to 0$ for any fixed $b,c>0$. In particular $e^{-c/t}<t^{b}$ for sufficiently small $t$. We know that $a_{n}\to 0$. So there ...

3

The areas for one, two, and three standard deviations about the mean are well known, as .68, .95, .997 roughly. But in general, there's no way to find the area under the standard normal distribution without using a computer, calculator, or a chart. edit1: most charts will give the area to the left of a given $z$-score. To find other values, use symmetry ...

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Let us evaluate the CDF of a Gaussian distribution: $$f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} \int_\mathbb{R} \exp \left( - \frac{(x-\mu)^2}{2\sigma^2}\right)\ dx \label{orig}\tag{1}$$ Let us make the transformation in $\ref{orig}$ by $\mu \mapsto N \mu$ and $\sigma \mapsto \sigma \sqrt{N}$. \begin{align*} f(x) &= ... 2 The variance of this random vector is \begin{bmatrix} 1 & \rho \\ \rho & 1 \end{bmatrix}. Level sets of the density are ellipses, which are circles only when the correlation \rho is 0. With a positive-definite symmetric matrix, one can always rotate the two coordinate axes to diagonalize the matrix; then one has two linear combinations of the ... 2 Notice how the function is odd. Therefore, \displaystyle \int_{-\infty}^{0} f(x)=-\int_{0}^{\infty} f(x)\implies \int_{-\infty}^{\infty}f(x)=0 If we can show that \displaystyle \lim\limits_{x\to0,\pm\infty}f(x) exists, which are where it can diverge, then the integral exists. As we can plainly see \lim_{x\to0 \text{or} \pm\infty} is 0. Therefore the ... 2 One hint that might help is to note that the function you are integrating is an odd function, so ... 2 No. The lognormal distribution doesn't have a moment generating function, so you can't use that approach. Instead, suppose Y = \log X. Then Y \sim \operatorname{Normal}(\mu, \sigma^2) by definition, and X = e^Y. Therefore for a positive integer k,\operatorname{E}[X^k] = \operatorname{E}[e^{kY}] = M_Y(k),$$where M_Y(k) is the moment ... 2 Suppose that$$\hat{f}(\xi) = \sqrt{\hat{g}(\xi)} \cdot h(\xi)$$for some function h: \mathbb{R} \to \{-1,1\}. Since \hat{f} is itself a characteristic function, we know that \tilde{f}(0)=1>0. Hence, h(0)=1. On the other hand,$$h(\xi) = \frac{\hat{f}(\xi)}{\sqrt{\hat{g(\xi)}}}$$is a continuous function as \hat{f}, \hat{g} are continuous ... 2 Cramér's Theorem even gives that if the sum X+Y of any two independent (not necessarily identically distributed) random variables X and Y has a normal distribution, then the summands X and Y themselves must be normally distributed. This is more than enough to cover your case: if X is independent of Y, it is also independent of -Y, and X-Y ... 2 You need to conduct a hypothis test for the variance not for the mean. The confidence interval 9\pm 1.5 is a confidence interval for the mean and will not help you to draw conclusions about the variance! Accordingly, you do not need the t-distribution but the \chi^2-distribution. Formally In order to conclude whether σ^2 has been reduced below 1.0 ... 2 The integral of any function over a set with measure 0 is equal to 0. 1 Let N be a normally distributed variable, with mean 0 and variance 1 (substract the mean and divide by the standard deviation in order be in this case). You look for a certain (smooth, increasing) f:\Bbb R\to[0,1] such as  f(N)  is uniform, that is:$$ P(f(N) \le q) = q $$for every q\in(0,1). Under regularity assumptions, this is$$ q = P(N \le ...

1

You did it all right, except for the final fraction. The joint event is: $$A \cap B \equiv (1 \le X \le 3) \cap (0 \le X \le 4)\equiv (1 \le X \le 3)\equiv A$$ (Notice that this holds because here $A \subset B$) Hence $$\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}=\frac{.383}{.6826}=0.561$$

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'The general procedure looks correct. After calculating the probability that a can is over $8$ ounces, you might want to use the normal to approximate the binomial. Your "$p$" for the binomial is not right. We want $\Pr\left(Z\gt \frac{0.04}{0.22}\right)$. This is much larger than $0.145$.

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From the rule of complement which says that $P(A^c) = 1 - P(A)$ it follows that: $P(z<-\frac{20.5}{\sigma}~~or~~z>\frac{20.5}{\sigma}) = 1 - 0.8 = 0.2$ From the symmetry of the normal distribution then, $P(z<-\frac{20.5}{\sigma}) = 0.1$ From here, either we look at a table of probabilities associated with different values of $z$ or we use calculus ...

1

More generally, if $X_1, \ldots, X_n$ are independent normal random variables with means $\mu_i$ and variances $\sigma^2_i$, and $c_1, \ldots, c_n$ are constants, $Y = c_1 X_1 + \ldots + c_n X_n$ is normal with mean $\mu_Y = c_1 \mu_1 + \ldots + c_n \mu_n$ and variance $c_1^2 \sigma_1^2 + \ldots + c_n^2 \sigma_n^2$. Here you're doing the case $n=3$ with ...

1

For every $\theta$, the change of variable $t=F_\theta(x)$ yields $$E(\psi_\theta(X)^2)=\int_{-\infty}^\infty\Phi^{-1}(F_\theta(x))^2\mathrm dF_\theta(x)=\int_0^1\Phi^{-1}(t)^2\mathrm dt.$$ The RHS does not depend on $\theta$ (and is equal to $1$) hence, differentiating, one gets ...

