Author Topic: S1 and S2 DOUBTS HERE!!!!  (Read 62849 times)

Offline Aadeez || Zafar

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Re: S1 DOUBTS HERE!!!!
« Reply #135 on: April 11, 2011, 06:52:08 pm »
Can someone please solve for me Question 3 in May 2009 with steps?
(i)
P(0 buses) = (1-0.16)11
P(1 bus) = 11C1(0.16)(1-0.16)10
P(2 buses) = 11C2(0.16)2(1-0.16)9

where X is number of buses passed

so

P(X<3)= P(0 Busus)+P(1 Buses)+P(2 buses)

(ii)

USE normal distribution


mean = np
        = 125(1-(0.16+0.2))
        = 125 (0.64)
       = 80
variance = npq
            = 125 (0.64)(1-0.64)
             = 28.8
let X be number of car passed

so

p(X>73)
= 1-P(X<=73)
=1 - ?((73-80)/root(28.8)
=1-?(-1.304)
=1-(1-?(1.304))
=1-1+?(1.304)
= ?(1.304)

use normal distribution tabel to get the value of ?(1.304)



Yeh BaaabuRaao ka style hein!

**RoRo**

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Re: S1 DOUBTS HERE!!!!
« Reply #136 on: April 11, 2011, 07:28:14 pm »
(i)
P(0 buses) = (1-0.16)11
P(1 bus) = 11C1(0.16)(1-0.16)10
P(2 buses) = 11C2(0.16)2(1-0.16)9

where X is number of buses passed

so

P(X<3)= P(0 Busus)+P(1 Buses)+P(2 buses)

(ii)

USE normal distribution


mean = np
        = 125(1-(0.16+0.2))
        = 125 (0.64)
       = 80
variance = npq
            = 125 (0.64)(1-0.64)
             = 28.8
let X be number of car passed

so

p(X>73)
= 1-P(X<=73)
=1 - ?((73-80)/root(28.8)
=1-?(-1.304)
=1-(1-?(1.304))
=1-1+?(1.304)
= ?(1.304)

use normal distribution tabel to get the value of ?(1.304)

I think you just solved a wrong question?
My question was about histograms.

Here' the link to the paper:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf

I want Question 3 please! :)

& if you don't mind can you please explain to me Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf

Thank you! :)

Offline Aadeez || Zafar

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Re: S1 DOUBTS HERE!!!!
« Reply #137 on: April 11, 2011, 07:29:15 pm »
ohhh!
ur edexcel studemt srry i thought u r a cie student



Yeh BaaabuRaao ka style hein!

Offline Aadeez || Zafar

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Re: S1 DOUBTS HERE!!!!
« Reply #138 on: April 11, 2011, 07:34:09 pm »
I think you just solved a wrong question?
My question was about histograms.

Here' the link to the paper:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf

I want Question 3 please! :)

& if you don't mind can you please explain to me Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf

Thank you! :)
widht would be::

  3*2/6 = 1 cm
and height is :

10*9/15 = 6
so 6/1 = 6 cm

SIMPLE



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Offline Aadeez || Zafar

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Re: S1 DOUBTS HERE!!!!
« Reply #139 on: April 11, 2011, 07:45:33 pm »
I think you just solved a wrong question?
My question was about histograms.

Here' the link to the paper:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf

I want Question 3 please! :)

& if you don't mind can you please explain to me Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf

Thank you! :)

Var(X) = Sigma(xi2pi)-(mmean)2
= = Sigma(xi2pi)-(xipi)2
=[02(0.4)+12(0.3)+22(0.2)+32(0.1)]-[(0(0.4)+(1(0.3)+(2(0.2)+(3(0.1)/4]2
=[0(0.4)+1(0.3)+4(0.2)+9(0.1)]-[(0+0.3+0.4+0.3)]2
=[0+0.3+0.8+0.9]-(1)2
=2-1
=1



Yeh BaaabuRaao ka style hein!

**RoRo**

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Re: S1 DOUBTS HERE!!!!
« Reply #140 on: April 11, 2011, 07:50:50 pm »
widht would be::

  3*2/6 = 1 cm
and height is :

10*9/15 = 6
so 6/1 = 6 cm

SIMPLE


Can you please explain to me what those numbers mean, what does the 3, 2, 6, 10, 9 and 15 stand for? :/

I'm sorry for the trouble, but this lesson has been annoying me since the IG's!

