I just discovered this- thank you SO much for taking the time to make this! Your working is really clear to follow (and you have quite a relaxing voice, so that's a plus)
Maybe you are thinking of finding the median? But the question is asking how many days 1 s.d above mean. You just need an estimate of days that had more than 10.32 hours of sunshine. No linear interpolation needed
@Mirco Ma I can't think of a better way to explain it than to say it's a proportion. (11-10.32)/3 is the proportion of days above 10.32 hours of sunlight. given there are 8 days in that group, multiply by 8 to get an estimate of the number of days with more than 10.32 hours of sunlight. An example: suppose you have 10 friends and I tell you that 5 are above 160cm, and 3 are above 180cm. Estimate how many are taller than 170cm. You would say 2 are between 160-180cm. So your estimate would be that one of those is taller than 170cm. Therefore in total you have 4 friends taller than 170cm. That's just a quick example, probably there are better ways to explain it. If it is still confusing, I know there are other walk throughs of this paper. Sometimes a difference explanation will clear up confusion.
I don’t understand the method being used for q9 at all. It seems like you’re equating moment forces and normal reaction forces when they aren’t the same thing. Could you please explain what’s going on I’ve been trying to wrap my head around this answer for hours now.
Not for this question. Class boundaries are used for discrete data with class intervals like 10-20, 21-30.. where you have a "gap" between 20 and 21. Then you might use a class boundary.
I do not understand how on question 1a how you got that height. I keep getting 8.8886666. I got the same width as 2.25 not 2.5 also. Can u help me please?
You can follow the method in the mark scheme if that makes more sense to you. It finds the frequency per 1cm^2 by dividing the frequency by total area. so for the third class total area is 8 x 1.5 = 12cm^2 (they give you those dimensions in the question) and frequency per 1cm^2 = frequency/area = 8/12 = 2/3. Then using that 2/3, if you have a frequency of 12, the area of the bar must then = 18cm^2. (because 12/(2/3) = 18). Then if the width is 2.5, the height will be 18/2.5 = 7.2 Personally I find the method using ratios more straightforward, but maybe you would prefer that method. Question 1a in this paper is exactly the same: ruclips.net/video/Dseb6W2tWq4/видео.html if you wanted to practise another one using a method you prefer.
In the question they say "when the boat is north-east of O". You need to understand what this means. I did my best to explain it briefly, but if you do not understand what "north-east of O" means, you will need to do some research into bearings. Once you understand what that means, you will see that "north-east of O" also means the i and j components are equal.
Yh I think I understand now , thanks a lot it’s because NE is in the middle and so it splits into 2 equal parts so magnitudes are equal to each other. Right?
when it says "projected" this basically means pushed. So there was an initial push to get it moving up the slope but no sustained force to keep it moving up other than momentum from that first push.
For me it's really in the name. If events are dependent, then the probability of A happening *depends* on if B happens or not. Therefore to work out P(A), you need to consider if B happened or not. This is called conditional probability. This is written as P(AlB). [The probability of A given B happened]. If you want to work out P(AnB), you need to work out the P(AlB) x P(B). This does not equal P(A)xP(B). This series on conditional probability might be helpful: ruclips.net/video/xAXtD36VXL8/видео.html
Maths Explained oh so if they’re dependent, the probs of A depends on if B happened or not. However when independent the probs of A doesn’t depend on whether B happened or not and therefore P(AnB) would be equal to P(AlB)????.
On Q3, I’m using an fx-CG50 calc and I got half the value for the z score, instead I got 0.9799 or 0.98 why is this? As it done the next step for me by multiplying 0.5 to it?
With that calculator you are finding the boundary x value, not the z-score. So yes the .9799 is the 1.96 x 0.5. If you do want a z-score, make sure to leave the mean and standard deviation and 0 and 1.
Maths Explained Thanks for the reply I’m really struggling with like Normal and Binomial Distributions, you make it so much easier than it is (practice). Like I get confused when u have to convert the binomial distribution to a normal one and visa versa. Distributions are the hardest part of stats for me especially when u add in hypothesis testing. Another question is how do I know when to use tables instead of calc like surely there must be a reason as to why u use tables instead of calculator because calc is so much easier.
