mightymiff

There seems to be a consensus here about this question (at least in this [ancient] thread), but--and unless I'm delirious due to sleep deprivation--it appears as though the credited response for this analogy is NOT disgrace. It's a tricky question... Revere is to treat with awe and devotion (honor and devotion being distinct vows). Disgrace means to be dishonored, shamed, and disfavored. This is base and worldly. I want something holier... Profane retains the spiritual sense of revere in that something that commands reverence is treated irreverently. REVERE:PROFANE Yeah?

Oh, gee DW, I actually find your post extremely helpful. I'm so happy you took the time to explain this to me in the clearest possible terms. Thank you! Humbly yours, mm

13/52 is not equal to 1/4? Methinks it is...right? In any case, thanks for helping out.

A standard formula, I'm not sure. z is years. total Y after z years=Y - .1Yz // (figure Y=10) 10 - .1 * 10 * 6 = 10 - 6 = 4 total X after z years=X - .1Xz // (X=5) 5 - .1 * 5 * 3 = 5 - 1.5 = 3.5 Almost easier for me to go 10, 9, 8, 7, 6, 5, 4 and 5, 4.5, 4, 3.5. Y is bigger than X. ooo, it is easier using V and 2V like BLUEKNIFE does. Thanks BK, that's pretty sweet.

Nobody can help drill this into my head more clearly? pretty please....

We've got to do some calculations, right? I mean, if you can't figure out how to do it without calculations, maybe we had better calculate? Average is not too hard. Sum of negatives is -23, positives is 48....average is thus 25/6=4 and 1/6. Standard deviation, as I understand it, is average distance from mean to any given number in the set. Keeping 25/6 in mind and looking through the set of numbers, it becomes clear in a hurry that the standard deviation is going to be quite a lot more than the mean. For every single number in the set, its distance to the mean exceeds the mean itself. A little under 8 to get from mean to 12, a little over 6 to get to -2 (from the mean)....and it just goes up from there. As for median, numbers are in order, middle two are -2 and 12, add 'em up, divide by 2. 10/2=5 ii) and iii) look false. i) looks true.

Thanks for the response. I'm getting closer to comprehension, but it's still not there. So for mutually exclusive events P(A and B)=P(A)*P(B|A)? I'm not sure how to apply this formula to the problem at hand (if it does at all), and I'm not sure how to calculate P(B|A)... Also, I'm still not clear on the difference between independent and not (in this example). Aces are pulled independent of clubs, but they overlap at the ace of clubs, right? Following this logic, I see (likely incorrectly, to be sure) pulling females as independent of pulling sophomores, with an overlap at female sophomores. Aces are of four kinds. High school females are of four grade levels. Aces can be clubs. Females can be sophomores. Aces can be other than clubs. Females can be other than sophomores...what am I missing? 40 cards are in a deck, 22 are clubs, 16 are aces, 10 are the ace of clubs. What's the probability I pull a card that is both a club and an ace? For the original card example (with a standard deck of cards), is P(A and B) potentially different than P(A given B)? I think that if someone is able to explain my question with formulas and or numbers, I might have a better chance of seeing what is going on. Thanks a bunch.

Not a math guy here, but this is bugging me... Given 40 students, 22 are female, 16 are sophomores, 10 sophomores are female. 10/40=1/4 students are sophomore and female. What's bugging me is I'm unclear of the definition of P(A and B) Doesn't P(A and B)=P(A)*P(B)? Elsewhere I've seen P(A and B)=P(A)*P(A|B), but I don't know what this means. Given the example above: P(sophomore and female)=16/40 * 22/40, which equals 11/50, which is incorrect. It seems to work with cards... 52 cards, 13 are clubs, 4 are aces, 1 ace of clubs. 1/52 is ace of clubs. P(ace and club)=4/52 * 13/52=1/52, which is correct. Obviously I'm missing something, I'd be glad if you could help me out. Thank you kindly.