# Tag Info

1

Given a CDF $F_X$, $P(a< X \leq b)=F(b)-F(a)$. You'll need a table or a calculator or something. Trying to integrate the PDF of the standard normal distribution will only bring you sadness.

0

Hint: what is the chance that the ticket loses the first draw? To not win a prize, the ticket needs to lose $365$ draws in a row. The chance of winning is one minus this. An approximate chance of winning is to compute the chance of winning on the first draw, then multiply by $365$. This gives too large a value (try it!) because this is the expected ...

2

Let the parameters of $X$ and $Y$ be $\lambda$ and $\mu$. The sum of independent Poisson with parameters $\lambda$ and $\mu$ is Poisson with parameter $\lambda+\mu$. By independence, the variance of $X+Y$ is the sum of the individual variances, so it is $5$. The variance of a Poisson is equal to the parameter, so $X+Y$ is Poisson with parameter $5$. Now ...

1

In the last line you seem to be writing $\hat{\boldsymbol{u}} \hat{\boldsymbol{u}}^\top = \mathbb{I}_D$, but this won't hold if the dimension of your points fulfils $D \ge 2$. You will have $\hat {\boldsymbol{u}}^\top \hat {\boldsymbol{u}} = \|\hat {\boldsymbol{u}}\|^2 = 1$, but not the other way round.

0

Don't forget that normal distributions are symmetrical! P[Y<= 64+a] = P[Y>= 64-a]. Also, you know that Pr[X<=1.65] = .95 if mean = 0 and std_dev = 1, and that Pr[X <= mean + constant*std_dev] is constant. The rest should be easy to figure out.

0

Use the CDF technique for $Y=1-X$. $$P(Y<y)=P(1-X<y)=P(X>1-y)= \int_{1-y}^1 \frac{\Gamma(\alpha +\beta)}{\Gamma(\alpha)\times \Gamma(\beta)} x^{a-1} (1-x)^{\beta -1} dx$$ Differentiate wrt $y$ using Leibniz's formula.

0

This is not negative. $$\Phi(1) + \Phi(0.5)-1=0.5318$$

0

The pmf of the Poission distribution is given by $p(x)=\frac{\mu^k}{k!}exp(-\mu)$ where $k \in \mathbb{N}_0$ and $\mu>0$. To answer a) just plug in the numbers. To answer b)-d) you need to realize that for a stochastic variable $X$ and some $a \in \mathbb{R}$ $1=P(X<a)+P(x\geq a) \Leftrightarrow 1-P(x<a)=P(x\geq a)$

1

I would not agree that this is correct. In particular, if $X$ ~ $N(\mu, \sigma^2)$ with pdf $f(x)$, and cdf $F(x)$, then the effect of the quality control mechanism is to doubly truncate the distribution, so that post-quality control, the distribution of nuts is say $g(x)$: $$g(x) = \frac{f(x)}{F(b)-F(a)}$$ where $b=445$ is the upper bound, and $a=420$ is ...

0

Part (a) is simply asking for the area under the Gaussian curve to the right of $z = \frac{163 - 143}{29} \approx 0.689655$. For parts (b) and (c), recall that if $n$ independent random variables each have a standard deviation of $\sigma$, their mean has a standard deviation of $\frac{\sigma}{\sqrt{n}}$.

