Recent content by ToxicBug

Hello, I'm trying to figure out how to program the Exponential Integral function, and then make it compute values for a vector of data. I have this: expint <- function(s){ integrand <- function(x) {exp(x)/x} integrate(integrand, lower = -Inf, upper = s)$integral } if I...

Hello, I've been breaking my head over this for a couple hours now. I'm trying to find the expected value of a Gumbel distribution. Integrate[ y*E^(-((y - a)/b))*E^(-E^(-((y - a)/b)))/b, {y, -Infinity, Infinity}, Assumptions -> b > 0] If I set a = 2 and b = 5, it works. Otherwise it gives...

The normal distribution table may give you numbers of P(Z < z), -inf to z. I think that you need to find P(.195 < Z < .4051), that would be P(Z < .4051) - P(Z < .195).

Look up the binomial distribution on wikipedia. You need to use its probability function to calculate the probabilities P(X = x). the answers are: << answers deleted by berkeman -- giving answers to homework problems are not permitted on the PF >>

I think that its like this: a) P(X = 0) + P(X = 4), X ~ Binomial(4, .5) b) This is an ordered sequence, so since p = .5 then it is just .5^7 c) P(X = 4), X ~ Binomial(7, .5) d) your question is incomplete. However, if the probability that A wins is p, then it is (P(X = 4), (X ~ Bin(4, p))) +...

I think that for an array of n numbers, there are n! possible permutations... so I guess that for your example the prob would be 1/(5!).

I don't think that you need a continuity correction here, unless the problem specifies that the machine can only dispense discrete amounts. (209 - 200)/15 = .6 P(Z < .6) = .7257 .7257 - (1-.7257) = .4514

I don't understand the question

I guess that you can represent it by a Poisson since N is large. I think that a Binomial( 1.859E-7, 19million) would describe X more accurately. If it were a binomial, then the expected value would be np = 19mil*1.859e-7. Your Poisson needs to have the same expected value, and since Poisson has...

P(A or B but not both) = P(A or B) - P(A and B) P(A or B) = P(A) + P(B) - P(A and B) P(A or B) - P(A and B) = P(A) + P(B) - 2P(A and B) I think you can continue from here.

But doesn't the question as for distinct birthdays, ie Xi would be 1 if there is only one birthday on day i and 0 if there is not only one birthday on day i?

Yeah, you're right, I didn't understand the question and this morning I thought about exactly the same thing that you're saying :)

I've answered your other topic.

Let the r.v. X be the amount of successes from n trials of an experiment that is a success with probability p. Then X ~ Binomial(n,p). Look up the binomial distribution on wikipedia, for the pgf, and the moments. You need to find the "expected value" of this distribution, E(X), the first moment.

I think that, since its distinct birthdays (from what I understand, distinct bdays are only the ones that don't coincide on the same day), you will have nCk bdays total, thus the expected number on any given day would be (nCk)/n since they're uniformly distributed. There will also be nC(n-k)...