# Tag Info

0

Actually the average IQ is 100 and its standard deviation is 15. Intelligence tests are scored in such a way the resulting IQ distribution conform to these properties. http://en.wikipedia.org/wiki/Intelligence_quotient

2

Since it is the "middle" $95\%$, meaning with with equal tails of $2.5\%$ on each side, the mean must be $\frac{60+140}{2}$. (Recall that the normal distribution is symmetric about the mean.) From tables of the standard normal, the point which has $2.5\%$ in the right tail is $1.96$ standard deviation units from the mean. So if $\sigma$ is the population ...

0

You can see that ties are not avoidable with a simpler example. Consider this election with three candidates and three votes: A>B>C (1 single vote) B>C>A (1 single vote) C>A>B (1 single vote) The result will be a cycle, the Smith set contains all candidates, and you will need to break the tie.

1

When the variance is unknown, you might want to estimate the variance $\sigma^2$ , or to estimate the standard deviation $\sigma$ and the square it. Sometimes the result is the same, sometimes it's not. For Maximum Likelihood it happens to be the same. But, anyway, it's good to state it clearly firsthand: which is the parameter I'm estimating? If it's ...

2

You need $$\frac{d}{d\sigma}\left(-n\log\sigma- \frac 1 2 \sigma^{-2} \sum_{i=1}^n x_i^2\right).$$ (I've written $x_i$ where you have $x_i-\mu$ since in this case it is known that $\mu=0$.) The derivative is $$\frac{-n}{\sigma} +\sigma^{-3} \sum_{i=1}^n x_i^2.$$ That differs from what you have. $$\frac{d}{d\sigma} \sigma^{-2} = -2\sigma^{-3}.$$

0

$$\sum (x_i-\bar{x})^2= \sum (x_i^2 -2x_i\bar{x} + \bar{x}^2) \\ =\sum x_i^2 -2\bar{x}\sum x_i + n\bar{x}^2= \sum x_i^2 -2n\bar{x}^2 + n\bar{x}^2=\sum x_i^2 -n\bar{x}^2$$ so$$\frac 1n\sum (x_i-\bar{x})^2=\frac 1n\sum x_i^2 -\bar{x}^2$$

0

Imagine that you work in a factory that makes thousands of cupcakes each day. The cupcakes are supposed to weigh 125g each and the distribution of the weight is supposed to be Normal with mean 125 and standard deviation 15. Now imagine that each day you randomly select 20 cupcakes from all of the cupcakes that were made that day. This is a sample of size ...

0

Gamma priors can also be used for the conjugate prior to the precision (inverse variance) as would be the case in a standard Bayesian regression model. In this case this prior would be considered vague where you have a reasonably number of parameters/observations that the variance parameters governs (more than 8 is usually enough for this prior to be vague). ...

0

It's correct. Note that you are using that the CDF of $Z$, $F_Z(z)=P(Z\leq z)$, is injective to conclude that $$F_Z(-1)=F_Z(-6/\sigma) \;\;\Longrightarrow \;\; -1=-6/\sigma.$$

0

As answered by Matt L., you face a nonlinear least square problem. This is not difficult but requires some reasonable estimates (you tell that you already have a rough idea of them). If you have a scatter plot of your data, you probably know already that $\beta$ is positive. Then an estimate of of $\alpha$ is the asymptotic value of the dependent ...

1

I would recommend finding the lat/lon differences in total or principal. Then with that you can get an average and variance. Then you can normalize and just use a standard z table. go here: http://stattrek.com/probability-distributions/chi-square.aspx and here: http://www.statlect.com/probability_exercises.htm

0

The $r$th event occurs before time $t$ if and only if the number of events before time $t$ is at least $r$. So $[X<t] = [Y(t)\ge r]$.

