AS AQA Mathematics Statistics 1B MS1B June 2014

The AS AQA Statistics 1B exam takes place on 06-06-14.

Feel free to use this thread to share revision sources and discuss the exam.

(edited 9 years ago)

Reply 1

NinjaMan009

I have this exam :smile:

Reply 2

muonfrisbee98

13

I've got this exam too :smile:

Do you know if we have to show our workings when finding the PMCC and the least squares regression line?

Reply 3

PulsarX

OP

Original post by muonfrisbee98

I've got this exam too :smile:

Do you know if we have to show our workings when finding the PMCC and the least squares regression line?

No, it's perfectly acceptable to punch the numbers into your calculator and write down the values for r, a and b.

Reply 4

PulsarX

OP

Also, here's some decent revision sources that I'm using for Statistics. It seems that they are few and far between...

http://www.mathsrevision.net/advanced-level-maths-revision/advanced-level-level-statistics

https://statistics.laerd.com/statistical-guides/normal-distribution-calculations.php

http://www.schoolworkout.co.uk/a_level.htm (this is a good one if you need revision notes)

It may be a bit late now but I've also been using the CGP Student Books for Maths all year and they are excellent. Check them out, especially if you're taking Maths in A2:

https://www.cgpbooks.co.uk/Student/books_as_maths

Happy revising :biggrin:

Reply 5

vee_wuvshugs

13

The S1 book is amazing. I wish I had it for core 2 :rolleyes:

Reply 6

muonfrisbee98

13

Original post by PulsarX

No, it's perfectly acceptable to punch the numbers into your calculator and write down the values for r, a and b.

Thanks

Reply 7

Pumuki63

10

Could someone explain confidence intervals vaguely and in the confidence intervals equation it says;
X - z (σ/√n) < μ < X + z (σ/√n).
What does X stand for?

Reply 8

PulsarX

OP

Original post by Pumuki63

Could someone explain confidence intervals vaguely and in the confidence intervals equation it says;
X - z (σ/√n) < μ < X + z (σ/√n).
What does X stand for?

Confidence interval - a range of values, calculated from sample data, with a specified probability of containing a population parameter.

Basically, it's a measure of how confident you are that the mean of a sample lies within two values of the population it was taken from. Think of it as placing a normal distribution graph for a sample over one for the entire population.

As for what X is in the equation, that is the sample mean.

Reply 9

Pumuki63

10

Original post by PulsarX

Confidence interval - a range of values, calculated from sample data, with a specified probability of containing a population parameter.

Basically, it's a measure of how confident you are that the mean of a sample lies within two values of the population it was taken from. Think of it as placing a normal distribution graph for a sample over one for the entire population.

As for what X is in the equation, that is the sample mean.

Thank you! that makes sense :biggrin:

, and how would you calculate z? (is that the table)

Reply 10

PulsarX

OP

Original post by Pumuki63

Thank you! that makes sense :biggrin:

, and how would you calculate z? (is that the table)

z = (x - mu) / sigma

So that's the x value minus the mean, all divided by the standard deviation.

Edit: z = (x − μ) / σ

(edited 9 years ago)

Reply 11

Pumuki63

10

Original post by PulsarX

z = (x - mu) / sigma

So that's the x value minus the mean, all divided by the standard deviation.

Ahhh thank you again!! :smile:

Reply 12

PulsarX

OP

Original post by Pumuki63

Ahhh thank you again!! :smile:

No problem :smile:

Reply 13

Photovoltaic

Original post by Mellyx

Hi, would anyone be able to help me with this question please. I've looked at the mark scheme and I am not sure on how to get the SD. Everyone seems to find it really easy :frown:

It's Jan 2013 Question 7

A machine, which cuts bread dough for loaves, can be adjusted to cut dough to any specified set weight. For any set weight, (mean symbol) grams, the actual weights of cut dough are known to be approximately normally distributed with a mean of (the mean symbol) grams and a fixed standard deviation of (standard deviation symbol) grams.

It is also known that the machine cuts dough to within 10 grams of any set weight.

