Mathematics: Post your doubts here!

[hassam]

now ,......how to know class boundaries here???
question says less than 45....less than 50.............bt ms writes this Class as 45-50 widout paying attention to the damn inequality signs............now why the hell were they there/....if we dun have to use them.....m cnfused.....is there any rule to detrmine class widths

 

[smartangel]

P(k-125/4.2 <Z< 128-125/4.2) =.7465
Fi(3/4.2) - fi(k-125/4.2) = .7465
fi(k-125/4.2) = .0158
since its lower than .5,
1-fi(125-K/4.2) = .0158
.9842 = fi(125-K/4.2)
Using normal table:
2.15 = 125-k/4.2
9.03 = 125-K
K = 115.97 = 116

how'd the k - 125 change to 125 - k???

 

[smzimran]

Assalamu alaikum,
Q.3 part 1 pls...
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_qp_61.pdf
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w11_ms_61.pdf

Total possible selections are 8C4 = 70.

If there are 1 girl and 3 boys in the team,
Selections = 5C1 * 3C3 = 5
P(1G) = 5/70 = 1/14

If there are 2 girl and 2 boys,
Selections = 5C2 * 3C2 = 30
P(2G) = 30/70 = 3/7

If there are 3 girls and 1 boy,
Selections = 5C3 * 3C1 = 30
P(3G) = 30/70 = 3/7

If there are 4 girls and 0 boys,
Selections = 5C4 * 3C0 = 5
P(4G) = 5/70 = 1/14

 

[Jaf]

now ,......how to know class boundaries here???
question says less than 45....less than 50.............bt ms writes this Class as 45-50 widout paying attention to the damn inequality signs............now why the hell were they there/....if we dun have to use them.....m cnfused.....is there any rule to detrmine class widths

Apparently, the mark scheme is wrong.

 

[Mattman]

So...how did the paper 6 go? Went somewhat badly..

 

[IsaacNewton]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

Question 3(i)
36 Total Values.

S = Sigma sign.

S(x-45) = -148
Open the brackets:
Sx-S45 = -148

Sx = sum of all the different values of x.
s45 = Sum of all the 45's. Since there are 36 total values, it means sum of 36 45's. So, S45 = 36(45) = 1620

Sx = -148 + 1620
Sx= 1472

Mean = Sx / 36 = 1472 / 36 = 40.9

S(x-45)^2 = 3089
Expand Bracket
S(x^2 -90x + 45^2) = 3089 Quadratic Expansion
Open brackets

Sx^2 -90Sx+ S2052 = 3089 90Sx = 90(1472) S2052 = 36(2052)

So Sx^2 = 3089 - 73872 +132480
Sx^2=61697

S.D^2 = Variance.

Var = (Sx^2 / 36 ) - (mean)^2
Var= (61697/36)- (40.9)^2
Var = 40.99
S.D = 6.4


(ii)

A value "29" was added to our Data. So our total values are now 37

This means that the sum of all the values now, is Sx + 29
So our New Sx = 1472 + 29 = 1501

We will also calculate our new Sx^2 in the same way.
Our new Sx^2 = 61697 + 29^2 = 62538

use the Variance Formula

(Sx^2 / 37) - (Sx/37)^2
(62538/37)-(1501/37)^2

Variance = 44.5
S.D = 6.67

 

[IsaacNewton]

um last minute problems.....:/
View attachment 10645View attachment 10646
Um just part ii of Q3 And part iii of Q7 thanku sooooooooooooooo much!!!

Question 7 (iii)

70 days at random, how many would have temperature more than 3 times the mean.

So first we calculate the probability, that a single day has, that its minimum temperature is more than 3 times the mean.


So, p(X > 3u )

p(Z > (3u - u)/2u )

p(Z > 2u/2u)

p( Z > 1 )
1 - P(Z<1)

after looking from the table:
1 - 0.8413
0.1587

So the probability of a single day having a minimum temperature more than 3 times the mean:
p = 0.1587

n, number of days = 70.

Expected value, = np, = 70(0.1587) = 11.109,
Therefore 11 days.

