!! HL Maths 2015 - predictions, guesses, Q & A, discussion ...

skippy1977

RoRo979 wrote: »

1/√N can still be used but next year will not be allowed.
Its easier and quicker to use than the 1.97√p-hat(.....)/n one

How official is this I wonder. We had kids come back with 'tips' from grinds schools (not that one) with detailed diagrams of things they said were coming up on Q4 and Q5 and ....no sign of them.
The Project Maths team and SEC have gone to a lot of effort this year implementing the new methods...in terms of in services for teachers on the new methods, printing sample questions and sending to all schools. I'm not saying that it's not the case but amazed that they wouldn't inform the rest of the country??

Funnily enough Examcraft's marking scheme this year had the 1/root n...so maybe.

MmmPancakes

so basically whenever you have to use error, you use the 1.96 one to be sure?

RoRo979

skippy1977 wrote: »

How official is this I wonder. We had kids come back with 'tips' from grinds schools (not that one) with detailed diagrams of things they said were coming up on Q4 and Q5 and ....no sign of them.
The Project Maths team and SEC have gone to a lot of effort this year implementing the new methods...in terms of in services for teachers on the new methods, printing sample questions and sending to all schools. I'm not saying that it's not the case but amazed that they wouldn't inform the rest of the country??

Funnily enough Examcraft's marking scheme this year had the 1/root n...so maybe.

ye i get where your coming from, im pretty sure its still excepted this year but sure might aswell use 1.96 to be safe

lala0815

dreading a probability question...anyone wanna give me any tips or should i just do exam questions and see where that leads me to !

is mise spartacus

RoRo979 wrote: »

attached pictures of q8 there however cant remember now if it is debs or examcraft, could some just check the first question and ensure I am right thanks

a(i)- marking scheme says 0.387
b(iii) according to the marking scheme...
step 1: p hat= 346/400=0.865
step 2: E= 1/root400=0.05
step 3: 0.865-0.05<p<0.865+0.05
0.815<p<0.915
step 4: Therefore the population proportion is within the confidence interval, no evidence to reject the company's claims

Fiona G

MmmPancakes wrote:

so basically whenever you have to use error, you use the 1.96 one to be sure?

Yeah that's how I remember it

Is anyone bothering with enlargements?

is mise spartacus

Fiona G wrote: »

Is anyone bothering with enlargements?

apparently they're due to come up ...

AnnaTorvFan

Fiona G wrote: »

Is anyone bothering with enlargements?

...what's an enlargement?

KatieGhomie

Help1997 wrote: »

Would it be ok if you could send me it aswell? Il pm you my email

Same here I am desperate I am so worried about tmro anything wud help

KatieGhomie

AnnaTorvFan wrote: »

...what's an enlargement?

hah u sound like me there's a whole chapter on enlargements in my book but I don't know it at all

is mise spartacus wrote: »

a(i)- marking scheme says 0.387
b(iii) according to the marking scheme...
step 1: p hat= 346/400=0.865
step 2: E= 1/root400=0.05
step 3: 0.865-0.05<p<0.865+0.05
0.815<p<0.915
step 4: Therefore the population proportion is within the confidence interval, no evidence to reject the company's claims

If only I knew what u were talking about

I am dreading tmro so much

simon360

Have a question from examcraft mock paper 2. Asks you to find the (new) approximate standard deviation of a new sample of 60 bulbs. Gives you the claimed (old) standard deviation of the company to be 3 months and mean 16 months. No idea what to do...

is mise spartacus

simon360 wrote: »

Have a question from examcraft mock paper 2. Asks you to find the (new) approximate standard deviation of a new sample of 60 bulbs. Gives you the claimed (old) standard deviation of the company to be 3 months and mean 16 months. No idea what to do...

3 divided by root 60
standard deviation over root n

RoRo979

simon360 wrote: »

Have a question from examcraft mock paper 2. Asks you to find the (new) approximate standard deviation of a new sample of 60 bulbs. Gives you the claimed (old) standard deviation of the company to be 3 months and mean 16 months. No idea what to do...

check the picture i put up, pretty sure i did that out

right quick question, write down the answer you think and justify it because im looking at the answer and aint gotta clue

Eight cards 1-8
1 and 2 are red,
3 and 4 are blue
5 and 6 are yellow
7 and 8 are black
Four,cards are selected at random

Find the probability that the four cards selected are:
.i) all of different colours
(II) two add-numbered .cards and two even numbered cards
(iii) all of different colours two odd-numbered and to even.numbered.

q2)
There are 30 days inJune. Seven people have their birthdays in June
The birthdays are independent of each other and all dates are equaly likely.
i) What is the probability that all seven students have the same birthday
(ii) What is the probability that all seven students have different birthdays?

Fiona G

AnnaTorvFan wrote: »

...what's an enlargement?

