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 

[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)??

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?/ (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.

[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! 

[ocean]

can someone please explain me the answer for this question..and why that method is used to solve! 

for this question we first find out the probabilities of the two groups seperately and multiply them. we divide each group's arrangement by the number of similar objects it has, so:

for the first group:
6!/4!2! = 15

for the second group:
6!/2!3! = 60

15x60=900

there are 900 possible arrangements

[candy]

Thanx a lot ocean!!

[saadkhan]

how do we do o/n 01 q7? is dere any oder method apart frm da 1 in solved pprx???

[night_buster]

Dude
u have to do
1/6 ----Beacuse out 6 paper clip in Box A They r asking probablity for picking Red
7/10 ----Beacuse out of 10 paper clip in box B Then r asking probablity of getting Reds will b in 7/10 ways
Therefore
Total Probablity of this Sub-part of the question is [(1/6 X 7/10)]
=(7/60)