Maths Explained I think I need to recap some of the equations tbh like I was never taught about the z score, Maybe cause I didn’t finish the course fully, I’m having to retake exams in 3 weeks
Maths Explained Think I’ve answered my own question because when X~N(n,p) needs to be changed to binomial distribution you do m=np then SD^2 = np(1-p) where SD=root of variance so it would be X~B(m,(root variance)^2) so when u put the value of SD in calculator u really are doing root variance, think I got confused in thing SD^2 was the value u put in calc. Where it’s meant to be SD or root of variance???
timestamps:
Q1 - 00:00:35
Q2 - 00:14:28
Q3 - 00:21:26
Q4 - 00:33:27
Q5 - 00:44:53
Q6 - 00:57:16
Q7 - 01:03:14
Q8 - 01:15:29
Q9 - 01:24:53
Q10 - 01:36:37
Pro trick : watch movies at kaldrostream. I've been using them for watching a lot of movies during the lockdown.
@Keaton Maurice yea, have been watching on kaldroStream for months myself :D
No way - you’re back! Just in time for me starting my A level - Christmas has come early.
I just discovered this- thank you SO much for taking the time to make this! Your working is really clear to follow (and you have quite a relaxing voice, so that's a plus)
Yess you're back! Hope you do more A level content
YOU ARE FINALLY BACK ! I REALLY WANTED TO SEE THIS VIDEO ! So nice seeing you around again !
I can't seem to find the mechanics half of this. Any idea where to get it from so I can use a paper copy before watching the video. Thanks.
YES YOU’RE BACK!
Exam tmrw but how come you can’t use ven diagram for question 4 a??
I think you need to use the conditional probability formula. I can’t see how to get the answer directly from a Venn diagram
34:57 (for myself)
You're awesome
heyy, thank you so muchh for thiss..they are really helpful for me to do in timed conditions, please keeping making more of this.
Question D - surely for linear interpolation is it 10.32-8/3 then multiplied by 8 ?
Maybe you are thinking of finding the median? But the question is asking how many days 1 s.d above mean. You just need an estimate of days that had more than 10.32 hours of sunshine. No linear interpolation needed
@Mirco Ma I can't think of a better way to explain it than to say it's a proportion. (11-10.32)/3 is the proportion of days above 10.32 hours of sunlight. given there are 8 days in that group, multiply by 8 to get an estimate of the number of days with more than 10.32 hours of sunlight.
An example: suppose you have 10 friends and I tell you that 5 are above 160cm, and 3 are above 180cm. Estimate how many are taller than 170cm. You would say 2 are between 160-180cm. So your estimate would be that one of those is taller than 170cm. Therefore in total you have 4 friends taller than 170cm. That's just a quick example, probably there are better ways to explain it. If it is still confusing, I know there are other walk throughs of this paper. Sometimes a difference explanation will clear up confusion.
I don’t understand the method being used for q9 at all. It seems like you’re equating moment forces and normal reaction forces when they aren’t the same thing. Could you please explain what’s going on I’ve been trying to wrap my head around this answer for hours now.
could be something wrong in my explanation. I openly admit mechanics is not my strong point and it took me a while to improve my explanations.
You're a legend
Could anyone tell me from where did we know the sample size of the strips (which is 10) in Q3? 27:00
From part (b)
QUESTION 1a) Do you not need to take the lower and upper bounds of the class intervals for a level?
Not for this question. Class boundaries are used for discrete data with class intervals like 10-20, 21-30.. where you have a "gap" between 20 and 21. Then you might use a class boundary.
37:25 Where did the 0.55 denominator come from, what does it mean?
P(B’) = 0.55. I was using the conditional probability formula there
I do not understand how on question 1a how you got that height. I keep getting 8.8886666. I got the same width as 2.25 not 2.5 also. Can u help me please?
You can follow the method in the mark scheme if that makes more sense to you. It finds the frequency per 1cm^2 by dividing the frequency by total area. so for the third class total area is 8 x 1.5 = 12cm^2 (they give you those dimensions in the question) and frequency per 1cm^2 = frequency/area = 8/12 = 2/3.