0

Hint: This pdf depends on two variables, so you should integrate with respect to both them to find the expected value (bivariate distribution) $$E[XY]=\int_{X}\int_{Y} xyf(x,y)\mathrm{dy\,dx}=\int_{-r}^{r}\int_{-\sqrt{r^2-x^2}}^{\sqrt{r^2-x^2}}\frac{xy}{\pi r^2} \mathrm{dy\,dx}$$ However, if you want to calculate the expectations $E[X]$ and $E[Y]$ of the ...

2

Hint: For $y\in\mathbb{R}$: $$P(Y\geq y)=(X_1\geq y,\ldots,X_n\geq y).$$ Without assuming anything about the joint distribution of $(X_1,\ldots,X_n)$ you can't say more. But if you're willing to assume that $X_1,\ldots,X_n$ are independent, then the probability above factors.

0

Assuming you have a random sample $X_1, X_2, \ldots X_n$ from a population with a (0,1) cauchy distribution, i.e. the $X_i$'s are i.i.d. with pdf $$f(x)=\frac{1}{\pi (1+x^2)}$$ for $x \in \mathbb R$. Then the likelihood function is $$\mathcal L(x;0,1)=\prod_{i=1}^{n}\frac{1}{\pi ... 0 This problem needs to be solved numerically. You can use a non-linear least-squares solver, because what you have is an overdetermined system of non-linear equations. There are 20 equations for 2 unknowns. The least-squares solver will minimize the quadratic error measure$$e=\sum_{i=0}^{19}[A_i-\alpha(1-e^{-\beta B_i})]^2where A_i and B_i are the ... 2 The first set denotes the set of all values for \theta for which the probability that X is at least Bin(n,\theta) is greater than \frac{\alpha}{2}. Similarly, the second set consists of all \theta for which the probability that X is at most Bin(n,\theta) is greater than \frac{\alpha}{2}. Since we have the intersection of these two sets, the ... 1 Your initial calculation shows there are 7^4-6^4=1105 combinations with largest number 6. This is not divisible by 7^3=343, not even by 7. The error is that combinations with more than one 6 get double counted in your approach. The logical value for K is 4, which selects which die is the 6. 4 \cdot 343=1372 But you are counting the ... 0 If you check the property of binomial distribution, you can find that the variance for a binomial distribution with mean np (in this case n = 4000, p=1/800, q=799/800) is npq=4000*(1/800)*(799/800) 0 Hint: Since the normal distribution is symmetric, you have P(W = X) = 1/2 and P(W = -X) = 1/2. But since the normal distribution is symmetric you also have P(X) = P(-X). 1 Any weighted average of the (X_i - \mu)^2's is an unbiased estimator. This is why you "should use \mu instead of \bar{x} and divide by n, if the true mean is known". Unfortunately, \mu is usually unknown. Of all the procedures that try to correct this problem by replacing \mu by a function of the X_i, the one that takes the replacement of ... 0 \begin{align*} \Pr\{X+Y\ne0\}&=\Pr\{X(1+Z)\ne0\}\\ &=\Pr(\{X\ne0\}\cap\{Z=1\})\\ &=\Pr\{X\ne0\}\Pr\{Z=1\}\\ &=p \end{align*} using the independence of X and Z and the continuity of the normal distribution. 1 @Stefanos Here is what I got. Did I go in the right direction? 02\sum(y_i-y_i^*)(y_i^*-\bar{y})= 2\sum[y_i(y_i^*-\bar{y})-y_i^*(y_i^*-\bar{y})]= 2\sum Ye_i - 2\bar{Y}\sum e_i= 0$$0 The density function for exponential distribution with mean 5 is: f(x)=1/5*e^{-5/x}. Then if you want to find the probability of receiving the call after waiting at least 7 minutes, you just integral the density function on the interval of [7,\infty]. lambda is just the inverse of your mean, in is case, 1/5. 0 Ordinarily, we say that the random variable X has exponential distribution with parameter \lambda if X has density function \lambda e^{-\lambda x} (for positive x). The mean of such a random variable X is equal to \frac{1}{\lambda}. It follows that if you are told that the mean is 5 minutes, then \frac{1}{\lambda}=5, and therefore ... 0 Your proposal to simulate is a good one. You can get a quick answer that way. There may be a clever way to get an analytic solution, and if so the users of this site will no doubt provide it. I suggest that in order to get some insight, you do two things. Do the simulation and study the distribution of the number of cards it takes to get the 5 of a kind. ... 1 The term$$\frac{\bar x}{nS_{xx}}$$is independent of i and can be taken out of the summation. Thus$$\sum_{i}\frac{\bar{x}(x_i-\bar{x})}{nS_{xx}}=\frac{\bar x}{nS_{xx}}\sum_{i}\left(x_i-\bar{x}\right)=$$where the latter term \sum_i \left(x_i-\bar{x}\right) is known to be equal to zero. But, for the sake of completeness$$\sum_i ...

