Statistics P6 Help needed

[parthrocks]

which variant??

62

 

[Adorkableme]

can someone plzzzz explain me in detail the CONTINUITY CORRECTION PART!!!!IN NOrmal distribustion?

 

[Adorkableme]

u have to convert the binomial to normal by approximation

first find the men by:
30 * 0.8
=24

then find variance:

30 * 0.8 *0.2
=4.8

since the question states fewer then so we subtract 0.5 from 25 and get 24.5

now just standardize (24.5-24)/2.19
=0.228

use the table and find phi of 0.228
you will get 0.590

thats the answer

 

[Khan_971]

You can find some Statistics Lectures here: http://www.khanacademy.org/math/statistics

Don't forget to check out the Probability Section too: http://www.khanacademy.org/math/probability

Please feel free to share tips and skills here.

Okay, to be honest, I have barely opened the book yet and I really need those marks to pass. Mechanics was hard, I lost 20 marks easily, and I left about 4 or 5 questions in P3, so I need to do my best in this paper. I only studied for P1 and P3 honestly..

Should I start by opening the book or by directly solving past papers? I need to achieve 30 marks or more (inshallah). The only things I have studied so far are measure of location and spread.

I need to study those still:

Normal Distribution
Binomial Distribution
Probability
Permutations and Combinations

So what are your suggestions? And I know, I did fail in maths this year.

Go for the books first. Try out a FEW starting questions and some of the last. Check yourself for clear concepts.
Then go for the past papers. Backwards from 2011 will be good.

 

[shanky631]

can someone plz solve and explain this question.

 

[Vinita Manek]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_61.pdf

qt numb 7 pls.....

 

[Knight]

can someone plz solve and explain this question.

Q6;(i) It is give that P(X<20)=0.1
P(X>20)=0.9 which from the table is 1.282
P(X<20)=z of -1.282
z=-1.282
so (20-mean)/standard deviation=-1.282
(20-mean)/0.8=-1.282
From this we get mean=21.0(i.e 3 significant figures)

Q6;(ii) P(21.5<X<22.5)=P(X<22.5)-P(X<21.5)
=((22.5-21.03)/0.8)-((21.5-21.03)/.8)
=(1.8375)-(0.5875)
From the table
=0.967-0.7217
=0.2453

Now this the probability for success(i.e length between 21.5-22.5)

P(X<2)=P(1)+P(0)
=4C1(0.2453)^1(1-0.2453)^3+4C0(0.2453)^0(1-0.2453)^4
=0.746

Hope you understand this....

 

[Knight]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_61.pdf

qt numb 7 pls.....

Q7;(i) P(answered correctly)=(0.7)+(0.2*0.95)
=0.89

(ii) Let
P(C1)=Peter answers correctly first time
P(C2)=Peter answers correctly second time
P(A)=Audience answers correctly
P(F)=Friend answers correctly
P(C1,2)=First two questions answered correctly

Then

P(C1,2)=P(C1*C2) + P(C1*A) + P(A*F) + P(A*C1)
=(0.7*0.7) + (0.7*0.2*0.95) + (0.2*0.95*0.2*0.65) + (0.2*0.95*0.7)
=0.7807
=0.781 (Three s.f)

(iii) Let
P(X)=P(Peter asked the audience)
P(C1,2)=First two questions answerd correctly

Then
P(X)=P(C1*A) + P(A*F) + P(A*C1)
=(0.7*0.2*0.95) + (0.2*0.95*0.2*0.65) + (0.2*0.95*0.7)
=0.2907
P(X/C1,2)=(P(X intersection C1,2))/P(C1,2)
=0.2907/0.7807
=0.37235
=0.372 (Three s.f)

 

[math]

can anyone help me with m/j 2011 question number 2....1st part


god bless u...plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

p=.8 q=.2 np=30*.8=24 npq=30*.8*.2=4.8
P(X<25)
P(Z<24.5-24/sqrt4.8)=.228=.590

*sqrt=square root
*.590 from the table

 

[math]

hey i want to ask u that if the z has a value of say 0.228 whose inverse cannot be found out from the table.228 ....thus in this case do we need to do 1-0.228 =0.772 from the value......that is something....???why bt

no .... we ddont hav to minus from 1
we hav to take ,228 value from inside the table i.e .5902
hopee u understood

 

[Jaf]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_62.pdf

In question 4(iv) shouldn't we be first finding out our class boundaries? The mark scheme directly used the values in the table for the histogram.
According to me the class boundaries should be:
-0.5≤x<19.5 19.5≤x<29.5 29.5≤x<39.5 39.5≤x<44.5 44.5≤x<49.5 49.5≤x<59.5 59.5≤x<69.5

The mark scheme specifically mentions a mark for 'correct bar widths starting at 20'.

Help?

 

[shanky631]

Q6;(i) It is give that P(X<20)=0.1
P(X>20)=0.9 which from the table is 1.282
P(X<20)=z of -1.282
z=-1.282
so (20-mean)/standard deviation=-1.282
(20-mean)/0.8=-1.282
From this we get mean=21.0(i.e 3 significant figures)

Q6;(ii) P(21.5<X<22.5)=P(X<22.5)-P(X<21.5)
=((22.5-21.03)/0.8)-((21.5-21.03)/.8)
=(1.8375)-(0.5875)
From the table
=0.967-0.7217
=0.2453

Now this the probability for success(i.e length between 21.5-22.5)

P(X<2)=P(1)+P(0)
=4C1(0.2453)^1(1-0.2453)^3+4C0(0.2453)^0(1-0.2453)^4
=0.746

Hope you understand this....

thanks a ton. i got the idea. grateful to u

 

[shanky631]

i don't understand when to choose the negative value of z.

 

[math]

same here...even i am confused
can anybody explain it

 

[math]

can anybnody solve qns no.7 o/n 2010
plz asap
thax

 

[parthrocks]

no .... we ddont hav to minus from 1
we hav to take ,228 value from inside the table i.e .5902
hopee u understood

How did u get it...can u explain me again

 

[parthrocks]

p=.8 q=.2 np=30*.8=24 npq=30*.8*.2=4.8
P(X<25)
P(Z<24.5-24/sqrt4.8)=.228=.590

*sqrt=square root
*.590 from the table

hey how did u get 0.590

 

[parthrocks]

i don't understand when to choose the negative value of z.

same here can anyone plzz help

 

[math]

hey did you get .228
then look at the distribution table
z values are on the borders of th table
.288 is z value
so take value of .288 you will get .590 (3 s.f)
hope you understood

 

[math]

i dont ndersstand permutation and combinatio
i am unable to solve pastpers qns
plz help
what should i do