Mathematics: Post your doubts here!

[GoodRobot]

thank you

alright here you go
i)When ever you have to find the original mean=E(x-a)/n +a
so here the mean is given as 28.325. Find the value of n from the equation by substituting the value of E(x-a). 28.325-a=E(x-a)/n where a=25 n=40
For the s.d you should remember that s.d isnt affected by constant shifts in the data such as -a.
assume that x-a =y so E(x-a)=E and hence E(x-a)^2=E^2
s.d=underoot(E^2/n-(E/n)^2)
s.d=9.11
ii)check your data booklet one of the formula for s.d is underoot(E(x)^2/n - (E(x)/n)^2). Since you know that the s.d is 9.11 and that the original mean is 28.325 substitute the given values and find the value of (E(x)/n )^2.

thanks dude :]

 

[mukki]

is there any book or site where i can get the concept of this formula ? thanks,by the way

yes the cambridge s1 book by steve dobbs

 

[Khan_971]

I used the alphabet y which is in brackets ()where u see the thumbs up icon

haha

 

[farrukh]

Please help me in the last part? how to calculate mid values?

 

[Khan_971]

hey check this question out.. in part 2 u do 4!/2! right? how to solve part 1 using this factorial method? it gets me confused when to use the factorials and when to use nPr.. :/

6C4 * 4!. Or it can be 6P4. If orders are important (prizes in some cases0 , Permutation is to be used. If not (order does not matter, example choosing people for commitee,etc)
In this question. order is important

 

[Khan_971]

Please help me in the last part? how to calculate mid values?

u have to inter-connect the intervals. First Interval: 1-100, 101-150.
subtract .5 from lower, add to higher.
1st interval: .5-100.5, 100.5-150.5
Mid-value = .5+100.5 / 2 = 50.5 and so on

 

[mominzahid]

6C4 * 4!. Or it can be 6P4. If orders are important (prizes in some cases0 , Permutation is to be used. If not (order does not matter, example choosing people for commitee,etc)
In this question. order is important

please explain the 4th and 5th part of the same questionn.. :/

 

[sea_princess]

can someone please explain how to solve question 6 (ii)
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_63.pdf
it doesn't ever come as 4 and 2

 

[Khan_971]

3 diff colors:
Red,red,not red, not red = 2C2 * (5 * 2C1) x 4!/2! since 4 pegs and 2 are same
Since the rest 5 colors have 2 pegs each.
Total= (5*2C1) * 4!/2! * 6
= 720

 

[usmiunique]

i wanted to know when to take the mean of the data when calculating the quartiles.​

for eg. in M/J/2010 paper 61 stats,​

in question 2, we get the following steam-and-leaf diagram.​

0|2 5 6 8 8 (5)​

1|2 4 6 7 7 9 (6)​

2|1 2 3 3 3 5 6 7 (8)​

3|1 5 (2)​

now in calculating the median, we will divide the total number of data by 2 which in this case is​

21/2=10.5 so, which value will we take? 10 th value or 11thvalue or the mean of the 10th and 11th value.​

also in calculating the lower quartile, we get 21/4=5.25 here also, do we take the 5th value or the 6th value or the mean of the 5t and 6th value?​

again in calculating the upper quartile, we get 21*0.75=15.75 here again do we take the 15th value or the 16th value or the mean of the 15th and 16th value?​

sometimes in the mark scheme i have seen them taking the mean and sometimes the value immediately next to it.. im confused as to what should i do?​

WILL appreciate any help!​

tomorow is the exam please help!​

 

[Pals_1010]

Using 2, 3 , 4, 5, 6

!) How many integers greater than 4000 can be formed, if there is no repetition?

2) How many integers which are multiples of 5 can be formed if there is no repetition?

Thank You

 

[Mattman]

No.4 http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s05_qp_6.pdf

What is the accepted way by CIE to find the upper and lower quartile with an even number of terms?
i got 5.4 for LQ and UQ 8.35 according to my teacher's method...but the marking scheme showed 8.3 and 5.4

 

[usmiunique]

No.4 http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s05_qp_6.pdf

What is the accepted way by CIE to find the upper and lower quartile with an even number of terms?
i got 5.4 for LQ and UQ 8.35 according to my teacher's method...but the marking scheme showed 8.3 and 5.4

i have the exact same doubt.. hope somone clears it up!

