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Statistics P6 Help needed

Jaf

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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_62.pdf

In question 4(iv) shouldn't we be first finding out our class boundaries? The mark scheme directly used the values in the table for the histogram.
According to me the class boundaries should be:
-0.5≤x<19.5 19.5≤x<29.5 29.5≤x<39.5 39.5≤x<44.5 44.5≤x<49.5 49.5≤x<59.5 59.5≤x<69.5

The mark scheme specifically mentions a mark for 'correct bar widths starting at 20'.

Help?
This! Someone help!!
 
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WHY THERE IS NO CONTINUITY CORRECTION???

thats because, the qs have already told us the data are normally distributed. In such case when qs mentions "normally distributed", we no need to normalise or lets say no need to do continuity checkings!
hope u understand this :)
 
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heybro and can u plzzz tell me the steps in detail for this....i know its easy bt a silly mistake here maybe.....IMAGINE......PLZZ
View attachment 10543

hmm for qs (i)
state that P(X>1002) = 1-P(X<1002)
you understand this step??
I have simply taken like inverse of original form ;)
for LHS, we have also given the probability = 225/900

now simply do the solvings doing LHS=RHS

finally u'll obtain ans: 997 (hopefully)

qs (ii).....
do the solvings binomially.
p=225/900=0.25
n=3

so, P(X=2) ...solve it (y)
 
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hmm for qs (i)
state that P(X>1002) = 1-P(X<1002)
you understand this step??
I have simply taken like inverse of original form ;)
for LHS, we have also given the probability = 225/900

now simply do the solvings doing LHS=RHS

finally u'll obtain ans: 997 (hopefully)

qs (ii).....
do the solvings binomially.
p=225/900=0.25
n=3

so, P(X=2) ...solve it (y)
AFTER GETTING 225/900 THEN WHAT DO U DO?
 
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i don't understand when to choose the negative value of z.
same here...even i am confused
can anybody explain it
same here can anyone plzz help
it's easy.....
if the probability has been given to you and it is less then 0.5, you can't read it's z value from the table......
Let x be the probability which is less then 0.5...
what we have to do is to minus the probability from 1(i.e 1-x)....
this calculated probability is obviously greater then 0.5....
Now read the z value for this calculated probability......
Now all you have to do is to include a negative sign in your z value......
Let
P(X<x)=0.25 (which is less then 0.5 )
P(X>x)=1-0.25
=0.75
From the table
z=0.674 for 0.75
z=-0.674 for 0.25

Hope you understand it.....and best of luck for your exams.....(y)
 
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