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seen Sep 13 '13 at 23:49

Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
i've been told that capital X and Y are random variables, I never knew you can obtain the actual value of X and Y by summing $X_k$ and $Y_k$. This summation method to obtain X and Y would not work if X and Y are, say, binomially distributed correct?
Nov
21
comment I don't understand why $X = \sum_{i=1}^n X_i$ and $Y = \sum_{i=1}^n Y_i$
i've been told that capital X and Y are random variables, I never knew you can obtain the actual value of X and Y by summing $X_k$ and $Y_k$. This summation method to obtain X and Y would not work if X and Y are, say, binomially distributed correct?
Nov
16
comment Does SSTR (sum of squares for treatments) = SSR (regression sum of squares)?
hold on a sec, your statement $SSR=\sum\nolimits_{i=1}^{N}(\hat{Y_{i}}-\bar{Y})^{2}=\sum\nolimits_{j=1}^{q}n_{‌​j}(\bar{Y}_{j\cdot}-\bar{Y})^{2}=SST$. means $SSR = SST$??!!!
Oct
25
comment What is the $P( |X-10| > 2)$ of a normal distribution when mean is 10, and standard deviation is 6?
for the second term are you sure it's $P(X<8)$ not $P(2<X<8)$?
Oct
17
comment Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$
is that correct?
Oct
17
comment Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$
$P(X=0) = {20 \choose 0}(0.5)^0(0.5)^{20}, P(X=1)={20 \choose 1}(0.5)(0.5)^{19}, P(X=2)={20 \choose 2}(0.5)^2(0.5)^{18}$ etc etc?
Oct
17
comment Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$
i think you've posted how $np(1-p)$ is derived, but can you show me how you'd substitute the numbers from this particular question (20 games 0.5 chance of winning). you don't have to write all 20, just write the first few then use "dot dot dot" ...
Oct
17
comment Calculating Variance of a binomial distribution using the standard formula $E(X^2) - \mu^2$
@Max you are correct. I made the correction. thank you
Oct
11
comment By convention $P[X=x] = 0$ for all x. How would you explain pdf $f(x) =3x^2$ (where x is between 0 and 1) when x =0.9
@AndréNicolas i understand that part. but do you know what does $f(0.9)$ yield?
Oct
11
comment By convention $P[X=x] = 0$ for all x. How would you explain pdf $f(x) =3x^2$ (where x is between 0 and 1) when x =0.9
@chris ok, i see. then what do i get when I plug in 0.9 for x?
Oct
11
comment By convention $P[X=x] = 0$ for all x. How would you explain pdf $f(x) =3x^2$ (where x is between 0 and 1) when x =0.9
ok, when $x =0.9$ then $f(0.9)= 3(0.9)^2 which \neq 0$. That's what I'm trying to ask
Oct
10
comment How can gender and class classification be dependent?
Ross, thank you for your answer! I upvoted everyone who contributed, honestly I still don't quite get it. but it's pretty late tonight. I'll think more on each answer when i wake up tomorrow morning.
Oct
10
comment How can gender and class classification be dependent?
oh no =( this is so convoluted. I started a post before on how I can identify a probability problem as a conditional probability problem(math.stackexchange.com/questions/205103/…). it seems like you're suggesting they're the same. this is still confusing =(
Oct
10
comment How can gender and class classification be dependent?
why would you think of this in terms of conditional probability? I'm thinking it's simple union and interects
Oct
9
comment How can gender and class classification be dependent?
If, however, the first die landed on 6, we would be unhappy because we would no longer have a chance of getting a total of 6. In other words, our chance of getting a total of 6 depends on the outcome of the first die; thus, E1 and F cannot be independent. So is there an explanation like this for class and gender?
Oct
9
comment How can gender and class classification be dependent?
@QiaochuYuan Suppose that we toss 2 fair dice. Let E1 denote the event that the sum of the dice is 6 and F denote the event that the first die equals 4. Hence, E1 and F are not independent. Realistically, the reason for this is clear because if we are interested in the possibility of throwing a 6 (with 2 dice), we shall be quite happy if the first die lands on 4 (or, indeed, on any of the numbers 1, 2, 3, 4, and 5), for then we shall still have a possibility of getting a total of 6.
Oct
7
comment Probability given a mean and standard deviation of a random variable
@DilipSarwate had the distribution been normal, i'd have been right?
Oct
7
comment Probability given a mean and standard deviation of a random variable
@DilipSarwate sorry, i take that back. I think i had assumed the distribution to be normal...
Oct
7
comment Why can we use poisson distribution in this case? the n is very small!!
are you sure $\begin{equation} \frac{n(n-1)\ldots(n-k+1)}{n^n} \approx 1 \end{equation}$? because (by looking at how it's derived) the denominator is $k!$
Oct
6
comment Is it possible to do this Poisson problem in Binomial?
ive added the second part of the problem, can you help convert that into binomial? Thank you very much!!