 

[mukki]

i wanted to know when to take the mean of the data when calculating the quartiles.​

for eg. in M/J/2010 paper 61 stats,​

in question 2, we get the following steam-and-leaf diagram.​

0|2 5 6 8 8 (5)​

1|2 4 6 7 7 9 (6)​

2|1 2 3 3 3 5 6 7 (8)​

3|1 5 (2)​

now in calculating the median, we will divide the total number of data by 2 which in this case is​

21/2=10.5 so, which value will we take? 10 th value or 11thvalue or the mean of the 10th and 11th value.​

also in calculating the lower quartile, we get 21/4=5.25 here also, do we take the 5th value or the 6th value or the mean of the 5t and 6th value?​

again in calculating the upper quartile, we get 21*0.75=15.75 here again do we take the 15th value or the 16th value or the mean of the 15th and 16th value?​

sometimes in the mark scheme i have seen them taking the mean and sometimes the value immediately next to it.. im confused as to what should i do?​

WILL appreciate any help!​

tomorow is the exam please help!​

dude if the total frequency is odd like 21 .. to find the median u add 1 to it and divide by 2. 21+1/2 so its the 11th value.
for a total that is even for example if its 22 u take both the 11th and 12th values from the data and half them.
here is a nice tip for the quartiles. if the median is odd find it and delete it from the list. now calculate the quartiles of the numbers before the median which should be q1 and of the numbers after the median which should be q3. if the median is an even number then take it with your calculation like if the median is 22 take the 11th value and values before it to calculate q1 and take the 12th value and values after it to calculate q3.

 

[Khan_971]

i wanted to know when to take the mean of the data when calculating the quartiles.​

for eg. in M/J/2010 paper 61 stats,​

in question 2, we get the following steam-and-leaf diagram.​

0|2 5 6 8 8 (5)​

1|2 4 6 7 7 9 (6)​

2|1 2 3 3 3 5 6 7 (8)​

3|1 5 (2)​

now in calculating the median, we will divide the total number of data by 2 which in this case is​

21/2=10.5 so, which value will we take? 10 th value or 11thvalue or the mean of the 10th and 11th value.​

also in calculating the lower quartile, we get 21/4=5.25 here also, do we take the 5th value or the 6th value or the mean of the 5t and 6th value?​

again in calculating the upper quartile, we get 21*0.75=15.75 here again do we take the 15th value or the 16th value or the mean of the 15th and 16th value?​

sometimes in the mark scheme i have seen them taking the mean and sometimes the value immediately next to it.. im confused as to what should i do?​

WILL appreciate any help!​

tomorow is the exam please help!​

I think u have to take (n+1)/2

 

[Esme]

Plz explain why the ans for (i) is 10C1+10C3+10C5+10C7+10C9. i knw they said 'odd numbr' of cups but whats the mathematical logic/reason behind it ??

 

[Khan_971]

Using 2, 3 , 4, 5, 6

!) How many integers greater than 4000 can be formed, if there is no repetition?

2) How many integers which are multiples of 5 can be formed if there is no repetition?

Thank You

Write it down. 4000-4999. 4P3
5000-5999: 4P3
6000-6999: 4P3
total : 72?

 

[Pals_1010]

Write it down. 4000-4999. 4P3
5000-5999: 4P3
6000-6999: 4P3
total : 72?

I got that as well but the answer in the book is 192

How is that possible?

 

[usmiunique]

dude if the total frequency is odd like 21 .. to find the median u add 1 to it and divide by 2. 21+1/2 so its the 11th value.
for a total that is even for example if its 22 u take both the 11th and 12th values from the data and half them.
here is a nice tip for the quartiles. if the median is odd find it and delete it from the list. now calculate the quartiles of the numbers before the median which should be q1 and of the numbers after the median which should be q3. if the median is an even number then take it with your calculation like if the median is 22 take the 11th value and values before it to calculate q1 and take the 12th value and values after it to calculate q3.

i dont exactly understand the quartiles part.. what do u mean by deleting them from the list? :-S

 

[Esme]

Write it down. 4000-4999. 4P3
5000-5999: 4P3
6000-6999: 4P3
total : 72?

no there will be more combinations... such as 20000 or above. 30000 and above and so on