7A))Estimate, with justification, a value for standard deviation . (2 marks)

The machine is set to cut dough to a weight of 415 grams.
She then asked him to calculate the mean and the standard deviation of his 15 recorded weights.
Dev subsequently reported to Sunita that, for his sample, the mean was 391 grams and the standard deviation was 95.5 grams.

B) Advise Sunita on whether or not EACH of Dec's values is likely to be correct. Give numerical support for you answers.
( 3 marks)

Confused me a little at first too, but it seems simpler than I at first thought.

It states "It is also known that the machine cuts dough to within 10 grams of any set weight. "

You know the machine is accurate to within 10 grams, this means it's accurate to +/- 10 grams of the mean.

A bell curve shows that in a normal distribution 99.9% occurs within +/- 3 standard deviations from the mean point.

So if we use only one side we can do 10g/3 = 3.33
You could alternatively use both sides and do the full 20g range divided by the full 6 standard deviation range, the result is the same obviously.

Not sure if you want help on part B or not, but as the mean of 391 lies outside of the lowest value in the range of 405g then this seems unlikely to be correct.
The standard deviation of 95.5 grams is also much greater than the value of 3.33 so is also unlikely to be correct

normal-distrubution-large.gif19.9KB

Reply 14

saad97

2

Hi guys, does anyone know whether we have to learn any equations for this exam or can they all be found in the formulae booklet? Thanks :smile:

Reply 15

PulsarX

OP

Original post by saad97

Hi guys, does anyone know whether we have to learn any equations for this exam or can they all be found in the formulae booklet? Thanks :smile:

Most of the equations you'll need are in there, but a few won't be. A few I can remember which we aren't given are as below. Does anyone know of any others???
z = (x − μ) / σ

μ +- z(σ / sqrt n)

(edited 9 years ago)

Reply 16

Photovoltaic

Original post by PulsarX

Most of the equations you'll need are in there, but a few won't be. A few I can remember which we aren't given are as below. Does anyone know of any others???
z = (x − μ) / σ

μ +- z(σ / sqrt n)

Yeah, those are the ones that sprung to my mind too, I'm not sure what others there are but may have a look through all the formulae soon and compare that with what's in the formulae booklet

Reply 17

CuteMissJessica

1

Can someone help me with normal distribution pleaseeee,
Right when you have something like p(x<60) and mean is 65 and SD is 20. I know you would have to do 60-65 divided by 20 and you get -0.25. So the you have to switch the signs around so it becomes >. But then how do you know when to substract the z score from 1.

What about when you have p(x>70) and you wouldnt end up with a negative number do still have to substact the z score from 1?? Im sooooooo confused, helpppp

Reply 18

Photovoltaic

Original post by CuteMissJessica

Can someone help me with normal distribution pleaseeee,
Right when you have something like p(x<60) and mean is 65 and SD is 20. I know you would have to do 60-65 divided by 20 and you get -0.25. So the you have to switch the signs around so it becomes >. But then how do you know when to substract the z score from 1.

What about when you have p(x>70) and you wouldnt end up with a negative number do still have to substact the z score from 1?? Im sooooooo confused, helpppp

Okay, so if you have P(X < 60) with a mean of 65 and SD of 20 then yep you do

\frac {60-65}{20} = -0.25

You have

P(z < -0.25)

due to the minus sign you swap to get

P(z > 0.25)

insead.

That gives you the probability that z is greater than 0.25, the tables however give you the probability that z is less than 0.25 from the 0.25 value of 0.59871.
In order to get the value for a greater than, this is when you need to do

1-p

so

1 - 0.59871 = 0.40129

That is the correct probability for P(X < 60)

The bit in bold is how you decide if you need to do 1 - or not, basically if P (z < whatever ) (z is less than) you do not need to do 1-, but if P (z > whatever ) (z is greater than) that is when you need to do 1 - the value as the tables give you the value for less than.

Hope that was helpful :smile:

Reply 19

Erelan

The only other formula I know of that hasn't been mentioned is the y=a+bx for the regression line which I thought I'd just note, just in case!
I am slightly terrified for this exam really!
I hope those are the only formulas I need to learn because I can't think of any others! Good luck to all in case I don't pop back after this post! :bunny2:

AS AQA Mathematics Statistics 1B MS1B June 2014

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you