Sure it's all in the tables lads...allll in the tables

is mise spartacus

RoRo979 wrote: »

check the picture i put up, pretty sure i did that out

right quick question, write down the answer you think and justify it because im looking at the answer and aint gotta clue

Eight cards 1-8
1 and 2 are red,
3 and 4 are blue
5 and 6 are yellow
7 and 8 are black
Four,cards are selected at random

Find the probability that the four cards selected are:
.i) all of different colours
(II) two add-numbered .cards and two even numbered cards
(iii) all of different colours two odd-numbered and to even.numbered.

q2)
There are 30 days inJune. Seven people have their birthdays in June
The birthdays are independent of each other and all dates are equaly likely.
i) What is the probability that all seven students have the same birthday
(ii) What is the probability that all seven students have different birthdays?

is i) 8 over 32?

RoRo979

is mise spartacus wrote: »

is i) 8 over 32?

unfortunately not, your close however, ill pop the answers in a few messages so people dont see it immediately if they wanna give it a go

OMGeary

RoRo979 wrote: »

check the picture i put up, pretty sure i did that out

right quick question, write down the answer you think and justify it because im looking at the answer and aint gotta clue

Eight cards 1-8
1 and 2 are red,
3 and 4 are blue
5 and 6 are yellow
7 and 8 are black
Four,cards are selected at random

Find the probability that the four cards selected are:
.i) all of different colours
(II) two add-numbered .cards and two even numbered cards
(iii) all of different colours two odd-numbered and to even.numbered.

q2)
There are 30 days inJune. Seven people have their birthdays in June
The birthdays are independent of each other and all dates are equaly likely.
i) What is the probability that all seven students have the same birthday
(ii) What is the probability that all seven students have different birthdays?

Can you give us the answers, cos I'm working it out and want to see if I'm getting them right first

Kremin

RoRo979 wrote: »

check the picture i put up, pretty sure i did that out

right quick question, write down the answer you think and justify it because im looking at the answer and aint gotta clue

Eight cards 1-8
1 and 2 are red,
3 and 4 are blue
5 and 6 are yellow
7 and 8 are black
Four,cards are selected at random

Find the probability that the four cards selected are:
.i) all of different colours
(II) two add-numbered .cards and two even numbered cards
(iii) all of different colours two odd-numbered and to even.numbered.

q2)
There are 30 days inJune. Seven people have their birthdays in June
The birthdays are independent of each other and all dates are equaly likely.
i) What is the probability that all seven students have the same birthday
(ii) What is the probability that all seven students have different birthdays?

Probability of red is 1/4, blue 1/4, yellow 1/4, black 1/4

1/4*1/4*1/4*1/4=1/256?
Unless when the cards are chosen they arent replaced? So then it would be like 2/8*2/7*2/6*2/5

is mise spartacus

RoRo979 wrote: »

unfortunately not, your close however, ill pop the answers in a few messages so people dont see it immediately if they wanna give it a go

ah it was worth a shot me and probability don't get on

BlueWolf16

Fiona G wrote: »

Sure it's all in the tables lads...allll in the tables

Where at? I think I had it one time, can't seem to find it now..

OMGeary

(2^4)/(8x7x6x5) = 1/105 (4!)
ans:8/35

RoRo979

answers:
q1)
5/35
18/35
3/35

q2)
1/30^6
2639/5625

i need help with part ii and iii of q1 and q2 part i

Kremin

OMGeary wrote: »

(2^4)/(8x7x6x5) = 1/105 (4!)
ans:8/35

Why do you multiply by 4 factorial?

MmmPancakes

Can someone explain the central limit theorem to me?

Kremin

MmmPancakes wrote: »

Can someone explain the central limit theorem to me?

From my understanding, when a sample goes over n=40, it is always normally distributed and the mean of the samples will equal the mean of the population.

http://www.gillmacmillan.ie/AcuCustom/Sitename/DAM/058/NCPM_5_Online_Supplement.pdf page 17

Fiona G

BlueWolf16 wrote:

Where at? I think I had it one time, can't seem to find it now..

That was a poor attempt at humour in my last post sorry there's nothing about enlargements in the tables :P Ypu don't need formulae to do the qs anyway

skippy1977

Fiona G wrote: »

That was a poor attempt at humour in my last post sorry there's nothing about enlargements in the tables :P Ypu don't need formulae to do the qs anyway

Couple of handy things to remember.

Scale Factor, k = image length/ object length.

Image area = object area x k^2

Image volume = object volume x k^3

skippy1977

Kremin wrote: »

From my understanding, when a sample goes over n=40, it is always normally distributed and the mean of the samples will equal the mean of the population.

http://www.gillmacmillan.ie/AcuCustom/Sitename/DAM/058/NCPM_5_Online_Supplement.pdf page 17

Think it's 30

JohnDunne

Anyone know how to tell if a question's asking hypothesis testing or P values?

MmmPancakes

Kremin wrote: »

From my understanding, when a sample goes over n=40, it is always normally distributed and the mean of the samples will equal the mean of the population.

http://www.gillmacmillan.ie/AcuCustom/Sitename/DAM/058/NCPM_5_Online_Supplement.pdf page 17

what does it mean by the mean of the samples? Say if we had a data array {1,2,3,4,5,6,7,8,9,10}

If I took a sample, say 7-10 and found the mean (8.5), and then 1-3, and so on for loads of samples, and plotted all the means, the curve would be normal?