0

I will assume that your discrete random variable takes integer values. I find it difficult to remember a bunch of rules, so I remember only one: That if $k$ is an integer, and we are approximating the discrete $X$ by a continuous $Y$, then $\Pr(X\le k)$ is often better approximated by $\Pr(Y\le k+0.5)$. Now $\Pr(2\le X\lt 9)=\Pr(2\le X\le 8)=\Pr(X\le ... 1 You have a standard deviation$\sqrt{T(1-T)}$for a sample of size$1$. If$0<T<1$then$\sqrt{T(1-T)}\le1/2$. But if it is known that$0.7<T$, then the standard deviation is$\le\sqrt{(0.7)(1-0.7)}\cong0.46$. The standard error for a confidence interval based on a sample of size$n$is therefore$\le0.46/\sqrt{n}$. The length of the interval is ... 1 You need to find the$5$percentile by first standardizing your variable. A standard normal table should help,http://www.mathsisfun.com/data/standard-normal-distribution-table.html , and see that the 5th percentile and 90th percentile are given by$z=1.28$and$z=-2.6$respectfully. Notice that this is the inverse of the case where we're given a$z$value ... 3 If cars arrive at a rate of 4 every minutes than an average of 8 arrives every two minutes. For the formula I use,$ p(y) = \displaystyle \frac{\lambda^y}{y!}e^{-\lambda} $where$\lambda$is the average amount over the interval. So if want the probability of eight cars passing then:$ p(8) = \displaystyle \frac{8^8}{8!} e^{-8} = .1396 $0 If the rate is 4 per minute, it is also 8 per two minutes. Therefore, you are looking at$P(X=8)$, where$X$is a Poisson random variable with mean 8. 0 The probability of rolling a certain number, can be easily calculated by listing the options available, or listing the probability of not rolling that specific number and subtracting the answer from 1. The probability of a certain event occurring is always a number between 0 and 1. 3 That's right. The easier approach would be to calculate the chance of not rolling a$6$- that's just$\frac56$for the first die, and$\frac56$for the second die, so by the product rule (as the events are independent), the probability is$\frac56 \cdot \frac56 = \frac{25}{36}$. Then the probability of rolling a$6$is$1$minus the probability of not ... 2 Yes indeed, you've got them all. So counting them, we get$11$of the possible$36$outcomes of which at least one$6$is rolled. Now simply express this probability as a fraction! 1 The probability of an event is the number of "successful outcomes" over the number of possible outcomes. Since you have correctly stated that there are$6^4$potential outcomes, let's try to count the number of outcomes that are "successful", i.e. contain two 1's and two 3's. If we roll four dice, the "successful" outcomes are$3,3,1,13,1,3,1$... 0 Hint: You are correct that the total outcomes are$1296$. The number of successful outcomes is the number of ways of listing two$1$'s and two$3$'s in a row. You can do that by hand, or find the number of ways to choose which two of the four positions have$1$'s in them. 2 The maximum entropy distribution is of the form$f(x) = \exp( \sum_k \lambda_k g_k(x))/Z$where$g_k(x)$are imposed by the restrictions (like Lagrange multipliers) and$Z$is the normalization factor. In our case, we have two restrictions (apart from the trivial one), which give the two functions$g_1(x)=x$(mean) and$g_2(x)=x^2$(second moment, or ... 0$1-H=(1-F)(1-G)$. Now show that$H$is increasing, and has limits 0 and 1 as required. 0 You can indeed compute the new mean easily as explained in the answer of Arkamis. However you cannot compute the new median: an assumption is missing for that. ex (defining the median as the average of the two median values for even numbers of values): you can take for group 1: 8 times '50', 60,70,90,9 times '120' group 2: 13 times '60',66,70,74, 14 ... 1 From a numerical point of view, your problem looks quite iteresting and I played with it, my concern being to get a quick estimate of the parameter "w" which has to be adjusted to your data. What I did was to start a Newton procedure at w=0. As a result, what I obtained is that a rough estimate of "w" can be obtained using the following formula for the ... 0 Obviously this is false!$F(x)$and$G(x)$are cdf's implies that $$\lim_{x\rightarrow\infty}F(x)=1,~~~~\lim_{x\rightarrow\infty}G(x)=1$$ Therefore$\lim_{x\rightarrow\infty}H(x)=2$, which itself implies that$H$is not a cdf. However$H(x)=\frac{F(x)+G(x)}{2}$is a cdf. This is because it is right-continuous and non-decreasing as$F$and$G$are ... 4 I'm posting two answers because one (this one) is very general, and the other only works nicely when the weights are positive. Rewrite the desired inequality through cross multiplication as $$n \sum_{i=1}^nw_ix_i < \sum_{i,\ j=1}^nw_ix_j.$$ On the right, every$w_i$is paired with each$x_j$exactly once. If we allow the indices to cycle for ... 4 Ok. I'm posting two answers because one is very general, and the other (this one) only works nicely when the weights are positive - but I wrote this one first because I thought of it first. Assume the weights are non-negative. Notice that if I multiply all the weights by a fixed positive number$k$, then $$\frac{\sum_{i=1}^nkw_ix_i}{\sum_{i=1}^nkw_i} = ... 3 I have studied both analysis and statistics to some extent. The short answer to your question is that a good understanding of an undergraduate book on real analysis (titles such as 'An Introduction To Real Analysis') should be enough. Attempting the exercises, understanding the answers and being able to talk about nearly everything in the (chosen) book with ... 2 Two events are mutually exclusive if P(A\cap B)=0. This means that there is no way for these two events to occur simultaneously. Two events are independent if P(A)\cdot P(B)=P(A\cap B), but P(A\cap B) isn't necessarily zero. In your example let E be the event of an emergency room claim and O be the event of an operating room claim. We are told ... 