1

What you have calculated for a and b is the solution for b. This is o.k. To get the formula for the excact probability, you have to use the binomial distribution: $$P(75 \leq X \leq 80)=\sum_{k=75}^{80} {100 \choose k} 0.7^k\cdot 0.3^{100-k}$$ $P(75 \leq X \leq 80) \color{red} = F_{Z}[\frac{80 + 0.5 - > 70}{\sqrt21}] - F_{Z}[\frac{75 - 0.5 - ... 1 It should be clear that$A=\left\lbrace Z\leq 106\right\rbrace \subset \left\lbrace Y\leq 106\right\rbrace=B$-- if$11$fit then so do$10$. What you want is the probability of$\left\lbrace Y\leq 106\right\rbrace\setminus\left\lbrace Z\leq 106\right\rbrace=B\setminus A$. Since \begin{equation*} \mathbf{P}(B)=\mathbf{P}(B\setminus A)+\mathbf{P}(B\cap A), ... 1 The answer is about right, but there are two errors that happen to cancel! The probability a tube is bad (too big or too small) is roughly$2(0.0062)$, that is,$0.0124$. And you need to multiply by$1000$, not by$100$, giving expected value about$12.4$. 1 Let$X_i$have mean$\mu_i$and standard deviation$\sigma_i$, and write$X_i = \mu_i + \sigma_i Z_i$where$Z_i$are independent standard normal random variables. The joint density of$Z_1, \ldots, Z_n$is $$f(z_1, \ldots, z_n) = (2 \pi)^{-n/2} e^{-(z_1^2 + \ldots + z_n^2)/2} = (2 \pi)^{-n/2} e^{-\|{\bf z}\|^2/2}$$ which is rotationally invariant, i.e. ... 1 This is a proof for$\mu = 0$. When$\mu \neq 0$, the proof can be done with some slight modification. You want to prove$\sqrt{n}(1-X_n^{-1}) - G_n = \sqrt{n}(2-X_n^{-1}-X_n)$converges to$0$in probability, which is equivalently that it converges to$0$in distribution. To do this, we can apply the second order delta method(for reference see this) with ... 1 No, the distribution is not necessarily normal if it satisfies the$68-95-99.7$rule. Take a normal distribution, then move all the events that are within$(\mu+0.25 \sigma, \mu + 0.75 \sigma)$, some of them to the mean and some of them to$\mu + 0.99 \sigma$Do the same for the ones within the range$(\mu - 0.75 \sigma, \mu - 0.25 \sigma)$in the other ... 1 You can use Wolfram alpha. For example, http://www.wolframalpha.com/input/?i=1%2Fsqrt%282pi%29+integral%28e%5E%28-x%5E2%2F2%29%2Cx%2C-infinity%2C-1.39%29 answers your first question. Let me point out that your the meaning of the area under the curve to the left of$-1.39$is the probability that a standard normal is at most$-1.39$. Make sure that you ... 1 First of all, define a normalized variable $$Z={X-\mu\over \sigma}$$ Then, on the normal distribution table, you need to look for a probability equal to about 0.8. The linked table gives$z_0\approx0.84$so we have $$P(Z<0.84)\approx0.80$$ Thus to get the right number of hotdogs, $${x_0-16036\over100}=z_0=0.84$$ so $$x=16036+840=16876\approx16900 ... 1 If X is any continuous random variable (e.g. normally distributed), then P(X=c) = 0 for all c in \mathbb{R}, no matter what distribution X has or what point c is. Points are too small to have non-zero probability of being hit by X. If by the quantity b you mean the variance -- normally, we write X \sim N(\mu,\sigma^2) -- then b is zero ... 1$$ f_{Z|Y}(z|y) = \frac{f_Z(z)f_{Y|Z}(y|z)}{\int_{-\infty}^{+\infty}{f_Z(a)f_{Y|Z}(y|a)da}} = \frac{\frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-z)^2}{2}}} {\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}} e^{-\frac{a^2}{2}} \frac{1}{\sqrt{2\pi}} e^{-\frac{(y-a)^2}{2}}da}} = ... 1 Let$ X_1 \sim N(23.95,7.44), X_2 \sim N(16.29,7.79) $, then, as mentioned by Dilip Sarwate in the comments,$ X_1 - X_2 \sim N(7.66,{\sqrt{7.44^2 + 7.79^2}}) $Thus the problem becomes $$\mathbb{P}(X_1<X_2) = \mathbb{P}(X<0),\ \ where \ \ X = X_1 - X_2$$ This can be calculated directly or reverting to the standard$ N(0,1) $distribution as ... 1 This cannot be true, as the example$X_n=Y_n=X=-Y$standard normal shows. If$X$and$Y$must be independent, try$X_n=Y_n=X$with$X$and$Y$i.i.d. standard normal. If$X_n$and$Y_n$must be independent, try$X_n=X=Y$and$Y_n=Z$with$X$and$Z$i.i.d. standard normal. 1 If the total number of hours worked$N=13500$, and the number of employees is$n=450$, then the mean number of hours worked per employee is given by: $$\mu = \frac{N}{n}=30\:\text{hours}$$ We are given that the standard deviation$\sigma = 7\:\text{hours}$, and therefore we have that the number of hours worked per employee is normally distributed$T \sim ...

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