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How was S1 Statistics 62?

Expected GT


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I did them all well except Q5b I got only 54 selections instead of 60 and I wrote all the 54 selections one by one and the weird thing is that my probabilities summed up to 1 that's why I didnt revise it so how many marks do you think I would get out of the 5 marks ??
 
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I did them all well except Q5b I got only 54 selections instead of 60 and I wrote all the 54 selections one by one and the weird thing is that my probabilities summed up to 1 that's why I didnt revise it so how many marks do you think I would get out of the 5 marks ??
There were 10 selections and 60 arrangements.... Note the difference.... I was about to make the same mistake until i read the qstn again... Anyways u will get 3/5marks if ur probabilities sum up to 1. So u'll lose only two marks.... 1 for mistake and 1 for wrong answer.... Rest marks you will get for defining S as (1,2,4), for making an attempt of finding probability(Method marks) and last for sum of probability to become 1
 
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Guys i just drew the table and didn't get time to complete so my probabilities didn't add up to 1
i listed a few of the possible selections
in the 7th question i forgot to take 0.80 as negative so my mean and s.d. came out wrong
in the 6th question none of my answers were correct but i think i gained method marks in each of the bits
can anyone tell me how many marks i lost?
 
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There were 10 selections and 60 arrangements.... Note the difference.... I was about to make the same mistake until i read the qstn again... Anyways u will get 3/5marks if ur probabilities sum up to 1. So u'll lose only two marks.... 1 for mistake and 1 for wrong answer.... Rest marks you will get for defining S as (1,2,4), for making an attempt of finding probability(Method marks) and last for sum of probability to become 1

Aha exactly 10 selections and 60 arrangements I got only 9 selections that's why I got only 54 arrangements. What I did exactly is that I wrote the arrangements under each selection but I didnt multiply by 3! I dont think there is a difference right ?? Anyway then I found that there are 30 selections out of 54 where the smallest number is 1 and then 18 out of 54 with smallest number 2 then only 6 out of the 54 with the smallest number 4 so apparently the number of arrangements I got where the smallest number is 2 or 4 is right but 1 is not right as it should have been 24 instead of 18 so yeah I hope I would get those 3 marks as you said because I really wish that I can get the star in the Math Alevel and el7amdullah all the other questions are totally right .
 
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Aha exactly 10 selections and 60 arrangements I got only 9 selections that's why I got only 54 arrangements. What I did exactly is that I wrote the arrangements under each selection but I didnt multiply by 3! I dont think there is a difference right ?? Anyway then I found that there are 30 selections out of 54 where the smallest number is 1 and then 18 out of 54 with smallest number 2 then only 6 out of the 54 with the smallest number 4 so apparently the number of arrangements I got where the smallest number is 2 or 4 is right but 1 is not right as it should have been 24 instead of 18 so yeah I hope I would get those 3 marks as you said because I really wish that I can get the star in the Math Alevel and el7amdullah all the other questions are totally right .
I dunno exactly what u did there.... But i guess u found arrangements first and then converted them back to selections.... Well if this is the case than u might get 4/5 too. Cheer up even i made a silly mistake. Accidentally multiplied 365days with the probability in last qstn :/
 
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Guys plz help me out... In Q5ii probability distribution table I did a mistake.. There were 10 selections so around 60 ways to arrange (6 for each selection) but I made a mistake and took it 30 and my probabilities were 18/30, 9/30, 3/30 but this gives the same answer as 0.6, 0.3, 0.1. Can anyone tell me how much marks will I loose
 
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Hey Guys,In the last part of last question,we had to make two equations right.But i made a very silly mistake in the first equation in which we had to change signs.I took phi of the probability given not phi inverse :( i did equation 2 perfectly correct,please tell me how many marks will i lose??
 
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Nearly 24 hours have passed so i'll just post the answers here for all to see:
1.0.222
2. Mean 45.8 sd 14.9
3.that was all diagram and stuff.
4. X=0.81 next 12/31
5. 0.6 was of even even odd and distribution was 1 0.6 2 0.3 4 0.1
6 20,60 and 1210
7.expected days 7 t=37 min and last mean 8.94 sd 2.42
for q6 I didn't get 1210. I calculated it as 13C8 * 2 .can I score atleast 1 mark bcz the question did not state it is necessary that either of the two people are to be selected in the group?
 
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Oh my gwad, yap!!!
But the method that stroke in ma mind first was the one we did... xD
Yup i did the same method,but the question was of 3 marks so i wondered why we had to take numerous selections but then i did that minus one and it gave same ans so i knew both were correct in the hall.In the exam i did that 3 selection addition one though ;)
 
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Yup i did the same method,but the question was of 3 marks so i wondered why we had to take numerous selections but then i did that minus one and it gave same ans so i knew both were correct in the hall.In the exam i did that 3 selection addition one though ;)
First of all, I didn't gave the exam. xD
But as being junior of one giving exams, they were disusing the questions, they were solving in book, I saw they are doing some long WRONG method, I showed them how to. :cool: :D
 
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Or quite simply 14C9(total selections) - 12C7(both together).although i used your method :p
Dude are we long lost brothers :eek:
Cox i did both methods and it confirmed me that ans was 1210 :p
Although I wrote 12C8+12C8+12C9
 
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