0

You have $$\sum_{i=1}^n \frac{\bar{x}(x_i-\bar{x})}{nS_{xx}} =\frac{\bar{x}}{nS_{xx}}\sum_{i=1}^n x_i -\frac{\bar{x}}{nS_{xx}}\sum_{i=1}^n \bar{x} =\frac{\bar{x}^2}{S_{xx}} -\frac{\bar{x}^2}{S_{xx}}=0,$$ since $$\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$$ and $$\bar{x} = \frac{1}{n}\sum_{i=1}^n \bar{x}.$$

0

I'd probably not solve it how your professor wants you to.... First, it seems you effectively have $13$ cards. You have $13^5$ states of having not succeeded (having drawn zero through four of each card type), and $1$ state of having succeeded. That's a 371,294 state transition matrix. http://en.wikipedia.org/wiki/Absorbing_Markov_chain See the section ...

1

Note that choosing a card from a well-shuffuled stack of infinitely many cards is mathematically the same as drawing a card from a stack of $52$, recording the card you get, and replacing your card into the stack. That is, we note that the probability of drawing any given card is $1/52$, regardless of how many or which cards have already been chosen.

0

Since you are asking for a hint, this is the Coupon collector's problem.

0

For part b, the strings can be of the following lengths - $0,1,2,3,4,5$ or $6$. Our requirement is that the required string must contain exactly 3 a's. Therefore, we can rule out strings of length $0,1$ and $2$. For strings of length $3$, as you have already figured out, you can have only $\underline{1}$ string that satisfies the given constraints, that is ...

1

You might think about orthogonal projection of the response vector $$y=\begin{bmatrix} y_1 \\ \vdots \\ y_n \end{bmatrix}$$ onto the column space of the "design" matrix $$X = \begin{bmatrix} 1 & \cos\theta_1 & \sin\theta_1 \\ \vdots & \vdots & \vdots \\ 1 & \cos\theta_n & \sin\theta_n \end{bmatrix}.$$ The projection, which is ...

0

Hint: for the first part, how many choices are there for each of the remaining six positions? for part b, you use the same logic as for part a. So for six letters, you choose the three places for a's (how many ways), then choose the other three letters (how many ways?)

0

With the given information you know that The two data sets have the same central tendency as it is expressed by the mean The two data sets have different dispersion as it is expressed by the standard deviation. In the first data set, the observations are located more closely around the mean (50) compared to the second data set, where they are more ...

0

The first is more precise. Given the question, that's the only difference in the data. Basically, the data in the second set is less precise, or more spread out.

1

For $E_2$, two die can sum to $5$ with the rolls $(4,1),(1,4),(2,3)$ and $(3,2)$. This is $\frac{4}{36} = \frac{1}{9}$. Now imagine you have to make bet on the probability of $E_1$ occurring and you win the bet if $E_1$ does occur. If you know that $E_2$ occurs, there is a $\frac{1}{2}$ probability that $E_1$ will occur. Otherwise, the probability is less ...

