statistics problematic questions

[adnanj]

Hi All

Any one can solve the below questions choosen from past papers (Answers should be matched with examiner report). i have discussed with three different teachers but the answers are different from the examiner report.

2008 paper 2 Question 7 (v)
2008 paper 1 Question 8 a(iii)
2003 paper 2 Question 3
2003 paper 2 Question 4
1995 paper 1 Question 9 (iv)

Regards

 

[OakMoon!]

2008 paper 2 Question 7 (v)
'they live in the same house with no other people'. This meant that once the first of the two people
had been chosen from among the 16 who lived in 'two-person households', the second person could not be
any one of the other 15, but only the one other person with whom the already-chosen person lived. So the second person is the ONE out of the 115 left.
So it goes like this 16/110 (This is the first person who lives in a 2 man house) * 1/109(The probability of the exact person out of the remaining that lives with the first person). Or 0.145*0.00917=0.00133


2008 paper 1 Question 8 a(iii)
Yar we will multiply the probability that a visitor calls to see a person and then the person not being at home and then add them.
For e.g for abdul we will multiply his probability that a visitor called him and he wasnt at home. So 0.3 * 0.2 (since they said to find the probability if not finding the person at home and that is 1-0.8=0.2)
Now take it out for the other 2 and add them
0.3*0.2+0.3*0.3+0.4*0.4=0.31 SOLVED


2003 paper 2 Question 3

The first 3 parts are straight forward Qs. Total number of marks= 225+165=390
Mean marks= total marks/number of pupils. Total pupils=25 so mean mark is 390/25=15.6
and then again sum of squars= 4495+2992=7487

Now the third part has given us a hint that we have to use that formula for the standard deviation that has the sum of square numbers since they have given that.
So sum of squares/N - mean ka square and then take the square root.
7487/25-15.6^2=56.12 And then take the square root which gives u 7.49.


2003 paper 2 Question 4

Yar in the first park. WE r talking about humans and they cant be halved. So the limits will stay as they are. And the interval will be 1999-1500=499 and mid point 1500+1999/2.
In the second part we have heights and have to convert the limits into true class limits hence increase both sides by 0.25 since the limits have an accuracy of 0.5. So it become 159.75-169.75. With interval 10 and mid pt. 164.75.


I dont have the 1995 paper.

If you have any other problem feel free to ask. I ll be glad to help, especially in mathematics questions.

 

[UxaiR]

Best possible explanations and answers of hamidali

 

[tahir03]

very well explained hamid

 

[adnanj]

i am really thankful to u... actually i was solving questions frm the past paper book of waris baig in which questions were wrongly printed.. but after looking to ur answers i checked on xtremepapers and found the correct questions...
thx alot for helping me out.. and i would be really grateful if u help me out with a few more questions in which i am stuck..

 

[OakMoon!]

Sure. I will try to answer them. You should do the past papers using Examiner reports and the original Papers. It gives u better practice.

 

[SAUD ASLAM]

Abdul Waris ke Answers tu galat hote hai ab question bhi Galat hai.Is se Achi Book bto Mein bana sakta hoo.Lol

 

[birnamz]

hamid need help in: 99 p2 q8 [iv] and [v]

 

[birnamz]

and also nov 02 p2 q3

 

[birnamz]

and hamid in case i need help in add maths ill asku here n this form iz this alrite?

 

[OakMoon!]

november p2 q3

Part (i) Yar they have asked about the particular student from the 100 students. The 100 students are chosen from 4 groups so one student can only be from one out of the four groups. So the particular student will have a probability of 1/4.

Part (ii) They say the total students frm each group are 25. And the total number of student in group D is 112. So the one particular student of group D chosen will be 25/112.

Part (iii) This is simple. Even though the group C and D have comparitively more students but their representaion is less.

Part (iv) Yar this is the simple ratio question. Take the ratios of each of the groups. Divide the number of students in group by the total and multiply by 100 (The number to be chosen)

Students in A= 88/400*100=22
Students in B= 96/400*100=24
Students in C= 104/400*100=26
Students in D= 112/400*100=28

Yar i dont have the 1999 paper since I never used the waris baig book. I prefer doing the past papers with examiner reports and marking schemes available online on sites such as xtremepapers.

You can make a new thread. Or you may ask the questions of Addmaths here. Addmaths used to be my favourite subject in Olevels. I ll try my best to help u as much as i can.

 

[SAUD ASLAM]

Qs: How systematic sampling can biased?

 

[OakMoon!]

Yar since every Nth person in the population is chosen it doesnt give the correct ratio of representation. Some sects and groups and genders may be totally ignored. Thus, leading the sample to be biased.

The danger in systematic sampling lies in the possible hidden periodicities of selection. If for some reason, we list the individuals in a certain order where, for example, every 10th individual belongs to a subgroup in the population, choosing every 10th out of the list will clearly be biased.

two similar answers.

 

[abdulwahabzia]

hey hamid do u have kind of notes for statistics

 

[Saad Qadir]

Yes pls hamid agr notes hai Or if u can jus make em within 1 2 dys it wud be reallllllllllllllllllyyyyyy helpful for us...

 

[OakMoon!]

Yar I can assure you that you dont need notes for STATISTICS. Pakistan Studies, Biology, Chemistry e.t.c are the subjects that require notes. Statistics is purely practice like maths and Addmaths. I didn't even attend a proper academy. Just solve past papers of the last 10 years. That's what I did and got an A* in one of the most difficult stats papers to date i.e. in November 2009. I solved past papers once and missed many questions as I didnt knew the concept. And then since i had time, I solved the past papers again. And the second time I didn't leave even one single question. Notes ka koi faida nahi. Just get a book from the market. And I would say keh even that isnt required. Only past papers can get you an A* or atleast an A.

 

[aaakhtar19]

Can any one help me out in Stats
2009 P1 Q9(vii), Q7(ii)
i am very bad at theory based Questions.

Pls help me out in these
then i ll tell u abt more pobz thnx

 

[OakMoon!]

Q9 (vii) The equation shows that the new gauge will give a reading 5MPa greater than the accurate reading given by the standard gauge so the new gauge calibration therefore needs to be adjusted 5 MPa downwards.
This is a clear answer given in the marking scheme. You could have seen that.

Q7(ii) Most of the lengths are close to that intended and so the process is quite precise.

Again this is from the marking scheme. The BEST answer. Nobody here can give you a better answer then the examiner

 

[birnamz]

thnx hamid ur the best

 

[birnamz]

i hv done past papers frm 1993 till 09 4 times and im proud but still nervous