Statistics P6 : post your doubts here!

[tanmaydube]

Guys please discuss P6 doubts here after P12 i guess i need to work hard for this paper.

The heights, x cm, of a group of young children are summarised by
Σ(x − 100)2 = 499.2. Σ(x − 100) = 72, The mean height is 104.8 cm.
(i) Find the number of children in the group.(Ans 15)
(ii) Find Σ(x − 104.8)2 .(Ans 153.6)

Please help me with part (ii)

Thank You!

 

[Jia Yean]

Standard deviation = [499.2/15 - (72/15)^2 ]^1/2 = 3.2
Mean = 104.8
Σ(x − 104.8)/15 + 104.8 = 104.8
Σ(x − 104.8) = 0.

Since std deviation is always the same,
Std dev = 3.2
[Σ(x − 104.8)^2 /15 - (0/15)^2]^1/2 = 3.2
Σ(x − 104.8)^2 = 153.6

 

[caro24]

can anyone help me for this question nov 10 p61 num 3ii)

 

[Kirabo Takirambudde]

I took a long route to show u why (o.o5) is the probability, inversely you could see that 1.645 is the z value and that (1-0.95) = 0.05

 

[caro24]

thanks but can u help with nov 10 p61 num 6 plz plz plz

 

[caro24]

plz is there no one to help me

 

[caro24]

definitely there is no one to help me

 

[babycry]

1) 6p4 since there are 6 pegs ad 4 holes, order matters
2) 4!/2 since the 2 red pegs are repeated
3) you can pick 4 pegs, lets say RBGO 6C4 times. then u can arrange them 4! ways in a line so 6c4* 4! gives 360
4) you can pick 3 pegs, lets say RRBG, RBBG and RBGG 6C3 times each and then since 2 are alike u can do 4!/2 as in part 2. then u multiply by 3 since there are 3 different ways of getting the pegs to give 720
5) you can pick 2 pegs say RRBB in 6C2 ways and then arrange them in 4!/2*2 ways, which is 6*6c2 which is 90. 1 different pegs is the same as part4. so the total is 720+90+360 which is 1170.

 

[Kirabo Takirambudde]

Please help with part (iii)

 

[caro24]

1) 6p4 since there are 6 pegs ad 4 holes, order matters
2) 4!/2 since the 2 red pegs are repeated
3) you can pick 4 pegs, lets say RBGO 6C4 times. then u can arrange them 4! ways in a line so 6c4* 4! gives 360
4) you can pick 3 pegs, lets say RRBG, RBBG and RBGG 6C3 times each and then since 2 are alike u can do 4!/2 as in part 2. then u multiply by 3 since there are 3 different ways of getting the pegs to give 720
5) you can pick 2 pegs say RRBB in 6C2 ways and then arrange them in 4!/2*2 ways, which is 6*6c2 which is 90. 1 different pegs is the same as part4. so the total is 720+90+360 which is 1170.

thank u vry mch

 

[babycry]

Please help with part (iii)

to find the probability you have to,
find the number of ways of arranging mrs brown. 3 seats and 1 person means 3P1
find the number of ways of mrs lin: there are 10 available seats, since 1 is taken by mrs brown in the front row. so 10P1
find the number of ways of seating students. there are 5 students and 1 seat since the student has to in front of mrs lin. so thats 5 ways
the remaining 9 people have to be seated in 11 seats so 11P9.

so multiply 3x5x10x11p9 and then divide by the total unrestricted ways which is 14P12. this gives 0.0687.

btw this was 1 of the most tricky questions ive seen man

 

[Iadmireblue]

Can someone help me with 4 part III from M/J/12 variant 63
Thanks

 

[Kumkum]

Please help me with question number 1

 

[caro24]

Please help me with question number 1

you must use the formula

where mean is calculated by (

)that is you are given

X is 645 then u divide by n and mean is 645 divide by 150..then u replace

x^2 and the mean in the formula above to find standard deviation (dont forget the square and all) the u hav standard deviation but remember standard deviation is also equal to

replace in standard deviation in this formula and n to find the answer..hope u undestand





,

 

[Kumkum]

you must use the formula

where mean is calculated by (

)that is you are given

X is 645 then u divide by n and mean is 645 divide by 150..then u replace

x^2 and the mean in the formula above to find standard deviation (dont forget the square and all) the u hav standard deviation but remember standard deviation is also equal to

replace in standard deviation in this formula and n to find the answer..hope u undestand





,

thanks!

 

[bamteck]

www.xtremepapers.com/papers/CIE/.../9709_s04_qp_6.pdf

HELP needed for no. 5

 

[rizwaan]

NOV 07 P6 NUMBER 1 AND JUNE 2012 P62 NUMBER 3 ??? PLEASE HELP ME WITH ALL THE CALCULATIONS.

 

[Kirabo Takirambudde]

Help with part (iii) please

Thanks

 

[Iadmireblue]

Help with part (iii) please

Thanks

Since the survey will definately include 1 house of style C and 1 from B,there are two houses remaining to chose from
Eliminating Style C and B houses you have 7 different kinds of houses to chose from.
So 1 of style B gives 2C1
1 of style C gives 3C1
and for the remaninig your'e left with 7C2
Putting these together you need 1 of B and 1 of C and two of something else other than B or C
so ----> 3C1 * 2C1 * 7C2

 

[yashi30]

www.xtremepapers.com/papers/CIE/.../9709_s04_qp_6.pdf

HELP needed for no. 5

June 2004 P6 nO 5
(i) For a three-course meal,
nO of choices=( 3C1 X 5C1*2 X 3C1 ) =90
From the starter course, you have to choose 1 from 3 so 3C1
From main course, you have to choose 1 from 5 so 5C1, but it's mentioned that main courses are either served with new potatoes of french fries, so from your choice of main course, you either have new potatoes or french fries, total = 5C1*2
from dessert course you have 1 choice from 3 so 3C1
add all and you get the answer.

(ii) If a customer chooses the
(a)Starter and Main course
number of choices= 3C1*5C1*2=30
(b) Starter and Dessert Course
Number of choices= 3C1 X 3C1= 9
(c) Main course and dessert
Number of choices= 5C1*2 X 3C1 =30

total number of choices= 30+30+9= 69