Statistics P6 Help needed

[Kangxidi]

the answer is right......but can you plz explain it.....
thanx

Assume we have B1 B2 B3 B4 B5 and G1 G2 G3 (They are all different people, hence the numbering )

We want the boys to sit together. As a result we have to group them as a single entity.

Now we have B1B2B3B4B5 as a group.

In the group, there are 5! ways in which they can be arranged, whether B1 is first followed by B2, or B2 is first followed by B1 and so on.

It is time to put this group amongst the girls.

We can have an arrangement of:

G1 [the boys group] G2 G3

Or,

[the boys group] G1 G2 G3

Or,

G2 [the boys group] G1 G3

And so on.

You should now notice that we have 4 different items to "play" with, hence 4!

Therefore 4! * 5!

 

[Kangxidi]

help needed urgently. 9709_w11_qp_62 Question 7 (ii), (iii) & (iv) Kindly help.

Download it, I wrote it on a piece of paper for you

 

[beeloooo]

http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w11_qp_61.pdf can someone help with question 5 pls ?

 

[Lujain M.]

Can you guys help me with Q1(ii) nov09 v62?

 

[syed1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf can someone help with question 5 pls ?

i) P(20-12<X<20+12)
P(8<X<32)=0.94
P(-12/SD<X<12/SD)=0.94
2Phi(12/SD)-1=0.94
2Phi(12/SD)=1.94
Phi(12/SD)=0.97
12/SD=invPhi(0.97)

And take it from there in part i...

part ii use SD from part i..

P(X>13)
P(z>13-20/SD)
P(z>-7/SD)
P(z<7/SD)

use the table..

part iii..

find P(X>32)

then use that in binomial .. with p= above n = 7 ..

And Answer would be P(r>=2) = 1-P(0,1)

That should give you the answer...

 

[Kangxidi]

Can you guys help me with Q1(ii) nov09 v62?

Q3 = 63.

Since Q3 is the 75th percentile of data, the probability of anything below Q3 is 0.75

From previous question we know that X~N(51,sd)

Therefore P(X<63) = 0.75

z = 0.674

P(Z < (63-51)/sd = 0.674

12 = 0.674 * sd

sd = 12/0.674 =17.8

 

[syed1995]

Q3 = 63.

Since Q3 is the 75th percentile of data, the probability of anything below Q3 is 0.75

From previous question we know that X~N(51,sd)

Therefore P(X<63) = 0.75

z = 0.674

P(Z < (63-51)/sd = 0.674

12 = 0.674 * sd

sd = 12/0.674 =17.8

haha .. I think I did that with Q1.. since I remember using 0.25!

 

[Kangxidi]

haha .. I think I did that with Q1.. since I remember using 0.25!

Any does not matter, as long as you got the logic right

 

[syed1995]

Any does not matter, as long as you got the logic right

yea i remembered that from the SD=17.8 answer

 

[Lujain M.]

Q3 = 63.

Since Q3 is the 75th percentile of data, the probability of anything below Q3 is 0.75

From previous question we know that X~N(51,sd)

Therefore P(X<63) = 0.75

z = 0.674

P(Z < (63-51)/sd = 0.674

12 = 0.674 * sd

sd = 12/0.674 =17.8

Omg thanks a lot. Appreciate it so much. ❤

 

[Lujain M.]

One more question, nov.10 v63. q7(iv) i dont get how to find p and q :/

 

[Kangxidi]

One more question, nov.10 v63. q7(iv) i dont get how to find p and q :/

Are you sure ON10 63 7 iv) ? I dont see why you need to find p or q there..?

 

[syed1995]

Guys please link the papers with your queries.. makes helping easier...

 

[Kangxidi]

Guys please link the papers with your queries.. makes helping easier...

<--- downloaded the paper xP

 

[syed1995]

<--- downloaded the paper xP

I am lazy...

 

[Lujain M.]

Are you sure ON10 63 7 iv) ? I dont see why you need to find p or q there..?

Yeah :/ p(x<16) and it's a continuity correction question but it say visits lasting lesss than 8.2 and 8.2 is the mean. So we have to find a new mew and standard deviation but for that we need p and q how do i find it?

 

[Kangxidi]

Yeah :/ p(x<16) and it's a continuity correction question but it say visits lasting lesss than 8.2 and 8.2 is the mean. So we have to find a new mew and standard deviation but for that we need p and q how do i find it?

Since 8.2 mins is the mean, probability of a visit LESS than 8.2 mins = 0.5

From the question, we can say that X~B(35,0.5)
Therefore, X~N(17.5 , 8.75) since np = 35 * 0.5 = 17.5 and npq = 35 * 0.5 * 0.5 = 8.75
P(X<16)
P(Z<= (15.5-17.5)/8.75^-0.5)

and so on

 

[syed1995]

Yeah :/ p(x<16) and it's a continuity correction question but it say visits lasting lesss than 8.2 and 8.2 is the mean. So we have to find a new mew and standard deviation but for that we need p and q how do i find it?

I think..

P(X<8.2)
P(z<8.2-8.2/SD) //SD from previous parts...
P(z<0)
p=0.5
q=0.5
n=35

Now use normal approximation

np= mean = 17.5 .. SD = √npq = √8.75

P(X<16)
P(X<15.5)
P(z<15.5-17.5/√8.75)
P(z<-0.6761)
1-Phi(0.6761)
P(X<16)=0.249 Answer

 

[Lujain M.]

Since 8.2 mins is the mean, probability of a visit LESS than 8.2 mins = 0.5

From the question, we can say that X~B(35,0.5)
Therefore, X~N(17.5 , 8.75) since np = 35 * 0.5 = 17.5 and npq = 35 * 0.5 * 0.5 = 8.75
P(X<16)
P(Z<= (15.5-17.5)/8.75^-0.5)

and so on

But i don't get it why it's 0.5?

 

[Lujain M.]

I think..

P(X<8.2)
P(z<8.2-8.2/SD) //SD from previous parts...
P(z<0)
p=0.5
q=0.5
n=35

Now use normal approximation

np= mean = 17.5 .. SD = √npq = √8.75

P(X<16)
P(X<15.5)
P(z<15.5-17.5/√8.75)
P(z<-0.6761)
1-Phi(0.6761)
P(X<16)=0.249 Answer

I think..

P(X<8.2)
P(z<8.2-8.2/SD) //SD from previous parts...
P(z<0)
p=0.5
q=0.5
n=35

Now use normal approximation

np= mean = 17.5 .. SD = √npq = √8.75

P(X<16)
P(X<15.5)
P(z<15.5-17.5/√8.75)
P(z<-0.6761)
1-Phi(0.6761)
P(X<16)=0.249 Answer

I think..

P(X<8.2)
P(z<8.2-8.2/SD) //SD from previous parts...
P(z<0)
p=0.5
q=0.5
n=35

Now use normal approximation

np= mean = 17.5 .. SD = √npq = √8.75

P(X<16)
P(X<15.5)
P(z<15.5-17.5/√8.75)
P(z<-0.6761)
1-Phi(0.6761)
P(X<16)=0.249 Answer

Ohhhhh okay. Thanks a lot man was so confused in the 0.5 part. C: