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A level Statistics doubt??Post your doubts here!

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profits can be eitehr -(24*20) or (450*10-24*200
which are -480 or 4020
its -480 if more than 1 firework fails to work since they get no revenue and only have the cost. From the prev question you found out the prob for failure of more than 1 firework to be 0.264
its 4020 profit when less than 1 firework fails to work since they get the revenue..In this situation the prob is (1-.2640
y -480 4020
p(y) .264 .736
do the E(y)=(4020*.736)+(-480*.264)=2832
 
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profits can be eitehr -(24*20) or (450*10-24*200
which are -480 or 4020
its -480 if more than 1 firework fails to work since they get no revenue and only have the cost. From the prev question you found out the prob for failure of more than 1 firework to be 0.264
its 4020 profit when less than 1 firework fails to work since they get the revenue..In this situation the prob is (1-.2640
y -480 4020
p(y) .264 .736
do the E(y)=(4020*.736)+(-480*.264)=2832
ohhh, makes sense. Thanks! :)
 
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i tried it that way, doesnt match, the examiners report says to use 0.97 as the probability :S
 
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as far as i know the first can be any colour so its prob is 1 and the one after that cant be the same colour so it will be 2/3 followed by 2/3 until the rest 7 slots are filled....as a result the ans is (2/3)^7
okei !! this seems more correct .. thanks for correcting me
 
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" The masses of articles produced in a particular workshop are normally distributed with mean 51.3 g and standard deviation 20.5 g. Find the symmetrical limits, about the mean, within which 75% of the masses lie."

someone PLEASE explain!:confused:
[ans: a=27.7 , b=74.9]
I'm not sure if your problem has already been addressed, but your question is too interesting to resist ;) So here goes..
Symmetrical limits means that both the values, let them be x and y are the same distance from the mean. That means x-mean and y-mean are the same but with opposite signs. like for 3 and 1c about 2.. 3-2=1 and 1-2=-1
So P(x< X < y) = .75
P[(x-51.3)/20.5< X < (y-51.3)/20.5] = .75
this brings us to
P(-a< Z <a) = .75
2*Fi(a)-1= .75
Fi(a) = .875
so a is approximately 1.152
(x-51.3)/20.5= 1.152 and x=74.9
(y-51.3)/20.5= -1.152 and y=27.7 :)
 
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I'm not sure if your problem has already been addressed, but your question is too interesting to resist ;) So here goes..
Symmetrical limits means that both the values, let them be x and y are the same distance from the mean. That means x-mean and y-mean are the same but with opposite signs. like for 3 and 1c about 2.. 3-2=1 and 1-2=-1
So P(x< X < y) = .75
P[(x-51.3)/20.5< X < (y-51.3)/20.5] = .75
this brings us to
P(-a< Z <a) = .75
2*Fi(a)-1= .75
Fi(a) = .875
so a is approximately 1.152
(x-51.3)/20.5= 1.152 and x=74.9
(y-51.3)/20.5= -1.152 and y=27.7 :)
could u please help me out with this one? it seems kinda odd http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf no.5 i
the examiners report says to use 0.97 as the probability... cant think of as to why :S would be very thankful!
 
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could u please help me out with this one? it seems kinda odd http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf no.5 i
the examiners report says to use 0.97 as the probability... cant think of as to why :S would be very thankful!
It IS .97.. but later on..
see the q starts the same way as your previous q. So I think you wont mind if I jump to
P(-a< Z < a) = .94
2*Fi(a) - 1= .94
Fi(a) = .97
I think its THIS .97 that they are referring to.. since taking .97 at the start does not lead to the right answer..
 
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It IS .97.. but later on..
see the q starts the same way as your previous q. So I think you wont mind if I jump to
P(-a< Z < a) = .94
2*Fi(a) - 1= .94
Fi(a) = .97
I think its THIS .97 that they are referring to.. since taking .97 at the start does not lead to the right answer..

I have the same doubt but i dont understand why do we have to do this step to get 0.97 ? Like what is clue for me to do this step?
 
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Hey guys good luck.for stats today..I hope we all can tackle it easily n get good marks...all the best :)
 
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I have the same doubt but i dont understand why do we have to do this step to get 0.97 ? Like what is clue for me to do this step?
There really isn't any clue ! Its just how u solve a question !
But to help yourself you can draw up a bell shaped curve
So u get something like
Phi(a)-phi(-a)
Which is
Phi(a)-(1-phi(a))
So expand brackets
2phi(a)-1=0.94
Now solve and find value of a
 
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kiara15 ASTAR syed1995 or any 1....

Help me knw how to know the lower class boundary from a cumulative frequency table ....

for example one in Qp 6 in here:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_61.pdf

i dnt understand 6 iv as in the ms! pls help!!! Reply asap!
i am sorry , just saw ur post.. ok if for example 2<x<7 then class boundry is 1.5<x<7.5... just simply sub 0.5 from lower one and add 0.5 to upper one .. got it?
 
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