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S1 and S2 - Silly Mistakes!

Hey Guys,

I'm sure the title is pretty obvious but I've got exams coming up in Maths and Further Maths. I really want to do well but S1 and S2 are holding me back. I understand how to do the calculations but I always make silly mistakes which is losing me marks or sometimes I misinterpret the question (frequently in Probability(S1) and Approximations and CDF AND PDF (S2)). Is there anyway I can improve in S1 and S2 and what I can do to avoid silly mistakes and not misinterpret the question? Do I need to understand the topic more? I work hard but sometimes it's annoying when I make mistakes but it makes me feel as if I'm working hard for nothing.
Reply 1
Original post by Logic938
Hey Guys,

I'm sure the title is pretty obvious but I've got exams coming up in Maths and Further Maths. I really want to do well but S1 and S2 are holding me back. I understand how to do the calculations but I always make silly mistakes which is losing me marks or sometimes I misinterpret the question (frequently in Probability(S1) and Approximations and CDF AND PDF (S2)). Is there anyway I can improve in S1 and S2 and what I can do to avoid silly mistakes and not misinterpret the question? Do I need to understand the topic more? I work hard but sometimes it's annoying when I make mistakes but it makes me feel as if I'm working hard for nothing.


Spend more time reading the questions. Instead of working out the answers immediately, convert words into equations as you read the question. So, for example, if you're reading "the height is normally distributed with mean 5 and standard deviation 8" write down HN(5,82)H \sim N(5, 8^2) and then continue reading the question, noting down other important words, etc...

Always check your answers to CDF's by plugging in the endpoint and making sure the answer you get is 1, etc...
Reply 2
Original post by Zacken
Spend more time reading the questions. Instead of working out the answers immediately, convert words into equations as you read the question. So, for example, if you're reading "the height is normally distributed with mean 5 and standard deviation 8" write down HN(5,82)H \sim N(5, 8^2) and then continue reading the question, noting down other important words, etc...

Always check your answers to CDF's by plugging in the endpoint and making sure the answer you get is 1, etc...


Thanks a lot for the advice! Sometimes when I read a question, there are 2 interpretations/calculations of doing the question which pop up in my head and sometimes I feel like I have to make a "lucky guess" as to which method I would use to answer the questions correctly.
Reply 3
Original post by Logic938
Thanks a lot for the advice! Sometimes when I read a question, there are 2 interpretations/calculations of doing the question which pop up in my head and sometimes I feel like I have to make a "lucky guess" as to which method I would use to answer the questions correctly.


Could you prove an example of this?
Reply 4
Original post by Zacken
Could you prove an example of this?


I'll message an S2 to you.
Reply 5
Original post by Logic938
I'll message an S2 to you.


It'd be better if you posted it here so others can benefit as well.
Reply 6
Original post by Zacken
It'd be better if you posted it here so others can benefit as well.


This was a past paper question I did yesterday:

On a typical weekday morning customers arrive at a village post office independently and at a rate of 3 per 10 minute period.

Find the probability that
- at least 4 customers arrive in the next 10 minutes
- no more than 7 customers arrive between 11am and 11.30am

The period from 11am to 11.30am next tuesday morning will be divided into 6 periods of 5 minutes each
- Find the probability that no customers arrive in at most one of these periods

The post office is open for 3.5 hours on Wednesday mornings.
-Using a suitable approximation, estimate the probability that more than 49 customers arrive at the post office next Wednesday morning.
Original post by Logic938
.....


Which part of this question did you struggle on?
Reply 8
Original post by iMacJack
Which part of this question did you struggle on?


a and b were fine but c and d screwed me.
Reply 9
Original post by Logic938
This was a past paper question I did yesterday:


This is what I'd think as I read the question, analysed bit by bit.

On a typical weekday morning customers arrive at a village post office independently and at a rate of 3 per 10 minute period.


Right away, I write down XPo(3)X \sim \text{Po}(3) where XX is the r,v "# of arriving customers"

Find the probability that


I'm not thinking that the next two parts are going to be me either using the tables to find cumulative probabilities or calculating it exactly using the formula.

- at least 4 customers arrive in the next 10 minutes


The words 'at least' tells me I need to have a \geq. I immediately write down P(X4)P(X \geq 4). Now that I've written it down, I think about it. This means I need P(X4)=1P(X<4)=1P(x3)\displaystyle P(X \geq 4) = 1 - P(X < 4) = 1- P(x\leq 3) and now I use the tables.

- no more than 7 customers arrive between 11am and 11.30am


The words 'no more' immediately tells me I need to have a \leq. I look at the time duration, they've specified it here for some reason and what's that? It's because it's no longer a 10 minute block. So I adapt my distribution:

XPo(3×3=9)\displaystyle X \sim \text{Po}(3 \times 3 = 9). I now want P(X7)P(X \leq 7) which is easy to do via tables.

The period from 11am to 11.30am next tuesday morning will be divided into 6 periods of 5 minutes each


Five minutes each. I immedaitely adapt my distribution:

XPo(32=1.5)\displaystyle X \sim \text{Po}\left(\frac{3}{2} = 1.5\right)

- Find the probability that no customers arrive in at most one of these periods


I first want the probability that no customer arrives in one period. This is obviously P(X=0)P(X=0). Now I have 6 periods. This is begging out for the use of a binomial distribution, so I immediately write down
Unparseable latex formula:

\Y \sim B(6, P(X=0))

.

I want "at most" one of these periods. Hence P(Y1)P(Y \leq 1). Again, tables.

The post office is open for 3.5 hours on Wednesday mornings.


3.5 hours - that's obviously 3.5×60=2103.5 \times 60 = 210 minutes which means I immediately adapt my distribution:

X(3×21)\displaystyle X \sim \left(3 \times 21\right).

-Using a suitable approximation, estimate the probability that more than 49 customers arrive at the post office next Wednesday morning.


I have poisson, and I want an approximation. This must mean it's a normal. Then I do continuity corrections, etc... and I'm there.
Reply 10
Original post by Zacken
...


This helped me - thank you!

Have you ever seen a tricky hypothesis testing question on past papers before? It always seems like plug and chug to me.
Reply 11
Original post by aymanzayedmannan
This helped me - thank you!

Have you ever seen a tricky hypothesis testing question on past papers before? It always seems like plug and chug to me.


And I need to question why I've spelt immediately wrong so many times in that post. :lol:

No worries. I think I've seen one or two on an S2 paper, mostly you questioning yourself whether you want a two or a one tailed test, like the very last question on my Jan one.

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