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A level Statistics doubt??Post your doubts here!

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Thanks a ton...could you plz elaborate what you mean by conversion from binomial to normal...i am sorry i can do questions just fine but i don't really get what is meant by binomial, normal etc...can yu plz just give me an example question...i ll be very thankful..
ummlook ill explain
in binomial we are taken random DISCRETE variable
while in normal we take CONTINOUS random variable
this difference in nature of binomial and normal causes a gap i the values of normal approx . hence to seal the gap we do it
 
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ummlook ill explain
in binomial we are taken random DISCRETE variable
while in normal we take CONTINOUS random variable
this difference in nature of binomial and normal causes a gap i the values of normal approx . hence to seal the gap we do it

really sorry for bothering you again and again but i don't get discrete and continuous either...:(
 
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thanks
2ii) for sd.=√ S(x − 100)^2/15 - (72/15)^2
=√10.24 = 3.2

Summer 12/61
mean is 8.34/9
for sd.= √ [(.76)^2 +(.85)^2+(.89)^2+(.91)^2+(.92)^2+(.94)^2+(.96)^2+(1.04)^2+(1.07)^2]/9 - (8.34/9)^2
=√(7.7984/9) - (8.34/9)^2 = .0882
for that in 12/61 i forgot to divide the x^2/9 thats why wasnt getting ans x.x
 
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salvatore
Sorry 4 the l8 reply though :(

s1_prob!.jpg

Hope i cud be helpful!
Notes on Prob distribution:
 

Attachments

  • stats_ch4_DISCRETE Probability distribution.pdf
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  • stats_ch7_7. Continuous Probability Distributions.pdf
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really sorry for bothering you again and again but i don't get discrete and continuous either...:(

Okay see, a continuous variable can take any value within a range. For example if you consider the weight of people in a population they can have a wide number of values between any 2 integers, between 64 and 65 kg there can be 64.2, 64.5 64.9 etc. So it is a continuous variable.
Discreet variable however can only take a specific value. Like for binomial the probability of success is fixed at a specific value. And there are only two outcomes of success or failure. If success probability is .4 failure is 1-.4 There's no range, so it's a discreet variable.
 
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Okay see, a continuous variable can take any value within a range. For example if you consider the weight of people in a population they can have a wide number of values between any 2 integers, between 64 and 65 kg there can be 64.2, 64.5 64.9 etc. So it is a continuous variable.
Discreet variable however can only take a specific value. Like for binomial the probability of success is fixed at a specific value. And there are only two outcomes of success or failure. If success probability is .4 failure is 1-.4 There's no range, so it's a discreet variable.

Thankyou so much...that is the explaination i was looking for...:)
 
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JazakAllah...these notes are really useful.
I hv a doubt though...Under the "guidelines for classes "(i.e. class widths) ...it says
  1. The classes must be equal in width. The exception here is the first or last class. It is possible to have an "below ..." or "... and above" class. This is often used with ages.
I don't quite get this...how could i compute class widths of say, histograms or CF, if the question involves ages?

Isnt it under Grouped frequency? i m not much clear with it too...
Its like we donot always get an equal class width! ryt! :confused:
http://people.richland.edu/james/lecture/m170/ch02-grp.html
 
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hey can anyone tell me tht whn do we have to use 0.5 in normal distribution in the value of X..... its costing a lot of marks .. asap plzz !!
 
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ummlook ill explain
in binomial we are taken random DISCRETE variable
while in normal we take CONTINOUS random variable
this difference in nature of binomial and normal causes a gap i the values of normal approx . hence to seal the gap we do it
hey same here plzz could u explain it
 
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hey i got it but could u plzz explain me about the 0.5 one ....... tht why in many cases we take the value X nearest to 0.5 and in some we dont ... plzz explain ... ?
if you are talking abt the normal approximation or the continuity correction(i think) then when you calculate the mean/s.d with binomial formula(np/rootnpq) and then do normal then if its greater than say 65 you have to consider it as 65.5 if its less than 65 then its 64.5 if its greater and equal then its 64.5 and if its less and equal its 65.5
hope i was clear enough
 
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if you are talking abt the normal approximation or the continuity correction(i think) then when you calculate the mean/s.d with binomial formula(np/rootnpq) and then do normal then if its greater than say 65 you have to consider it as 65.5 if its less than 65 then its 64.5 if its greater and equal then its 64.5 and if its less and equal its 65.5
hope i was clear enough
but in many cases thy take the same value as give n like 64 thy take it as 64 only
 
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but in many cases thy take the same value as give n like 64 thy take it as 64 only
when you dont need to use binomial distribution for the normal distribution then you dont need to do the approximation...in those case you just take 64
 
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when you dont need to use binomial distribution for the normal distribution then you dont need to do the approximation...in those case you just take 64
so tht means if we have to calculate mean using (np)and s.d using (rootnpq) thn we have take 0.5 into account ryt ?
 
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yes dat is wat i said it is done when u calculate the mean n variance using dat formula or calculate
the probability using Bionomal!!
Visit here for more info:
http://people.richland.edu/james/lecture/m170/ch07-bin.html

y
a i m talking about normal approximation .. i know how to calculate mean and all but in many markschemes and all thy take the value of X 0.5 less thn the exact value of X or 0.5 more thn the exact value of X .
 
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