Then using that 2/3, if you have a frequency of 12, the area of the bar must then = 18cm^2. (because 12/(2/3) = 18). Then if the width is 2.5, the height will be 18/2.5 = 7.2
Personally I find the method using ratios more straightforward, but maybe you would prefer that method. Question 1a in this paper is exactly the same: ruclips.net/video/Dseb6W2tWq4/видео.html if you wanted to practise another one using a method you prefer.
@@mathonify thank u I understand it now. the ratio method Is much easier
Starting A level maths this year after getting a 9 at GCSE with your help,keep the videos coming and welcome back! :D
Btw, is A levels maths hard. I will be doing A maths this year.
parkour .11Parkour yes it’s hard
Hi I don’t understand q8 part c, how did you get the equation 0.6-0.05t=0.35t, why would the i and j components be equal to each other? Many thanks.
In the question they say "when the boat is north-east of O". You need to understand what this means. I did my best to explain it briefly, but if you do not understand what "north-east of O" means, you will need to do some research into bearings. Once you understand what that means, you will see that "north-east of O" also means the i and j components are equal.
Yh I think I understand now , thanks a lot it’s because NE is in the middle and so it splits into 2 equal parts so magnitudes are equal to each other. Right?
@@Hhxhx531 exactly!
For question 7 it says the particle moves up the plane so surely there must be a force acting up the slope right?
when it says "projected" this basically means pushed. So there was an initial push to get it moving up the slope but no sustained force to keep it moving up other than momentum from that first push.
@@mathonify thanks
You know If P(A) x P(B) doesn’t equal P(AnB) why are they not independent?
For me it's really in the name. If events are dependent, then the probability of A happening *depends* on if B happens or not. Therefore to work out P(A), you need to consider if B happened or not. This is called conditional probability. This is written as P(AlB). [The probability of A given B happened].
If you want to work out P(AnB), you need to work out the P(AlB) x P(B). This does not equal P(A)xP(B).
This series on conditional probability might be helpful: ruclips.net/video/xAXtD36VXL8/видео.html
Maths Explained oh so if they’re dependent, the probs of A depends on if B happened or not. However when independent the probs of A doesn’t depend on whether B happened or not and therefore P(AnB) would be equal to P(AlB)????.
Maths Explained Also thanks for the link I’ll watch it.
For part 48:47 why is it more than or equal to 5?
As there are 10 trays “at least half” in the question means 5 or more.
@@mathonify ahh makes sense thank you 😁
If fd = 8/3 how can that also equal 8?
frequency density is 8/3, the height of the bar (the actual measured height of the bar in centimetres) is 8cm. two different scales.
@@mathonify Okay cheers
On Q3, I’m using an fx-CG50 calc and I got half the value for the z score, instead I got 0.9799 or 0.98 why is this? As it done the next step for me by multiplying 0.5 to it?
With that calculator you are finding the boundary x value, not the z-score. So yes the .9799 is the 1.96 x 0.5. If you do want a z-score, make sure to leave the mean and standard deviation and 0 and 1.
Maths Explained Thanks for the reply I’m really struggling with like Normal and Binomial Distributions, you make it so much easier than it is (practice). Like I get confused when u have to convert the binomial distribution to a normal one and visa versa. Distributions are the hardest part of stats for me especially when u add in hypothesis testing. Another question is how do I know when to use tables instead of calc like surely there must be a reason as to why u use tables instead of calculator because calc is so much easier.
Maths Explained I think I need to recap some of the equations tbh like I was never taught about the z score, Maybe cause I didn’t finish the course fully, I’m having to retake exams in 3 weeks
Maths Explained Think I’ve answered my own question because when X~N(n,p) needs to be changed to binomial distribution you do m=np then SD^2 = np(1-p) where SD=root of variance so it would be X~B(m,(root variance)^2) so when u put the value of SD in calculator u really are doing root variance, think I got confused in thing SD^2 was the value u put in calc. Where it’s meant to be SD or root of variance???
Maths Explained how come your calculator gave u double mine tho cause it was finding the z score instead but I don’t see why