Mathematics: Post your doubts here!

[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

 

[Khan_971]

5 The weights of letters posted by a certain business are normally distributed with mean 20 g. It is
found that the weights of 94% of the letters are within 12 g of the mean.
(i) Find the standard deviation of the weights of the letters. [3]

This is from Oct/Nov 2011, paper 61. I only have a doubt with how the question is phrased. It says 94% are WITHIN 12 g of the mean. So if we say that 94% of the weights lie between, per say, 26 g and 14 g, shouldn't that also be right? In the marking scheme they use 32 g. Shouldn't they stick to saying that 94% of the weights are below a value which is 12 g above the mean instead of writing within?
Thanks.

Lower boundary, 20-12 = 8, upper 20+12 = 32.
P( 8<X<32) =.94

 

[Khan_971]

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

then won't that go on??

 

[mukki]

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

u know that 0.028 = 0.85r -.82q
the possible values of r is 2 or 3 and for q 3 or 4. try each value for both and ul realize that r=4 and q=2 for the iqr to be 0.028

 

[Pals_1010]

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

You're right!

When you consider 5-digit numbers, No of integers = 5! = 120

Total no of integers = 72 + 120 = 192

Thanks

 

[Khan_971]

Pals_1010
I got it.

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

How is that possible?

if theres a 5 digit number. and without repetition. it will be 5! as all those 5 numbers can be either way. 5! + 72 = 192

 

[Pals_1010]

Pals_1010
I got it.
if theres a 5 digit number. and without repetition. it will be 5! as all those 5 numbers can be either way. 5! + 72 = 192

Thanks a lot . It clicked in my mind at the same time as yours it seems

 

[usmiunique]

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

How is that possible?

yes it can b either a 4 digit number or 5 digit numer..
for 4 digt numer u found out 72 possible ways. for 5 digit number, all numbers will be greater than 4000 so that is 5! numbers..
so, the total is 72+5!=192

 

[mukki]

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

i mean that circle the median and dont use it in your calculation its for un grouped data. look if the median is odd make a mark on each side of the median and calculate the MEDIAN for all the values before the median which will result as q1 and calculate the MEDIAN for all the values after the median which will result in q3. TRY IT OUT BUT PLZ DONT INCLUDE THE MEDIAN IF IT IS FROM AN ODD NUMBER.

 

[mominzahid]

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

i dont get itt.. :'(
i think i have problem in the basic concept of applying these factorials and permutations and combinations.. :/

 

[GoodRobot]

yes the cambridge s1 book by steve dobbs

found it.thanks xD

 

[azambezi]

In stats 1 when drawing the cumulative graph do u draw a curve or one with straight lines?

 

[Aarjit]

can you plz do part iii as well! thanks

[Q4] (iii) Find the number of different selections of 4 letters from the 9 letters of the word HAPPINESS which contain no Ps and either one or two Ss.

Discard all the Ps and keep the Ss aside:

H A P P I N E S S

H A I N E ........... S S

1st Criteria: 1S no Ps

S _ _ _ ......... [Note that you need to fill the 3 _ _ _ with the remaining 5 (the other S is discarded with P since we need only 1S) letters 'H A I N E' and S remains fixed; the order does not matter since we are 'selecting']

n = 1 x 5C3 = 10

2nd Criteria: 2S no Ps

S S _ _ ......... [Now there are merely 2 _ _ for the 5 letters 'H A I N E' and the 2 Ss remain fixed]

n'= 1 x 5C2 = 10.... [If you're wondering why I used '1 x' again instead of '2 x' or '2!' , it's because both the letters (S) are the same and it won't make a difference if we swap them. Again, the order does not matter since we are 'selecting']

∑ n = n + n' = 10 + 10 = 20...... Q.E.D.

 

[Khan_971]

In stats 1 when drawing the cumulative graph do u draw a curve or one with straight lines?

kind of sigmoid curve

 

[smartangel]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s11_qp_63.pdf
Q6

 

[usmiunique]

i mean that circle the median and dont use it in your calculation its for un grouped data. look if the median is odd make a mark on each side of the median and calculate the MEDIAN for all the values before the median which will result as q1 and calculate the MEDIAN for all the values after the median which will result in q3. TRY IT OUT BUT PLZ DONT INCLUDE THE MEDIAN IF IT IS FROM AN ODD NUMBER.

oh... thanks a lot!