 

[IsaacNewton]

So...how did the paper 6 go? Went somewhat badly..

When did you give Paper 6??
Which variant, and what came in it ?
I still have the paper in a few hours.

 

[smzimran]

When did you give Paper 6??
Which variant, and what came in it ?
I still have the paper in a few hours.

Its wrong dude, don't ask what is to come!

 

[hassam]

So...how did the paper 6 go? Went somewhat badly..

r u kidding....cos paper 62 is taken most early....and there are still 5-6 hours in that

 

[hassam]

Its wrong dude, don't ask what is to come!

well Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other?
ms does 8!/3! -2*7!/3!....i cnt understand why he is multiply by 2

 

[hassam]

?????

 

[smzimran]

leadingguy , Mustehssun Iqbal , Khan_971 , hassam , user , RGBM211 , alphabravocharlie
and
@Everyone:
Some important points....

1. If a box and whisker plot is more stretched towards the lower values, it is said to be positively skewed.

2. If a box and whisker plot is more stretched towards the higher values, it is said to be negatively skewed.

3. An outlier is a value that is more than 1.5 times the interquartile range.
(a) A high outlier is given by = Q3 + 1.5 (Q3 - Q1)
(b) A low outlier is given by = Q1 - 1.5 (Q3 - Q1)
*Q1 and Q3 are lower quartile and upper quartile respectively.

4. Remember to include +/- 0.5 in class boundaries when drawing histograms, although they are not necessary for cumulative frequency curves.

5. Conditional probabilty; when asked probability of B given that A is occuring, the probability is given by
P (B | A) = P (A and B) / P (A)
This can be rewritten as
P (A and B) = P (A) * P (B | A)

6. Independent events are events that have no effect on one another. For two independent events A and B ,
P (A and B) = P (A) * P (B)

7. If A1 , A2 , ... , An are n independent events, then
P (A1 and A2 and ... and An) = P (A1) * P (A2) * ... * P (An)

8. Two events which have no outcomes in common are called mutually exclusive events. The addition law of mutually exclusive events is given by
If A1 , A2 , ....... , An are n mutually exclusive events, then
P (A1 or A2 or .... or An) = P (A1) + P (A2) + ... + P (An)

 

[Diamona151294]

first attachment is Q3

second attachment is Q7

hope it helps
gonna have my math paper 6 later.. wish me luck!

Thanku soooooooooooooo much!!!!!! gdluk for ur paper!!
oh um 1 thing, can u exp why we multiply the probability by 70?

 

[smzimran]

well Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other?
ms does 8!/3! -2*7!/3!....i cnt understand why he is multiply by 2

Im not getting that also,
Which year ?

 

[hassam]

its w11 qp 62

 

[Diamona151294]

Question 7 (iii)

70 days at random, how many would have temperature more than 3 times the mean.

So first we calculate the probability, that a single day has, that its minimum temperature is more than 3 times the mean.


So, p(X > 3u )

p(Z > (3u - u)/2u )

p(Z > 2u/2u)

p( Z > 1 )
1 - P(Z<1)

after looking from the table:
1 - 0.8413
0.1587

So the probability of a single day having a minimum temperature more than 3 times the mean:
p = 0.1587

n, number of days = 70.

Expected value, = np, = 70(0.1587) = 11.109,
Therefore 11 days.

thanku!! got it.... ugh i just dont know how im suposed to conc on every single word in the paper.....:/

 

[smzimran]

its w11 qp 62

Im not getting this method but i managed to do it with the other method mentioned in the mark scheme!
That 6! * (7! / 3!)

 

[alphabravocharlie]

Im not getting this method but i managed to do it with the other method mentioned in the mark scheme!
That 6! * (7! / 3!)

Can't we do it like: No.of arrangements - No G together - All G together
[9!/(3!*3!)] - [(7C3*6!)/3!] - 7!/3!

 

[Diamona151294]

um in this question, im finding the z value for 0.94, but the answer's wrong, somehow they are giving the z value for 0.97..... can sm1 pls explain whats going on here?