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1

You need the standard error of the mean $\frac{\sigma}{\sqrt n}$ not $\sigma$


1

If $\sigma = 1.5$, then $P\{X < c\} = P\{\frac{X-\mu}{\sigma} < \frac{c-\mu}{\sigma}\} = P\{Z < \frac{c-4}{1.5}\} = 0.35,$ where $Z$ is standard normal. Then from normal tables $(c - 4)/1.5 \approx -0.3853.$ Solve for $c.$ If $\sigma^2 = 1.5$ as you say, then $\sigma = 1.224745$ and adjust the denominator accordingly before solving. If $\sigma = ...


1

See my comments under the question. What I write below will assume my guesses there are right and that $a=1$. Your symmetric matrix $C$ will have one row corresponding to each unordered pair $\{i,j\}$. The entry in row $\{i,j\}$ and column $\{k,\ell\}$ will be $2$ if $\{i,j\}=\{k,\ell\}$. Those are just the diagonal entries in the matrix. In row ...


1

If $X$ and $Y$ are not independent, I doubt that there's much that can be said. There are just too many ways to obtain $Z$ as the sum of $X$ and $Y$. For example, here are some ways to get an arbitrary distribution as $Z = X + Y$ with $X$ and $Y$ identically distributed. Let $B$ be Bernoulli($1/2$) and independent of $Z$. Let $f$ be an arbitrary Borel ...


0

Hint: The sum of independent normally distributed random variables is a normally distributed random variable with (1) its mean being the sum of the means, (2) its variance being the sum of the variances. You want to find $n$ such that $\mathsf P((\sum_{k=1}^n B_k) - C \geq 0)\gt 0.2$, where $\{B_k\}$ are the weights of boxes and C is the capacity of the ...


0

All you have to do is to count an integral. Normal distribution with mean 140 and standard deviation 10 has density function $f(x)= \frac{1}{\sqrt{200 \pi}} e^{\frac{-(x-140)^2}{200}}$. You want to count the probability that $X \in (110; 150)$ (because of position of the net). So the solution is $\int^{150}_{110} \frac{1}{\sqrt{200 \pi}} ...


0

As @DillipSarwate says $V(aX+b) = a^2V(X).$ So your comment is right about (1). For (2), the covariance is linear in both arguments. Let $X$ and $Y$ be independent, and thus have null covariance: $$Cov(X, aX + bY) = aCov(X, X) + bCov(X, Y) = 3V(X)$$ also $$Cov(aX + bY, X) = aCov(X, X) + bCov(y, X) = 3V(X)$$ and so on. In particular $Cov(X, 3X) = 3V(X).$ ...


0

Here is something to get you started: You need to find $a$ such that $P(X\leq a)=0.8$, where $X$ follows a normal distribution with a mean of 38 and standard deviation of 6. Now you need to standardize it. Remember that if $X\sim\mathcal{N}(\mu,\sigma)$ then $\dfrac{X-\mu}{\sigma}\sim\mathcal{N}(0,1)$.


1

I agree with the comment by @eigenchris for 'well-known' distributions encountered early on in a probability course. However, one does not have to venture too far into the study of probability distributions to find examples in which knowing the population mean and variance does not easily specify the distribution. It is useful to make a distinction between ...


1

I think the particular values of $m$ and $s$ you are using give rise to unexpected computational difficulties. Perhaps it is just too much to wish for two-place accuracy retrieving $s$ with a million simulated values. Also, taking the ordinary SD of lognormal data may not be the optimal way to estimate parameter $s$ (especially when it is small). I'm not ...


0

The intersection of $\mathbb{Q}$ with the unit interval $[0,1]$ contains infinitely-many discontinuities (Dirichlet's famous "pathological" function and many similar functions behave strangely/pathologically by taking advantage of $\mathbb{Q}$ as an everywhere dense but incomplete set). It is not Riemann integrable. It is, however, integrable using e.g., ...


0

For part b), you are asked what the 95 percentile is. You know the mean in a normal ( or symmetric) distribution is the 50 percentile. Now you need to see what z-value will give you the additional 45% of the data between the mean and the 95 percentile. By symmetry, this should be the z-value associated with the 90 percentile. By the 1-2-3 or 68-95-99 rule ...


10

The best approximation in the $L^2$ sense is given by the value of $\alpha\in\mathbb{R}^+$ for which: $$ \frac{d}{d\alpha}\left(\frac{1}{2\pi}\int_{|x|\geq \alpha}e^{-x^2}\,dx + \int_{-\alpha}^{\alpha}\left(\frac{1-|x/\alpha|}{\alpha}-\frac{e^{-x^2/2}}{\sqrt{2\pi}}\right)^2\,dx\right)=0,$$ i.e. by minimizing the $L^2$ norm of the difference between the pdf ...


