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4

You want to calculate $$ 0.4 = 0.2 \cdot 0.08 + 0.8x \iff x = \frac{0.4-0.2 \cdot 0.08}{0.8} = 0.48. $$ The amount of points you need for the exam is then $0.48\cdot 400 = 192$. Good luck!


3

Add an $(n+1)$-th draw. Glue the ends of the interval together to form a circle, and cut the circle at the location of the $(n+1)$-th draw to form an interval again. The remaining $n$ draws are independently uniformly distributed on that interval. By symmetry, the $n+1$ segments between the $n+1$ draws all have the same expected length. Thus the expected ...


2

Hint 1: Note that the hypothesis can be characterized in the following form: $$H_0 : \theta = 1, \quad \text{vs.} \quad H_1 : \theta = 2,$$ where $$X \sim \operatorname{Beta}(\theta,1), \quad f_X(x) = \theta x^{\theta-1}, \quad x \in [0,1].$$ So what we have here is a simple (point) hypothesis for the value of the parameter $\theta$ for a test of a fixed ...


2

If $f(x)$ represents a legal probability density function (which you should check based on the support of the distribution), then for the median we find an $m$ such that $$\int_0^m f(x) dx = \frac{1}{2}$$ $$\iff \int_0^m (3(x+x^2))/14 dx = \frac{1}{2}$$ $$\iff \frac{3}{14}\int_0^m x+x^2 dx = \frac{1}{2}$$ $$\iff \int_0^m x+x^2 dx = \int_0^m x+x^2 dx = ...


2

Assuming that you scored $40\%$ of the homework correctly: If you scored $8\%$ in the homework worth $20\%$, then you will need to score $32\%$ in your $80$% final. $\frac{32\%}{80\%} \,\times 400=160 $ marks needed.


2

This is purely a matter of definitions: recall that $Var(X)=E(X^2)-E(X)^2$. Try to conclude from there based on what you know. Also they might have meant that variance is actually $b^2$, not $b$.


1

Hint: If you cut the circle along the first placed point, you can see that the situation is equivalent to taking the interval $[0,1]$ and placing $n-1$ points uniformly at random into the interval.


1

The probability for $j$ particular variables to be at least $\frac1n$ is $$ \left(1-\frac jn\right)^{n-1}\;, $$ so by inclusion-exclusion the probability for exactly $k$ variables to be at least $\frac1n$ is $$ \sum_{j=k}^n(-1)^{j-k}\binom nj\binom jk\left(1-\frac jn\right)^{n-1}\;. $$ I'm not aware of any way to simplify this.


1

$$F(x)=\int_{-\infty}^x f(t)\,dx=\int_1^x\frac2{t^3}dt+...\;,\;\;t\ge1$$


1

I think you meant standard deviation is b? Then Variance=$b^2$ Because $Var(X)=E[X^2]-(E[X])^2$, and $E[X]=\mu=a$ So $E[X^2]=a^2+Var(X)=a^2+b^2$


1

The language "you want to measure at some confindence intervel some hypothesis about the difference of their means" confuses confidence intervals and hypothesis testing. You probably want to say 'significance level' instead of 'confidence interval'. In the first method, your notation seems to conflate population and sample variances. If $Var(X)$ and ...


1

Both "Cov" and "Var" are used to represent the covariance matrix of the vector $\bf X$. See for example this Wikipedia remark: Nomenclatures differ. Some statisticians, following the probabilist William Feller, call the matrix $\Sigma$ the variance of the random vector X, because it is the natural generalization to higher dimensions of the 1-dimensional ...


1

As commented on the question, the assumption has to be made that Kyle is the same number of standard deviation units from the mean. Therefore my calculation is correct. If that assumption cannot be made then the probably is not possible to solve with the given information.


1

Since this is a discrete distribution, $Y=0,1,2$ If $Y=0$, then it means two heads are next to each other. So, $P(Y=0|X=2)=\frac{n-1}{{n\choose 2}}=\frac{2}{n-2}$ (since there are n-2 ways to choose the two consecutive heads, and $n\choose 2$ ways to randomly choose 2 positions for 2 heads among n heads) $Y=1$ is not possible, because you cannot have one ...


1

The proofs of simple versions of the central limit theorem (for instance, for a sample that's drawn iid from some distribution) use techniques involving characteristic functions or moment generating functions, that can be shown using undergraduate real analysis. These can be found in most books (for instance, the book Statistical Inference by Casella and ...


1

We have to show if $X\sim \mathcal N(\mu, \sigma^2)$ and $Z=\frac{X-\mu}{\sigma}\sim \mathcal N(0,1)$ then $P(X\leq w)=P(Z\leq \frac{w-\mu}{\sigma})$. $Z=\frac{X-\mu}{\sigma}\Rightarrow Z\cdot \sigma+\mu=X$ $P(X\leq w)=P(Z\cdot \sigma+\mu\leq w)=P(Z\cdot \sigma\leq w-\mu)=P(Z\leq \frac{w-\mu}{\sigma})$.


1

In linear regression with Gaussian (and heteroscedastic) noise, our model assumes that for $n$ observations of data, for each $i \in [n]$, $$Y_i = \beta X_i + \epsilon_i,$$ where $\epsilon_i$ is our ERROR term for the $i$th observation (note that residual $e_i$ is an estimator of $\epsilon_i$) Such that $\epsilon_i \sim N(0,\sigma^2_i).$ NID means ...


1

Every correlation matrix is a non-negative-definite (or "positive semidefinite") $n\times n$ matrix (for some $n$) in which every diagonal entry is $1$. Every $n\times n$ matrix that is non-negative-definite and in which every diagonal entry is $1$ is a correlation matrix. The first bullet point above is easy to prove by using the definition of ...


1

A correlation matrix must be not only symmetric but positive definite. The same goes for the inverse. Updated: As noted in the comments, a true correlation matrix is a normalized covariance matrix, with ones in the diagonal (put in other way, it does not include the variances, only the cross-covariances). So you should take that into account. I didn't ...


1

For the first question, you need to condition on whether the fruit is chosen from the red box, the green box, or the blue box and use the law of total probability. For the second question, you wish to find $P(\text{green} \mid\text{orange})$. Here Bayes' formula (and your answer from the first question) will be helpful. For the last question, you need to ...


1

With the well deserved criticism of the problem in my Comments, let's see how one would solve it if we ignore some of the unnecessary clauses in the problem and we make it work by changing one of the inputs. Let's say only 60 customers buy cake at Cafo (to make the numbers work with the other conditions), and the problem doesn't say stupid things like "each ...


1

So, I will explain a solution that does not introduce a binomial random variable. Concretely, there is only one road that will lead him to the mall and three other roads that will not lead him to the mall, but back to the starting position. The problem is asking: what is the probability $A$ that in his first attempt, he chooses one of these three incorrect ...


1

I realize this is a late answer to this post, but it still makes the top two to three results on Google for "ensemble average" and an answer has not yet been officially accepted. For posterity, I figured I would try to answer it to the best of my ability in the way that the question has been phrased. First, it is important to have a broad understanding of ...



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