Author Topic: statistics CIE AS level doubts!!  (Read 2700 times)

Offline Tohru Kyo Sohma

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statistics CIE AS level doubts!!
« on: May 19, 2011, 06:56:03 pm »
ok....i actually didnt know where to post my statistics doubts.......so i opened a new topic for it(dont ban me if i did a mistake).....so i had doubts in;
1. may/june 2007 qns1
2.oct/nov 2005 qns 1 and 6 part 2
3.may/june 2006 qns 1 and 2 part1 and 5 part4 and 6 part4
4.may/june 2005 qns 5 part3 and 6 part2 and 2 part1
5.oct/nov 2004 qns6
6.may/june 2004 qns 6 part3 and 7 part2
....ok....i guess that is alot to ask....and i'd be glad if u help me in some atleast!!!

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #1 on: May 19, 2011, 06:59:47 pm »
ok...i got qns 1 answer!!

Offline Ghost Of Highbury

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Re: statistics CIE AS level doubts!!
« Reply #2 on: May 19, 2011, 07:05:58 pm »
1. https://studentforums.biz/math-146/s1-doubts-here!!!!/msg413139/#msg413139

2. best to go with 2 pie-charts
    males : Young = 40/100  360 = 144o
              Middle 35/100 * 360 = 126o
              Elderly 25/100 * 360 = 90o

    Do the same for the females

. 6ii) Profit = 0, means he hits on the fourth throw or the fifth throw
        4th = doesn't hit for the first three + hits in the 4th = (4/5)^3 * 1/5 = 0.1024

         or 5th = doesn't hit for the first 4 + hits in the 5th = (4/5)^4 *1/5 = 0.08192
total = 0.184

3. q1)The data is skewed as there is an outlier = 352 (way above the rest)
   Thus, go for median, arrange them in ascending order, and spot the middle one = 47 = 47000
   q2 ii ) burgers for suppers = 0.5
             meaning burgers after swimming, or not.
             0.2*0.75 + 0.8*x = 0.5
              x = 0.4375 ~ 0.438
    5iv) 24 were b/w 25-30, 88 were>30
          (24/110)(88/110) = 0.273
    6iv) use ur answer to iii to determine mean and variance
          mean = (1*16 + 2*8 + 3*4 + 4*2 + 5*2) / 32 ~ 1.94
          variance = sum (xi^2/total f) - 1.94^2 = 166/32 -  1.94^2 ~ 1.43

        
4. 5iii) Independent if P(A) * P(B) = p(A and B)
         ain't the case here, so not independent.
       6ii) P(1.9-b<X<1.9 +b) = P(1.9-b-1.9 / 0.15 < Z < 1.9+b-1.9 / 0.15) = 80% = 0.8
             P(-b/0.15 < Z < b/0.15) = 0.8
           @(b/0.15) - (1-@(b/0.15) = 0.8
          2@(b/0.15) -1 = 0.8
         @(b/0.15) = 0.9

          b/0.15 = 1.282
            b = 0.1923

thus the limits are 1.9-b = 1.71 and 1.9+b = 2.09

  2i) sqrt(sum(xi^2fi)/26 - mean^2 = (5^2*2 + 15^2 *9 + ....)/26 - 27.5^2) = 16.1

gotta sleep now, will do the rest 2morrow
« Last Edit: May 19, 2011, 07:33:54 pm by Ghost Of Highbury~ »
divine intervention!

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #3 on: May 19, 2011, 08:37:56 pm »
1. https://studentforums.biz/math-146/s1-doubts-here!!!!/msg413139/#msg413139

2. best to go with 2 pie-charts
    males : Young = 40/100  360 = 144o
              Middle 35/100 * 360 = 126o
              Elderly 25/100 * 360 = 90o

    Do the same for the females

. 6ii) Profit = 0, means he hits on the fourth throw or the fifth throw
        4th = doesn't hit for the first three + hits in the 4th = (4/5)^3 * 1/5 = 0.1024

         or 5th = doesn't hit for the first 4 + hits in the 5th = (4/5)^4 *1/5 = 0.08192
total = 0.184