& again, you solved the wrong question, I wanted part e not d! ;)

**RoRo**

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Re: S1 DOUBTS HERE!!!!
« Reply #141 on: April 11, 2011, 10:59:47 pm »
Okay, I've compiled my questions here in this post to make it easier! :)

1. Question 3 of June 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf

2. Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf

3. Question 7c of June 2008 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20June/S1%202008-june-qp.pdf

4. Questions 6b and c of January 2008 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20Jan/6683_01_que_20080115.pdf

With explanation please, I have the marking scheme, so I know the answers, but I can't understand the working/how to do it! :/

Thanks alot! :)

Offline Arthur Bon Zavi

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Re: S1 DOUBTS HERE!!!!
« Reply #142 on: April 12, 2011, 09:38:02 am »

Can you please explain to me what those numbers mean, what does the 3, 2, 6, 10, 9 and 15 stand for? :/

I'm sorry for the trouble, but this lesson has been annoying me since the IG's!

& again, you solved the wrong question, I wanted part e not d! ;)


Formula for

(i) frequency density (height) :

    frequency  
= ------------
    class width

(ii) class width (interval) :

        frequency
= ------------------
     frequency density


Now for your question :

3 (a)

x = 10 - 15 (9.5 - 15.5); frequency = 15.

class width = (15.5 - 9.5) = 6. This is taken in histogram as 2 cm.
For 16 - 19, width is (19.5 - 16.5) = 3.

In the histogram, the width of the 16 - 18 class will be :  (3 X 2)/6 = 1



(b)

Height (or frequency density) = (15 / 6) = 2.5. This is taken in histogram as 5 cm.
For 16 - 19, height is (9 / 3) = 3.

In the histogram, the height of the 16 - 18 class will be : (3 X 5)/2.5 = 6

Tell me if you still didn't get.


Continuous efforts matter more than the outcome.
- NU

**RoRo**

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Re: S1 DOUBTS HERE!!!!
« Reply #143 on: April 12, 2011, 12:40:46 pm »
Formula for

(i) frequency density (height) :

    frequency  
= ------------
    class width

(ii) class width (interval) :

        frequency
= ------------------
     frequency density


Now for your question :

3 (a)

x = 10 - 15 (9.5 - 15.5); frequency = 15.

class width = (15.5 - 9.5) = 6. This is taken in histogram as 2 cm.
For 16 - 19, width is (19.5 - 16.5) = 3.

In the histogram, the width of the 16 - 18 class will be :  (3 X 2)/6 = 1



(b)

Height (or frequency density) = (15 / 6) = 2.5. This is taken in histogram as 5 cm.
For 16 - 19, height is (9 / 3) = 3.

In the histogram, the height of the 16 - 18 class will be : (3 X 5)/2.5 = 6

Tell me if you still didn't get.



Thanks a lot! :)

Alpha

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Re: S1 DOUBTS HERE!!!!
« Reply #144 on: April 12, 2011, 05:56:55 pm »
1. Question 3 of June 2009

Frequency density= Freq/ Class width
The freq. density will be the height of your histogram in case of unequal class widths.

x :10 – 15
Frequency= 15
Class width= 15 - 10 = 5

Frequency density= 15/5 = 3

Now, use proportions.

Class width,  5 : 2
Height, 3 : 5

For 16 – 18:

Class width=18 - 16 = 2
On histogram -> 2/5*2= 1/5

Frequency density=9/2 = 4.5
On histogram ->(5/3)*(9/2) = 15/2 = 7.5


I don't find the MS from your link... so, please check it.




« Last Edit: April 12, 2011, 06:00:10 pm by ~Alpha »

**RoRo**

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Re: S1 DOUBTS HERE!!!!
« Reply #145 on: April 12, 2011, 07:25:32 pm »
1. Question 3 of June 2009

Frequency density= Freq/ Class width
The freq. density will be the height of your histogram in case of unequal class widths.

x :10 – 15
Frequency= 15
Class width= 15 - 10 = 5

Frequency density= 15/5 = 3

Now, use proportions.