1

According to Bayes' Theorem, $$P(A | B) = \frac{P(B|A) P(A)}{P(B)}$$ denote "has a headache" by A and "has a fever" by B then by a straight forward application of Bayes' we have $$P(A | B) = \frac{P(B|A) P(A)}{P(B)} = \frac {0.4 * 0.01}{0.02} = 0.2$$

0

Another method : The maximum resistance of the two R. in pll. is $(2+0.1)/2 = 1.05$ The minimum resistance of the two R. in pll. is $(2-0.1)/2 = 0.95$ So, the resistance is between 0.95 and 1.05 Comparison to the single resistor which resistance is between 0.9 an 1.1 shows that it is not better. Note : the method shown by L__ is more general and better ...

1

Parallel resistance is $$R_p = \frac{R_1 R_2}{R_1+R_2}$$ Uncertainty can be calculated by using differentials and a linear model. $$dR_p = \frac{\partial R_p}{\partial R_1}dR_1 + \frac{\partial R_p}{\partial R_2} dR_2= \frac{R_2^2}{(R_1+R_2)^2}dR_1+\frac{R_1^2}{(R_1+R_2)^2}dR_2$$ Hence, the total max ...

3

Hint: First step: You should recognize that the above functions are pdfs of normal random variables. Second step: Determine the parameters of the normal distributions. For the random variable $X$ they are $μ_Χ=1$ and $σ_Χ^2=2$. Third step: Use the fact that the sum (or difference) of independent normal random variables has again the normal distribution ...

0

Hint: $$A(tx+(1-t)y)+b=t[Ax+b]+(1-t)[Ay+b],$$

1

If you knew the mean value of your distribution, the variance should be divided by the number of samples $n$. On the other hand if you extract the mean value from your data, you are fixing a relation on your $n$ samples (their sum is $n\bar X$) so you are left with the equivalent of $n-1$ samples.

0

Hint: for i, you need the lifetime to be more than one standard deviation high. for ii, it needs to be more than 2 standard deviations low.

1

For your first question, suppose $X$ and $Y$ are independent random variables. The statement is that for any Borel measurable functions $f$ and $g$, $f(X)$ and $g(Y)$ are independent. In fact, independence of $X$ and $Y$ is equivalent to (and in some formulations is defined as) the events $X \in A$ and $Y \in B$ being independent for all Borel subsets $A$ ...

0

I'm not sure I understand your questions, but let's see if this can answer part of it. $$\begin{bmatrix} X_1 \\ \vdots \\ X_n \end{bmatrix} = \begin{bmatrix} \bar X \\ \vdots \\ \bar X \end{bmatrix} + \begin{bmatrix} X_1-\bar X \\ \vdots \\ X_n-\bar X \end{bmatrix}$$ Suppose $X_1,\ldots,X_n \sim \mathrm{i.i.d.}\ N(\mu,\sigma^2)$, Then $\bar X$ is ...

2

This question is a Bernoulli trial (more specifically a binomial distribution)(an action is performed and only one of the two possible outcomes is happens at once). Bernoulli's trials are a foundation and will be heavily used later in your class and I highly recommend you attempt this question by yourself. Here are a few tips to get you started: (1) Define ...

0

For roll $i$, let $X_i=1$ if roll $i$ of the die is 1 and 0 otherwise. Hence, $$R_1 = \sum\limits_{i=1}^R X_i$$ The dice cannot roll 6 in the first $R$ rolls, so $X_i=1$ with probability 1/5 and 0 with probability 4/5. Hence, $$E[R_1|R] = \sum\limits_{i=1}^R E[X_i|R] = \frac{1}{5}R$$ $$Var[R_1|R] = \sum\limits_{i=1}^R Var[X_i|R] = \frac{1}{5}\frac{4}{5}R$$ ...

-2

look at the table of binomial probabilities. N=15 P(710)= .003+.014+.043+.103=.163

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