5

It depends in what sense you want your triangular distribution to "approximate" the normal distribution. The normal distribution is symmetric about $0$ and unimodal, so you probably want your triangular distribution to be symmetric about $0$ and unimodal as well. In order for your triangular distribution to be a probability distribution, the area under the ...


2

As $X$ and $Y$ are independant, $X-Y\sim\mathcal{N}(0,1)$, so $V(|X-Y|)=V(|Z|)$ where $Z\sim\mathcal{N}(0,1)$. $V(|Z|) = E(|Z|^2)-E(|Z|)^2 = E(Z^2)-E(|Z|)^2$. $E(Z^2)=V(Z)=1$. Now all you have to do is finding $E(|Z|)$ where $Z\sim\mathcal{N}(0,1)$ (calculate the corresponding integral for example).


1

Note that $-Y$ is also $N\left(0,\frac12\right)$. Thus, $X-Y$ is the sum of two $N\left(0,\frac12\right)$ variables, which is a $N(0,1)$ variable. The absolute value of a $N(0,1)$ variable has the PDF $$ \sqrt{\frac2\pi}e^{-x^2/2}[x\ge0] $$ where $[\cdot]$ are Iverson Brackets. Using the substitution $t=x^2/2$, we get $$ \begin{align} ...


1

Yes, you've done it correctly. However, there is some inaccuracy in your answer due to the floating-point roundoff error which happens because you are subtracting $P(Z \leq 8)$, which is close to 1, from 1: > 1-pnorm(8) [1] 6.661338e-16 Instead of this, a better method is to use the option `lower.tail=FALSE' to give the upper tail directly: ...


0

Let $I = \int_{-\infty}^{\infty} \frac{|x|^r}{x} \phi(x) dx; \quad r>0$. Since the integrand is an even function through out the domain, we can write it as, $$ \begin{eqnarray} I &=& 2\int_{0}^{\infty} \frac{x^r}{x} \phi(x) dx\\ &=& 2\int_{0}^{\infty} x^{r-1} \phi(x) dx\\ &=& \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} x^{r-1} ...


0

In an answer to this question, I showed that if $X$ and $Y$ are independent zero-mean normal random variables with the same variance $\sigma^2$, then, with $\hat{X}$ and $\hat{Y}$ being random variables whose joint density is $4f_{X,Y}(x,y)\mathbf 1_{\{x>0, y> 0\}}$, $$F_{\hat{X}+\hat{Y}}(\alpha) = P\{\hat{X}+\hat{Y} \leq \alpha\} = ...


-1

Part b: \begin{align} \mathbb{P}(|X|=|Y|) &= \mathbb{P}(X=Y\ {\rm or}\ X=-Y) = \mathbb{P}(X=Y \cup X=-Y) = \mathbb{P}(X=XZ \cup X=-XZ) \\ & \hspace{5mm} = \mathbb{P}(X=XZ) + \mathbb{P}(X=-Y)-\mathbb{P}(X=XZ \cap X=-XZ) \\ & \hspace{10mm} = \mathbb{P}(Z=1)+\mathbb{P}(Z=-1)+0=\frac{1}{2}+\frac{1}{2} = 1 \end{align}.


1

This is easy if you look for the probability distribution of $a-b$. As $a \sim \mathcal{N}(\mu_a, \sigma_a^2)$ and $b\sim \mathcal{N}(\mu_b, \sigma_b^2)$ are independant, you have that $$a-b = Z \sim \mathcal{N}(\mu_a-\mu_b, \sigma_a^2+\sigma_b^2)$$ Then you can calculate $P(Z<0)$ as usual


0

In my experience the expression 'true standard deviation' is often used to mean 'population standard deviation', as distinguished from 'estimated standard deviation' or 'sample standard deviation'. Then, in the formula $\bar X \pm 1.96\sigma/\sqrt{n},$ you have $\bar X = 4.85,\; \sigma = 0.75$ and $n = 20.$ This gives the CI $(4.521, 5.179)$ in agreement ...


1

Different books have different formats for normal tables. It will go something like this. The normal table has cut-off points in the margins and probabilities in the body. Look in the vertical margin for 1.6. Under the column headed .00 you will see .0548. That is the probability corresponding to cutoff 1.60. But you want the probabilities corresponding to ...


1

Using a standard normal table, which you can find a good example of one here: https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf, for $P(x < 31.5) = 0.05$, that means the area to the left of the z-score is $0.05$. Because we know that the distribution is normal, we would go into the middle of the table and find the closest value to $0.05$ ...