3. q1)The data is skewed as there is an outlier = 352 (way above the rest)
   Thus, go for median, arrange them in ascending order, and spot the middle one = 47 = 47000
   q2 ii ) burgers for suppers = 0.5
             meaning burgers after swimming, or not.
             0.2*0.75 + 0.8*x = 0.5
              x = 0.4375 ~ 0.438
    5iv) 24 were b/w 25-30, 88 were>30
          (24/110)(88/110) = 0.273
    6iv) use ur answer to iii to determine mean and variance
          mean = (1*16 + 2*8 + 3*4 + 4*2 + 5*2) / 32 ~ 1.94
          variance = sum (xi^2/total f) - 1.94^2 = 166/32 -  1.94^2 ~ 1.43

        
4. 5iii) Independent if P(A) * P(B) = p(A and B)
         ain't the case here, so not independent.
       6ii) P(1.9-b<X<1.9 +b) = P(1.9-b-1.9 / 0.15 < Z < 1.9+b-1.9 / 0.15) = 80% = 0.8
             P(-b/0.15 < Z < b/0.15) = 0.8
           @(b/0.15) - (1-@(b/0.15) = 0.8
          2@(b/0.15) -1 = 0.8
         @(b/0.15) = 0.9

          b/0.15 = 1.282
            b = 0.1923

thus the limits are 1.9-b = 1.71 and 1.9+b = 2.09

  2i) sqrt(sum(xi^2fi)/26 - mean^2 = (5^2*2 + 15^2 *9 + ....)/26 - 27.5^2) = 16.1

gotta sleep now, will do the rest 2morrow
hehe....thanks alot anyways....sleep tight...dont let the bed bugs bite ;D

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #4 on: May 19, 2011, 09:41:37 pm »
ok...i have 1 more doubt....may/june 2007 qns 3 (b)

Offline Ghost Of Highbury

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Re: statistics CIE AS level doubts!!
« Reply #5 on: May 20, 2011, 08:58:18 am »
5. 6i) 1 2 3 4 5 - 10 outcomes - 123 345 124 245 125 235 134 234 135 145
     odd number sum - 4/10 =  0.4
6ii)  largest number 4 - 124 134 234 = 3/10 = 0.3

6iii) L can take values 3 4 5

     use the same method as done previously (in ii), list from the possible outcomes and calculate the probability
 
6iv) E(X) = sum (l*p)
      Variance = Sum(li^2pi) -( (E(x))^2

6. 6iii) P(A|B) = P(AnB) / P(B) = 0.65*0.1 / (your answer to part ii) = 5/19

   7ii) The probability of atleast one damaged tape = 1 - P(no damaged tape)

        probability of no damaged tape = 1-0.2 = 0.8

         1 - (0.8 )^n >= 0.85

        n = 9

divine intervention!

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #6 on: May 20, 2011, 02:47:14 pm »

6. 6iii) P(A|B) = P(AnB) / P(B) = 0.65*0.1 / (your answer to part ii) = 5/19

   7ii) The probability of atleast one damaged tape = 1 - P(no damaged tape)

        probability of no damaged tape = 1-0.2 = 0.8

         1 - (0.8 )^n >= 0.85

        n = 9


its 0.8^n
how can u find n......i didnt get it.....is there a method......pls do tell........and thanks for helping too

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #7 on: May 20, 2011, 03:09:22 pm »
ok......i have a doubt...not pastpaper this time....how do we know when to give a plus or minus sign in normal distribution qns......like when they ask to find standard deviation or mean....then im never sure when "z" should be + or -
HELP ME

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #8 on: May 21, 2011, 02:35:10 am »
ok...there is this qns in o/n 2009 p61 qns 3 part 2
they use 0.33 instead of 0.34.............i dont get it...y??

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #9 on: May 21, 2011, 03:14:57 am »
ok...i have 1 more (endless doubts ;D)....oct/nov 2009 p61 qns 5

Offline Tohru Kyo Sohma

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Re: statistics CIE AS level doubts!!
« Reply #10 on: May 25, 2011, 09:07:34 pm »
comeon guys pls help me....my teacher had an accident so she wont be able to help me.....pls help someone!!!