S1 and S2 DOUBTS HERE!!!!

[Aadeez || Zafar]

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.
=1-?(-1.304)
=1-(1-?(1.304))
=1-1+?(1.304)
= ?(1.304)

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

[**RoRo**]

(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.
=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!

[Aadeez || Zafar]

ohhh!
ur edexcel studemt srry i thought u r a cie student

[Aadeez || Zafar]

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

[Aadeez || Zafar]

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

[**RoRo**]

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**]

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!

[Arthur Bon Zavi]

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.

[**RoRo**]

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]

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.

[**RoRo**]

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.

[Arthur Bon Zavi]

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 ?

[Arthur Bon Zavi]

Doing the rest.

[Arthur Bon Zavi]

Okay, I've compiled my questions here in this post to make it easier!

Here they follow.

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

Done.

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

[Arthur Bon Zavi]

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