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A level Statistics doubt??Post your doubts here!

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P(X > 2µ) = 0.1016


3µ = 7σ^2
µ = 7/3σ^2

P(X>2u)
P(X>2u-u/SD)
P(z>u/SD)
P(z>(7 SD^2/3 / SD) =0.1016
P(z>7/3 * SD)=0.1016
z<-(7SD/3 )=-invPhi(1-0.1016)
7SD/3 = invPhi(0.8984)
SD = 3*invPhi(0.8984) / 7

Now Mean will be (7*(SD)^2)/3
 
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long method means more time consumption it just leaves less time for other mind draining tasks

naw .. not really :p it just consumes pages for me.. I just hope in the main paper I don't find the time short to solve all the questions!

My target has always been 1 hour for S1. but hardly meet that target:p AND PLEASE PLEASE.. NO PLOTTING OF GRAPHS in the coming paper.. -,- especially the CF.
 
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naw .. not really :p it just consumes pages for me.. I just hope in the main paper I don't find the time short to solve all the questions!

My target has always been 1 hour for S1. but hardly meet that target:p AND PLEASE PLEASE.. NO PLOTTING OF GRAPHS in the coming paper.. -,- especially the CF.
cF rocks but if it doesnt then they will give another question that will make me mad
 
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9 diff fruit pies
3 ppl ---- all gets odd no.

List of odd no. 1 3 5 7 _(9 not possible otherwise all will not get pies!)

Now , lets list possible ways the no. can be divided

1st per 2nd per 3rd per

1 3 5

1 5 3

1 7 1

7 1 1

1 1 7

3 1 5

3 3 3


(See that all combinations add to 9)

Now u c ms says:


1 1 7 = 9C1× 8C1× 7C7 (oe)× 3C1 =216 In above table:here c only 1 out of the three cn get 7 pies so 3C1.
1 3 5 = 9C1× 8C3× 5C5(oe)× 3! = 3024 In above table:here c either can hv 5 or 3 or 1 pie in any order so 3!
3 3 3 = 9C3× 6C3× 3C3 (oe) = 1680
In above table:here c its all getting 3 pie n order doesnt matter!
so you add all n get d final ans!
Hope i cud xplain! Though i myself is not much clear abt y we use 3C1 instd of 3! in the first part ...Bt dat was d explanation we were told @class . A little help there with this Alice123 or syed1995 ASTAR or any1!
 
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