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can u post paper link and question number??DUde A star i can also access to the MS....but i need an explanation :/
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can u post paper link and question number??DUde A star i can also access to the MS....but i need an explanation :/
can u post paper link and question number??
okI gave a worked example for him one post back = o see if you can get the proper question so I can help him solve it
thanx yar...but i need a little explanation here..question 6 part 23alaikumissalam.
Work out the P(X > mean + sd)
I can't remember which year that is, but I will just give you a quick example.
If the question is X~N(10,5^2) (10 mean and 5 sd) and you are looking for lengths MORE THAN 1 sd from the mean, that means you need to work out P(X > 10 +5)
That is P(X>15).
Whatever answer you get, multiply it by 1000.
Hope that helps.
this makes binomial expansion invalid. yup took me 3 minutes to notice what i find odd in this questionThings to note, remember that they are not independent and therefore the probability changes each time, be careful.
It is very straightforward really, I in fact understand the probability one and can't understand the combination way.
He took 4 biscuits without replacement. Find that wrapped is EXACTLY 2.
Therefore, it can be in .. (W = wrap) (U = unwrap)
P(U) * P(W) * P(U) * P(W)
Or
P(W) * P(U) * P(W) * P(U)
Or
P(W) * P(W) * P(U) * P(U)
So on.
Counting the amount of ways this can be done, it is 4C2 ways.
Take any of the arrangement above ^, multiply it by 4C2 give you the answer.
thank you ♥1st of all when converting binomial to normal :
1. np and npq should be greater than 5 to be suitable to normal dist.
2. X~N(np ,npq) will ne the distribution of X.
3. Use continuity correction as we have to be accurate :
if p(X>5) ... We will calculate p(X>5.5)..... if its p(x<5) we will calculate p(X<4.5)
if its p(X≥5)... it will be p(X≥4.5) .... if its p(X≤ 5 ).... it will be p(X≤5.5).....
4. then solve remember npq is variance so under root it to calculate s.d while substituting it in formula to find Z
one thing wrongly saidthank you ♥
actually its a rule that within 1 sd of the average 2/3 of the values liethanx yar...but i need a little explanation here..question 6 part 2
It is very straightforward really, I in fact understand the probability one and can't understand the combination way.
He took 4 biscuits without replacement. Find that wrapped is EXACTLY 2.
Therefore, it can be in .. (W = wrap) (U = unwrap)
P(U) * P(W) * P(U) * P(W)
Or
P(W) * P(U) * P(W) * P(U)
Or
P(W) * P(W) * P(U) * P(U)
So on.
Counting the amount of ways this can be done, it is 4C2 ways.
Take any of the arrangement above ^, multiply it by 4C2 give you the answer.
Lol.. I understand the combination way better.
Thanks for you help. Just one more thing, is the no. of ways 4C2 because you are choosing 2 wrapped biscuits from 4?
lolYes Oh I'm surprised people find the (personally chinese) combination method easier wow glad to see the different approaches
a) 8! x 2!find the no of ways 4boys and 5 girls can be arranged in a row if
a) a particular boy and a particular girl must be next to each other
b)the 1st n last persons are of the same sex
syed1995 littlecloud11 please elp
Explaina) 8! x 2!
b) 4 x5 x 7!
so much angerExplain
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