Search Results

Is X binomial distributed? The professor's answer was: Yes, because there is only right or wrong answers. My answer: No, because each question has a different "success-probability" p. … As I did understand a binomial distribution is just a series of Bernoulli-experiments, which each have a simple outcome (success or failure) with a given success-probability p (and all are "identical" …

B is the binomial distribution. … Note that starting with $E(X|X\geq1)$ does not give the proper answer, as knowing that a specific child is Italian violates the exchangeability assumed by the binomial distribution. …

Your professors answer assumes that all questions have same probability of "success" and are independent, since binomial is a distribution of a sum of $n$ i.i.d. Bernoulli trials. … The distribution could be approximated with binomial, or Poisson, but that's all. Otherwise you're making the i.i.d. assumption. …

There are $Binomial[2n,n]/(2n-1)$ ways for the couple to achieve parity for the first time with $n$ boys and $n$ girls. These are twice the Catalan numbers. … So the expected value would be $$\sum \frac{Binomial[2n,n]}{(2n-1)2^{2n}}2n $$ but the sum is divergent: the expected number of children is infinite. …

There are a few crucial distributions everyone must know: Normal, Binomial, Beta, Chi-Squared, F, Student's t, Multivariate Normal. …

To answer your first question we just need to use Bayes' Theorem to update our binomial likelihood with the beta prior. … Now, let $x_i\sim\text{Binomial}(N_i,\theta)$ and $\theta\sim\text{Beta}(\alpha,\beta)$. …

Then the number of heads has *binomial distribution with mean $(900)(1/2)=450$, and standard deviation $\sqrt{(900)(1/2)(1/2)}=15$. … In the normal approximation to the binomial, we get a better approximation to the probability that the binomial is $\ge 490$ by calculating the probability that the normal is $\ge 489.5$. …

It's unclear what "risk of extinction category" means, but it appears the question asks to compute the distribution of the sum of 15 independent binomial variates having the given expectations. …

. $$ As you see we do not need the hypothesis that the variables have a binomial distribution (except implicitly in the fact that the variance exists) in order to derive this estimator. …

.$$ This is a one-sided binomial proportion test. $H_a$ is the statement that, if it were true, would need to be strongly supported by the data we collected. …

$Pr(A=2)$ is a simple binomial calculation: $A\sim Binom(6, 1/3)$, so $Pr(A=2) = {6\choose 2}(1/3)^2(2/3)^4 = 80/243$. …

library(bbmle) #successes in first column, failures in second Y <- matrix(c(1,2,4,3,2,0),3,2) #predictor X <- c(0,1,2) #use glm fit.glm <- glm(Y ~ X,family=binomial (link=logit)) summary(fit.glm) # …

However, if (oblivious to a continuity correction) you take the key probability to be $P(X < 24),$ round excessively, and do a normal approximation to binomial using printed tables, you get $0.8264.$ …

Therefore, the number of heads thrown in 900 tries, $X$, has a ${\rm Binomial}(900,\frac{1}{2})$ distribution under the null hypothesis of a fair coin. … We know that the mass function for $ Y \sim {\rm Binomial}(n,p)$, is $$ P(Y = y) = \binom{n}{y} p^y (1-p)^{n-y} $$ I'll leave it to you to calculate $p$-value you seek. …

.$$ Consequently, the number of females in the group has a binomial distribution: $$\dot{X} \equiv \sum_{i} X_i \sim \text{Bin}(12, \tfrac{1}{2}),$$ and the conditional probability of interest is: $$\mathbb …