Class width,  5 : 2
Height, 3 : 5

For 16 – 18:

Class width=18 - 16 = 2
On histogram -> 2/5*2= 1/5

Frequency density=9/2 = 4.5
On histogram ->(5/3)*(9/2) = 15/2 = 7.5


I don't find the MS from your link... so, please check it.






I'm sorry, your answers are wrong, check Fidato's post above for the right answer.

Offline Arthur Bon Zavi

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Re: S1 DOUBTS HERE!!!!
« Reply #146 on: April 13, 2011, 11:10:52 am »
1. Question 3 of June 2009

Frequency density= Freq/ Class width
The freq. density will be the height of your histogram in case of unequal class widths.

x :10 – 15
Frequency= 15
Class width= 15 - 10 = 5

Frequency density= 15/5 = 3

Now, use proportions.

Class width,  5 : 2
Height, 3 : 5

For 16 – 18:

Class width=18 - 16 = 2
On histogram -> 2/5*2= 1/5

Frequency density=9/2 = 4.5
On histogram ->(5/3)*(9/2) = 15/2 = 7.5


I don't find the MS from your link... so, please check it.

Alpa, there you went wrong. Check the intervals before 16 - 18. They were 10 - 15, so in order to built a histogram, you need to amend the values to make them continuous (so you subtract 0.5 from the lower limit and add 0.5 to the upper limit).

Like take two class intervals, (10 - 15) and (16 - 18). There's gap of 1 between 15 and 16, so how can you make a histogram ?
« Last Edit: April 13, 2011, 11:13:05 am by Fidato »

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Offline Arthur Bon Zavi

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Re: S1 DOUBTS HERE!!!!
« Reply #147 on: April 13, 2011, 11:14:18 am »
Doing the rest.

Continuous efforts matter more than the outcome.
- NU

Offline Arthur Bon Zavi

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Re: S1 DOUBTS HERE!!!!
« Reply #148 on: April 13, 2011, 11:52:01 am »
Okay, I've compiled my questions here in this post to make it easier! :)

Here they follow.

Quote
1. Question 3 of June 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20June/6683%20June%202009.pdf

Done.

Quote
2. Question 3e of January 2009 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2009%20Jan/S1%202009-01.pdf

After 3 games Rohit has scored 6.
2 games are left and he has to get 4 points or more to win the prize.

List all the outcomes of the points he can get (see the image below).
You can see that there are 16 possible outcomes and six of which fetch him 4 points or more.
You know that :
P(0) = 0.4
P(1) = 0.3
P(2) = 0.2
P(3) = 0.1

So to win a price he should get points with the outcomes as follows (as circled in image).

{ (1,3) (2,2) (2,3) (3,1)(3,2) (3,3) }

Multiply and add the probabilities :

(0.3 X 0.1) + (0.2 X 0.2) + (0.2 X 0.1) + (0.1 X 0.3) + (0.1 X 0.2) + (0.1 X 0.1)

= 0.03 + 0.04 + 0.02 + 0.03 + 0.02 + 0.01
= 0.15
« Last Edit: April 13, 2011, 04:08:50 pm by Fidato »

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Offline Arthur Bon Zavi

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Re: S1 DOUBTS HERE!!!!
« Reply #149 on: April 13, 2011, 12:45:20 pm »
Quote
3. Question 7c of June 2008 on this link:
http://www.xtremepapers.net/Edexcel/Advanced%20Level/Mathematics/2008%20June/S1%202008-june-qp.pdf

(7) (c)

For this you need to find out the probability that the bag weighs more than 53 kg.


(a)

More than 53, so (53 + 0.5 = 53.5)

        53.5 - 50
Z =  ------------
             2
= si (1.75)
= 0.9599

Now, it's greater then 53, so :

   1 - { si (1.75) }
= 1 - 0.959
= 0.0401


(c)

2 bags greater than 53 kg and 1 less than 53 kg .

Probability that any bag is greater then 53 kg = 0.0401
and
Probability that it is less then 53 kg = 0.959

So :

3C2 X (0.401)2 X (0.959)
= 0.004630586
= 0.0046

Continuous efforts matter more than the outcome.
- NU