- Messages
- 681
- Reaction score
- 438
- Points
- 73
Aww ... I was close to messing up that part ... thanks to the tension that Q5 built up .. but thank god i got it in the end
I sort of panicked in the end when I realised how little time is left :|
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Aww ... I was close to messing up that part ... thanks to the tension that Q5 built up .. but thank god i got it in the end
same at my end ..I sort of panicked in the end when I realised how little time is left :|
if there was time i could get some around 70 because in checking i would have figured out the vectors and another mistake but now i am looking below 60 it sucks. i mean such a long exam definitely the lengthiest from 2002-2012I sort of panicked in the end when I realised how little time is left :|
if there was time i could get some around 70 because in checking i would have figured out the vectors and another mistake but now i am looking below 60 it sucks. i mean such a long exam definitely the lengthiest from 2002-2012
i had to walk run and ride a three wheel driven by a human(rikshaw) for over 40mins in the rain and then entered the exam hall. half the energy was gone there oh well got s1 and a good as mark to look forward tooI just needed 5 more minutes..just 5!
i had to walk run and ride a three wheel driven by a human(rikshaw) for over 40mins in the rain and then entered the exam hall. half the energy was gone there oh well got s1 and a good as mark to look forward too
no i still don't get it i am sorry ! once again with a little more detail please .it is from 8 to 20 so P(8<X<20)=.96 so therefore P(-12/r<Z<12/r)=.96 which is 12/r=1.96/2 therefore r = 5.84
sorry did something wrong in the first time took prob as .96 when it actually is .94 really sorry bout thatno i still don't get it i am sorry ! once again with a little more detail please .
yea...Let me know how did you do that? did you get x=2a?
Well for my school many got stuck after one point
when they ask you for approximation you are supposed to use 6.5. Also in many sums they dont ask you those and still you got to use 6.5. As far as i know if you have to get the mean/standard deviation from the binomial formula(np and Rootnpq) then you need to use 6.5 incase of less than 7.can someone tell me if when we are suposed to use 6.5 incase of lesser than 7 in case of normal approximation/normal distribution?
Because sometimes they use 7 as whole and i don't seem to get the difference between these type of questions!
when they ask you for approximation you are supposed to use 6.5. Also in many sums they dont ask you those and still you got to use 6.5. As far as i know if you have to get the mean/standard deviation from the binomial formula(np and Rootnpq) then you need to use 6.5 incase of less than 7.
And when just for normal when you got the mean/s.d or you have to find out using the normal then you shall use 7
in part 2, why arent we arent we taking p as o.35/0.83???PART i)
The question said that only Rhesus + is taken, so if we note down the probabilities for all the Rhesus+=
A+= 0.35
B+= 0.08
AB+= 0.03
O+= 0.37
Now, what you do is, you draw up a probability distribution table for the Rhesus + values. So first of all calculate the total probability of the Rhesus + values= 0.35+0.08+0.03+037= 0.83
So what will the probability of selecting a person with the blood group of A+? It will be 0.35/0.83. For B+ 0.08/0.83. For AB+ 0.03/0.83 and for O+ 0.37/.83.
The question asks to find the probability that fewer than 3 are group O+. Now, you have think. It can either be O+, or it can't be O+. So the immediate thought that should hit you is that you have to use a binomial approximation. So the probability of a success (The probabolity of selecting a person who is O+) is 0.37/.83. The probability of a failure (Not selecting a person who's O+) is (0.35+0.08+0.03)/0.83= .46/.83
So P(X<3) means that you have to use the binomial expansion for 0, 1 and 2.
9C0*(.37/.83)^0*(0.46/.83)^9 + 9C1*(.37/.83)^1*(0.46/.83)^8 + 9C2*(.37/.83)^2*(0.46/.83)^7
=0.1555= 0.156
PART ii)
Now the question asks you to find a probability that involves a huge number. Using a binomial approximation can be very time consuming. You could use a normal approximation (np>5)
You have to find the probabily that more than 60 people are group A+. For using normal approximation, firstly, you have standardize 60.5. For which you need the mean and the standard deviation.
To find the mean we use np (total number* probability), and to find the variance we use npq (total number*probability*(1-probability))
So mean= np= 150*.35=52.5
And variance= npq= 150*.35*(1-.35)= 34.125
So standard deviation= square root of variance= 5.84
So now you have to standardize 60 i.e P(X>60). However, the questions asks you to find the probability of MORE than 60 people. so you have to take P(X>60.5).
Now first of all standardize 60.5 using (60.5-mean)/(standard deviation)
(60.5-52.5)/5.84
This give yous the Z value which is 1.369
Now you are to find P(X>1.369) using the normal distribution table. The value corresponding to 1.369 is 0.3145. Since you have to find the probability greater than 0.329, you have to subtract 1 from the probability
So 1-03145= 0.0855
Hope this helps!
oh umm right i'll edit it in the original post as wellOh thank you so much !! but one more question please . which mean is used then for the standard deviation calculation ?
Thanks for your help n support throughout..I pray for all of u
littlecloud11
PhyZac
Esme
iKhaled
Anika Raisa
Dug
and everyone who helped me n whose name i forgot to mention
Goodluck to all of u!!!
Thanks for your help n support throughout..I pray for all of u
littlecloud11
PhyZac
Esme
iKhaled
Anika Raisa
Dug
and everyone who helped me n whose name i forgot to mention
Goodluck to all of u!!!
not as good as expected... i screwed up Qs 9 n 10Best of luck 4 awl xams! Hw ws P3?
it went terrible... the 1st 7 questions made me feel like i wud get full marks n then ....... felt like murdering the a***hole who made the question
not as good as expected... i screwed up Qs 9 n 10
(i) okay umm in the first part you found that n was 10 by solving o.75^n < 0.06
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now