Qualification > Math
STATISTICS
candy:
Where did the answer by louis and mac for my question go??? ??? ???
u guys do we need to know this formula??
Median= lower class boundary +( 0.5n-c.f of class below/class frequency) x class width
is in the syllabus?? i didnt come accross any question in the pastpapers so far...but this is formula is stated in the book i have used!
do i need to know it????? :-[ :-[
sweetsour:
no...,i don't think so we need that formula in exams... ;)
but it is always good to have knowledge about it :D
mac:
hey u guys should have a look at: http://www.schoolworkout.co.uk/a_level.htm
its for another uk board but its helpful
I came across this example from some site (not the above 1). Isn't there something wrong in part (b)?? ???
--- Quote ---Example C3: In a club with 8 males and 11 female members, how many 5-member committees can be chosen that have
(a) 4 females?
Solution: Since order doesn’t matter, we will be using our combinations formula.
How many ways can the 4 females be chosen from the 11 females in the club?
C(11, 4) = 11!/[(11 – 4)! ?4!] = 11!/(7! ?4!) = (11?10?9?8)/ (4?3?2?1) = 330 ways.
But this gives us only the 4 female members for our 5-member committee. For a 5-member committee with 4 females, how many males must be chosen? Only 1. How many ways can the 1 male be chosen from the 8 males in the club?
C(8, 1) = 8!/[(8 – 1)! ?1!] = 8!/(7! ?1!) = 8 ways.
Thus we have 8?330 = 2640 ways to select a 5-member committee with exactly 4
female members from this club.
(b) At least 4 females?
Solution: In part (a) we found that there are 2640 ways to select exactly 4 females for our 5-member committee. If we need at least 4 females, we could have 4 females or 5 females, couldn’t we? How many ways can we select 5 females from the 11 females in the club, given that order doesn’t matter?
C(11, 5) = 11!/[(11 – 5)! ?5!] = 11!/(6! ?5!) = (11?10?9?8?7)/ (5?4?3?2?1) = 462 ways.
So to have at least 4 females, we could have C(11, 4) ?C(8, 1) or C(11, 5). Since there are 330 ways to achieve the first and 462 ways to achieve the second, we have
330 + 462 = 792 ways
to select a 5-member committee with at least 4 female members from the club.
--- End quote ---
ocean:
yea in this question they took 330 instead of 2640. the correct answer will be:
2640+462=3102
candy:
can someone please explain me the answer for this question..and why that method is used to solve! :'( :'(
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