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The sum of i.i.d. uniform random variables follows the Irwin–Hall distribution.


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Because the posterior (or, more precisely, the posterior mean that you state here) is formed by weighting over the contribution from the prior and the observations. Hence, if $n$ increases, the contribution of any particular "head" to the total estimate decreases. When you move from $n=1$ to $n=2$ the first observation still is pretty important to form the ...


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Well the probability that you are asking for is not computable, at least not using frequentist statistics. Observe that in general, a conditional probability is computed as $$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$$ But here, $P(B)=$ I flipped the coin n times seeing x occurrences of heads. And for this probability we would need to know the value of $p$. You ...


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Yes, basically you've just changed the objects: modem $\rightarrow$ doctor customer $\rightarrow$ patient And, since $n,c,p$ are the same, your probability is $\approx 0.04$ as well.


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I see a couple of problems in your approach. For one, it seems you are using techniques typically seen in t-tests (pooled estimates of variance, using pt, etc). Here is the solution to the problem. Our hypothesis is that the fault probability in the two factories is the same. Written mathematically $$ H0: \pi_1-\pi_2=0$$ $$HA: \pi_1 - \pi_2 \neq 0 $$ ...


1

$$ P(2k, n-2k| \Theta) = \frac{2^{n-2k} \binom{\Theta-k}{n-2k} \binom{\Theta}{k}}{\binom{2\Theta}{n}}.$$ Denominator - Unconstrained no. ways to choose socks This is given exactly by the binomial coefficient to choose $n$ objects from $2\Theta$: $$\binom{2\Theta}{n}.$$ Numerator - No. ways to choose $k$ pairs This is again a binomial coefficient: we ...


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A "significant" intercept is one whose estimated value is "significantly" different from 0. In a logistic regression, that means different from equal outcome group probabilities when the predictors are at reference levels (categorical) or at 0 (continuous). So just centering a continuous predictor or changing the reference level of a categorical predictor ...


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As @gunes (+1) has said: Taken exactly as stated, you have the same question either way. $X \sim \mathsf{Binom}(n=100, p=0.1)$ is the probability a modem/doctor will be busy at a given time. And you seek $P(X > 15) = 1 - P(X \le 15) = 0.0399.$ In R, where pbinom is a binomial CDF, the computation is shown below: 1 - pbinom(15,100,.1) [1] 0.03989053 ...


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The issue here is scaling by the standard deviation. The tail points are not any further out in absolute terms when $p$ is extreme, but they are further out as a multiple of the standard deviation. The standard deviation is $\sqrt{p(1-p)}$, so being nearly 1 unit away is further away in standard deviations as $p$ gets extreme. The scaling for the ...


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Here is an intuitive explanation that works for me: $Binomial(n, p)$: When repeating a Bernoulli trial with $p$ probability $n$ times. The chance of exactly $k$ successes is: $$Binomial_\mathit{pmf}(\pmb{k}, n, p) = {n\choose \pmb{k}} p^{\pmb{k}} (1-p)^{n-\pmb{k}}$$ $Beta(n, k)^*$: For a fixed $n$ and $k$, given probability $p$, calculate the probability,...


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Binomial PDF and CDF in R. First, let's look at pbinom and dbinom in R. If $Y \sim \mathsf{Binom}(n = 5, p = 1/2),$ then $P(Y = 2) = {5\choose 2}(1/2)^5 = 0.3125.$ In R you can compute the formula for yourself or use the binomial PDF function dbinom. choose(5, 2)/2^5 [1] 0.3125 dbinom(2, 5, 1/2) [1] 0.3125 Now, if you want $P(Y \le 2) = P(Y=0) + P(Y=1) + ...


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Regarding how to model your data, I guess you may divide your map into a grid (i.e. you create spatial point of absences) and have indicator of presence. From that way, you may have only one model. Maybe there is an additional challenge in that altitude may matter a lot here and it might be difficult to create a grid. You will be interested in a fixed-...


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Yes, binomial is the correct interpretation here. Note that, without a calculator, it could be easier to calculate $\sum_{k=6}^8P(X=k)$ directly (Let $X$ be the number of pairs that treated kangaroo is faster). Note also that, due to the symmetry of the problem, this is also equal to the case where untreated kangoroo is faster in more than 5 trials.


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