1

The terminology is a weighted mixture of normal distributions or, in short, a mixture distribution. For a mixture of two normals ($N(\mu_1,\sigma_1^2)$ and $N(\mu_2,\sigma_2^2)$) the density is $$f(x)=\frac{w \exp \left(-\frac{\left(x-\mu _1\right){}^2}{2 \sigma _1^2}\right)}{\sqrt{2 \pi } \sigma _1}+\frac{(1-w) \exp \left(-\frac{\left(x-\mu ...


0

The inverse Mill's ratio is defined as $$ \lambda(x)=\frac{\phi(x)}{\Phi(x)}. $$ For answering the question, it suffices to show that $\lambda'(x)<0$. Note that the p.d.f. of standard normal $\phi$ is differentiable. Thus we can apply quotient rule to it $$ ...


0

Maybe there is a simpler way, but here is a thought. Use the relations $ \Phi(x)' = \phi(x)$ and $\phi(x)'= -x\phi(x)$ $$\left(\frac{\Phi(x)}{\phi(x)}\right)' = \frac{\Phi(x)'\phi(x) - \Phi(x)\phi(x)'}{\phi(x)^2} = \frac{\phi^2(x) + x \Phi(x)\phi(x)}{\phi(x)^2} = 1 + x \frac{\Phi(x)}{\phi(x)}$$ If $\left(\frac{\Phi(x)}{\phi(x)}\right)' \geq 0$ we are ...


2

On the line you don't understand, you are using the standardization of your random variable $X$ to find your standard deviation value $\sigma$. Recall that in order to standardize a Normally distributed random variable, you subtract its mean, and divide by its standard deviation. Essentially your formula is $$ Z = \dfrac{x-\mu}{\sigma} $$ where $x$ denotes ...


0

(a) and (b) are correct. By the way, the probability you got in (b) isn't actually too small. Think about it this way, $\mu$ is $6$ so the probability that the sample mean ($\bar{X}$) is less than $-4$ will be really small. (c) You forgot to divide by $\sigma^2$ on the right hand side of the inequality: $P(S^2<4)=P(X<\dfrac{4(16-1)}{4})=P(X<15)$, ...


0

You have $\displaystyle \bar X \sim N\left( 20, \left(\frac 2 3 \right)^2 \right)$ approximately. (Here you may have been slightly misled. Is this a reasonably close approximation? An answer to that would have to rely on information beyond what is specified in the problem. For some highly skewed distributions, $144$ would be nowhere near a big enough ...


0

note that, given to conditional probability: $$ X_t = Ax_{t-1} + N $$ i think if we know the $x_{t-1}$ , then $$f_{{X_t}|{X_{t-1}}}(x_t|x_{t-1})$$ will be two dimensional Gaussian distributed. I am very busy currently, if it does not help you comment me.


1

For $X\sim N(\mu,\sigma^2)$ the first quartile is $$q_{0.25}=\sigma\cdot \Phi^{-1}(0.25)+\mu$$ where $\Phi$ is the CDF of the standard normal r.v. because $q_{0.25}$ is the solution of $$P\{X\le q_{0.25}\}=P\left\{\frac{X-\mu}{\sigma}\le \frac{q_{0.25}-\mu}{\sigma}\right\}=\Phi\left(\frac{q_{0.25}-\mu}{\sigma}\right)=0.25$$


1

If I have understood the question correctly, you have a list of n elements {$x_i$} some of which are "special" (exceed some level you set) and the rest of which are non-special. You want to define two probabilities p and q where p is the probability that a given special element is chosen and q is the probability that a given non-special element is chosen. ...


3

The sum of independent normal random variables is normal. That assumption of independence is very important: don't leave it out! The mean of the sum is the sum of the means (expected value is always additive), the variance is the sum of the variances (variance is additive for independent random variables).


2

First observe that the required probability $<\ \approx 0.68^2=0.4624$ Then observe that you might as well assume $\sigma_1=\sigma_2=1$, then your question becomes what is the probability that a point lies inside the unit circle. Then proceed by converting to polars, when: $$ p=\int_{r=0}^1 f_R(r)\;dr=\int_{\theta=0}^{2\pi} ...


0

Well, you can just throw away some of the choices by using the $68-95-99.7$, a.k.a, 1-2-3 rule. So , by symmetry of the normal, the z-value must be negative. And, by the $1-2-3$ rule, $z(-1)$ corresponds with the percentile $50-34.3$..... Remember that the $1-2-3$ -$68-95-99.7$-rule for $1$ deviation of the mean, (around) 95% is within (plus/minus) $2$ ...


1

The highlighted "at least one credit card" corresponds to the "at least one credit card" on lines 1-2. And the $X$ you have been using is the number of persons, not number of cards.