0 The number p is fixed. The union is "smallest" if A=B=C, in which case the probability of the union is p. To make the probability of the union "big," separate A,B,C as much as possible. As you saw, the answer is sort of 3p. More precisely, it is 3p if 3p\le 1, that is, p\le 1/3, and 1 if p\gt 1/3. "Formulas" are not of great help in ... 0 You could also use that E[(X-Y)^2]=E[X^2-2XY+Y^2]=E[X^2]+2E[XY]+E[Y^2] 1 Fifty two cards are labeled with the numbers 1 through 52 . Note that there are 13 spades amongst the cards. For the first spade, We may view these as separating the remaining 39 cards into 14 groups of non-spades - those appearing before the ﬁrst spade, between the ﬁrst and second, etc. Each of these groups is equally likely to appear ﬁrst, so 39/14 ... 0 Integrate (x-y)^2f(x,y) over the plane. 0 Your confusion was probably the probability density function:$$ f(y)=\left\{ \begin{array}{lr} \frac{1}{\beta}e^{-\frac{y}{\beta}} & y >0\\ 0 & \text{otherwise} \end{array} \right. $$If you integrate it, you get cumulative distribution function F(y)=\int_{0}^{y}f(t)dt=1-e^{-\frac{y}{\beta}}, hence what you wrote as the definition of your pdf ... 0 The proposed strategy is far too complicated. Let D be the region above the curve y=x^2. You want$$\iint_D f(x,y)\,dy\,dx.$$This in a sense finishes things if we do not know f. You can express the integral, if you wish, as an iterated integral, y goes from x^2 to \infty, x then goes from -\infty to \infty. 1 Label the non-spades 1 to 39. Let X_i=1 if non-spade with label i is drawn before the fifth spade, and 0 otherwise. Then the total number Y of cards drawn up to an including the fifth spade is given by$$Y=5+X_1+X_2+\cdots+X_{39}.$$I think you know how to find E(X_i). The number of relevant cards is 14. 0 The 95 percent confidence interval for a percentage is classically obtained by adding and subtracting to the proportion p an interval computed using the formula 1.96 Sqrt[(p(1-p)/N], where N is the total number of observations, and 1.96 is the standard number of SDs extending from the mean of a normal distribution and encompassing 95% of the population. ... 1 Probabilities are between 0 and 1, so you should be able to spot that 7.5 is certainly wrong. Try dividing by the total area of your rectangle. 1 The main idea is notice that:\Pr[X=Y]=\Pr[\bigcup_{i} (X=i \land Y=i)] =\sum_{i} \Pr[X=i \land Y=i] and use the definition of independence. 1 The central limit theorem applies only to strongly stationary processes, where all of the \rho_i's are equal to the variance. In this situation, the one formula reduces to the other. 1 First, describe your system as matrix multiplication$$A{\bf x} = {\bf b}$$where A is a 300\times 7 matrix, {\bf x} is your 7\times 1 vector of variables, and {\bf b} is a 300\times 1 vector. Let A^T be the transpose of A. To find a least squares solution to A{\bf x}={\bf b} you should solve the system$$A^TA{\bf x} = A^T {\bf b}.$$2 Clusters are essentially partitions of your sample such that each cluster has a mean similar to the entire population, then you randomly sample from a radomly chosen subset of the clusters. Thid double randomization allows you to get a represntative sample with fewer samples. See this link Stratified sampling is done when there is a heterogenous population ... 1 If your pre-condition interval is a fixed constant x, the process, once triggered, lasts x+1, and then there is a post-condition period of x+2 seconds long, then the probability of two processes overlapping can be calculated from the start times of two sequential instances of the process (i.e., T_1, T_2) using conditional probability: P_{(T_2\space ... 0 Just figured out! Factorize it in another way! { \left( \frac { 1 }{ \theta } \right) }^{ n }{ I }_{ \left[ { Y }_{ 1 },\theta \right] }\left( { y }_{ n } \right)\prod _{ i=1 }^{ n }{ {I}_{\left[0,\infty\right)} }({x}_{i}) 1 Means follow the rule of linearity . E[X+Y] = E[X] + E[Y] It doesn't matter whether the events are independent or not . So for any linear combination of random variables you can take the mean of the individual random variables and then combine them . You are asking for the intuition for X+Y . These are two events whose outcome can be defined by ... 0 The definition of the "standard deviation of a set of values" is not at all what you write. A definition would be like "Let A be a set of cardinality n containing as elements real numbers, denoted x_1,...,x_n, ordered or unordered. Define the magnitude "Sample mean", denoted \bar x, by$$\bar x = \frac 1n\sum_{i=1}^nx_i$$Then the "standard ... 1 Let v_i denote the ith value. For the 20 values, the mean is given by$$85 = \mu = \frac{v_1+v_2+v_3+\cdots+v_{20}}{20}.$$You don't know the individual amount for each value, but you can easily figure out what the sum of the values is:$$20\times 85 = v_1+v_2+\cdots + v_{20}.$$Now, do the same thing for the 30 values,$u_1$to$u_{30}\$. Add those ...

1

The obvious alternative to the standard deviation as a measure of dispersion is the mean absolute deviation. In the 18th century Abraham de Moivre wrote a book called The Doctrine of Chances. ("The doctrine of chances" is 18th-century English for the theory of probability. De Moivre wrote in English because he had fled to England to escape the persecution ...

1

This involves a random variable...that is, a real-valued function defined on the space of outcomes. In this case, a person plays five times, a certain outcome happens, and a real number (the casino's profit) emerges. A random variable has a mass function, which gives the probability that the random variable will take a certain value. From this, we can ...

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