0

Okay, so we know that $X\sim N(\mu,\sigma)$, looks like this: $$ Pr(X<x) = F_X(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(\frac{t-\mu}{\sigma})^2}dt $$ Next observe that $$ Pr(\frac{X-\mu}{\sigma}<x)=Pr(X<x\sigma+\mu)=\int_{-\infty}^{x\sigma+\mu}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-(\frac{t-\mu}{\sigma})^2}dt $$ We then let $y = ...


0

One straightforward argument is to calculate the moment generating function of $(X - \mu) / \sigma$, $$ \begin{align} M_{(X - \mu) / \sigma}(t) &= M_{X - \mu}(t / \sigma) \\ &= M_{X}(t / \sigma) M_{-\mu} (t / \sigma) \\ &= e^{t \mu / \sigma + t^2 / 2} e^{ - t \mu / \sigma} \\ &= e^{t^2 / 2} , \end{align} $$ which is the standard normal ...


3

The transformations in $\frac {X -\mu}{\sigma}$ are just translation and scaling, i.e., this is a change of coordinate from $X$ to $\frac {X-\mu}{\sigma}$ given by the function $X'=\phi(X)=\frac {X-\mu}{\sigma}$. More formally, you can use the Jacobian to study the effect of the change of variables in the density function: ...


2

Just write out the explicit formula for the gaussian (https://en.wikipedia.org/wiki/Normal_distribution) and then perform the transformation. The algebra shouldn't be difficult.


1

In your book $\Phi$ is the cumulative distribution of $Y \sim \mathcal{N}(0,1)$ And $X$ doesn't follow (approximatively) a $\mathcal{N}(0,1)$, but follow (approximatively) a $\mathcal{N}(5000,50^2)$ So $$P( 4900 \leq X \leq 5100 ) =P( -2 \leq \frac{X-5000}{50} \leq 2 ) $$ And now $\frac{X-5000}{50} \sim \mathcal{N}(0,1)$, so $$P( 4900 \leq X \leq ...


0

I get that $\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right) - 1 \approx \frac{3}{x^2} $ for large $x$. Since $N(x) =\int_{-\infty}^x n(t)dt $, $1-N(x) =\int_x^{\infty} n(t)dt $. Call this $M(x)$. The inequality becomes $\left(\frac{n(x)}{M(x)}-x \right)\left(\frac{2n(x)}{M(x)}-x\right)\geq 1 $. Multiplying by $M^2(x)$, I get $(n(x) - ...


2

I think this is the "continuity correction". You observe single units occurring, so you are in the middle of a value. Hence add or subtract 0.5 to actually switch to the next block and make the result more appropriate.


1

For a standard normal distribution, you should use $\Phi$ for the cumulative distribution function, and $\phi$ for the density. I also suspect you want $-\frac{x-\mu}{\sigma}$ trather than $\frac{-x-\mu}{\sigma}$. Making both changes will give you the answer in the book, or close enough. Since a normal distribution $N(\mu,\sigma^2)$ is symmetric about the ...


1

Since the normal distribution is symmetric, what the question is essentially asking is: Can you find a $z$-score such that the area between $0$ and $z$ is $0.6 / 2 = 0.3$? Since a $z$-score is for a standardized normal distribution, you'll have to then convert this back to unstandardized form by setting $z$ equal to one of the given inputs for the cdf; note ...


0

If $\mu=0.5$ and $\sigma=0.002$, $P(0.496<X<0.504)=P(\frac{0.496-0.5}{0.002}<Z<\frac{0.504-0.5}{0.002})=\Phi(2)-\Phi(-2)\approx 0.9545$, where $Z$ is the standard normal r.v. and $\Phi(\cdot)$ its cumulative density function. If $\mu=0.499$ and $\sigma=0.002$, ...


1

It is immediately obvious that there is insufficient information to obtain the desired probability: you have, for a given constant, three parameters that may vary: the common marginal mean $\mu$, the common marginal variance $\sigma^2$, and the correlation $\rho$; yet you have only one condition which is that when the constant is zero, $$\Pr[X < 0 \cap ...


1

In part (a) you calculated $P(T > 36 \mid Y^\complement),$ where $Y^\complement$ is the event that the computer does not have a hardware problem and $T$ is the time taken to assemble the computer. In part (b) you calculated $P(T > 36 \mid Y),$ where $Y$ is the event that the computer has a hardware problem and $T$ (as before) is the time taken to ...


0

To begin, I checked your computations in (a) and (b), using R statistical software. They are correct: 1 - pnorm(36, 30, 3) 0.02275013 # part (a) 1 - pnorm(36, 50, 7) 0.9772499 # part (b) If you know about the Empirical Rule you can get very close to the answer to (a) without tables or extra computation: The probability of a result more than ...



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