binomial dist : ( (2 Viewers)

Luukas.2 · Oct 2, 2023

synthesisFR said:
View attachment 40118 View attachment 40119 View attachment 40120
Im pretty sure we need to know all this except for the third image right?
Apparantly the teacher skipped B)Sampling without replacement (third image)
and Also apparently skipped continuity correction because you dont need it for the hsc????

All of the content you have quoted appears in various texts and trial papers, though without the detail.

The last image may sound unfamiliar but I'd bet you've done examples of it by drawing a tree diagram for selections without replacement.

Suppose you toss a bag has 6 red balls and 4 green balls, and you select 3 of them (without replacement. You get a probability distribution where

$P(\text{0 red}) = P(\text{GGG}) = \cfrac{4}{6 + 4} \times \cfrac{3}{6 + 3} \times \cfrac{2}{6 + 2} = \cfrac{2}{5} \times \cfrac{1}{3} \times \cfrac{1}{4} = \cfrac{1}{30}$

$\begin{align*} P(\text{1 red}) &= P(\text{RGG or GRG or GGR}) \\ &= \left(\cfrac{6}{6 + 4} \times \cfrac{4}{5 + 4} \times \cfrac{3}{5 + 3}\right) + \left(\cfrac{4}{6 + 4} \times \cfrac{6}{6 + 3} \times \cfrac{3}{5 + 3}\right) + \left(\cfrac{4}{6 + 4} \times \cfrac{3}{6 + 3} \times \cfrac{6}{6 + 2}\right) \\ &= \left(\cfrac{3}{5} \times \cfrac{4}{9} \times \cfrac{3}{8}\right) + \left(\cfrac{2}{5} \times \cfrac{2}{3} \times \cfrac{3}{8}\right) + \left(\cfrac{2}{5} \times \cfrac{1}{3} \times \cfrac{3}{4}\right) \\ &= \cfrac{1}{10} + \cfrac{1}{10} + \cfrac{1}{10} = \cfrac{3}{10} \end{align*}$

$\begin{align*} P(\text{2 red}) &= P(\text{RRG or RGR or GRR}) \\ &= \left(\cfrac{3}{5} \times \cfrac{5}{9} \times \cfrac{1}{2}\right) + \left(\cfrac{3}{5} \times \cfrac{4}{9} \times \cfrac{5}{8}\right) + \left(\cfrac{2}{5} \times \cfrac{2}{3} \times \cfrac{5}{8}\right) \\ &= 3 \times \cfrac{1}{6} = \cfrac{1}{2} \end{align*}$

$P(\text{3 red}) = P(\text{RRR}) = \cfrac{3}{5} \times \cfrac{5}{9} \times \cfrac{1}{2} = \cfrac{1}{6}$

From this distribution we can calculate:

$E(X) = 0 \times \cfrac{1}{30} + 1 \times \cfrac{3}{10} + 2 \times \cfrac{1}{2} + 3 \times \cfrac{1}{6} = \cfrac{3 + 10 + 5}{10} = \cfrac{9}{5}$

$E(X^2) = 0^2 \times \cfrac{1}{30} + 1^2 \times \cfrac{3}{10} + 2^2 \times \cfrac{1}{2} + 3^2 \times \cfrac{1}{6} = \cfrac{0 + 3 + 20 + 15}{10} = \cfrac{19}{5}$

$Var(X) = E\left(X^2\right) - [E(X)]^2 = \cfrac{19}{5} - \left(\cfrac{9}{5}\right)^2 = \cfrac{95 - 81}{25} = \cfrac{14}{25}$

$Var(X) = \sigma^2 \quad \implies \quad \sigma = \sqrt{Var(X)} = \sqrt{\cfrac{14}{25}} = \cfrac{\sqrt{14}}{5} \approx 0.748...$

So, choosing 3 balls, you expect $\mu =1.8$ red (or average) with a variance of $\sigma^2 = 0.56$

Using the normal as a guide, $(\mu - \sigma,\ \! \mu + \sigma) \approx (1.05,\ \! 2.55)$ and $(\mu - 2\sigma,\ \! \mu + 2\sigma) \approx (0.30,\ \! 3.30)$ . These are the intervals for $|z| \le 1$ and $|z| \le 2$ , and we can compare these to the discrete probabilities:

P(1 or 2 red) = 0.8, which is a little more than the 0.68 expected for the $|z| < 1$ , but this uses a wider interval
P(1 or 2 or 3 red) = 0.966..., which is a little more than the 0.95 expected for the $|z| < 2$

These are in good agreement considering we don't actually have a normal distribution.

mmmmmmmmaaaaaaa · Oct 2, 2023

scaryshark09 said:
How many people do you reckon got it in the state?
Highest aligned mark last year was 99 which would have been 97-99 raw

12e or 13e? they were pretty chill

Luukas.2 · Oct 2, 2023

scaryshark09 said:
wait this is new to me and i did accelerated advanced last year. I got 98 also and never learnt this

Grove (Maths in Focus), Yr 11 Advanced, ,Chapter 10
Cambridge Yr 11 (Advanced), Chapter 11
Cambridge Year 11 (Extension 1), Chapter 13

synthesisFR · Oct 2, 2023

hey guys for this question when we get to 3.125
Pr of 3.12 is 0.99910
whilst pr of 3.13 is 0.99913
our tutor wrote it as 0.99910 so i assume that we have to round down to the nearest decimal on the table?

Luukas.2 · Oct 2, 2023

synthesisFR said:
hey guys for this question when we get to 3.125
Pr of 3.12 is 0.99910
whilst pr of 3.13 is 0.99913
our tutor wrote it as 0.99910 so i assume that we have to round down to the nearest decimal on the table?

Being picky about it, Is there any number that is less than 3.125 that rounds to 3.13 (2 dec. pl.)?

Alternatively, one could argue that the answer is 0.99910 < P < 0.99913 and so P must be 0.9991 (4 sig. fig.)

synthesisFR · Oct 2, 2023

now im confused loll is whats written there fine?

Luukas.2 · Oct 2, 2023

synthesisFR said:
now im confused loll is whats written there fine?

You had P( z < 3.125). If z is less than 3.125 then it can't be 3.13 to 2 decimal places, is my point.

Masaken · Oct 2, 2023

synthesisFR said:
@Masaken sorry u probably know this ^^
does the hsc accept these or use these?

this is sample proportion stuff right?? I've used those before in sample qs when it's appropriate to

synthesisFR said:
View attachment 40118 View attachment 40119 View attachment 40120
Im pretty sure we need to know all this except for the third image right?
Apparantly the teacher skipped B)Sampling without replacement (third image)
and Also apparently skipped continuity correction because you dont need it for the hsc????

we also skipped content in the third image lol
continuity correction was taught in my school (prob cos it was in the Cambridge textbook lol) but (someone corroborate me on this) apart from like 2 qs I've never had to apply it or use it unless I've been doing it wrong

synthesisFR · Oct 2, 2023

Masaken said:
this is sample proportion stuff right?? I've used those before in sample qs when it's appropriate to

we also skipped content in the third image lol
continuity correction was taught in my school (prob cos it was in the Cambridge textbook lol) but (someone corroborate me on this) apart from like 2 qs I've never had to apply it or use it unless I've been doing it wrong

so do we not have to know continuity correction? bc im pretty sure sir said we dont need to.
and looking at one of the question from last years hsc, they gave four sample answers and only one was cont correction. so i assume not?

synthesisFR · Oct 2, 2023

Luukas.2 said:
You had P( z < 3.125). If z is less than 3.125 then it can't be 3.13 to 2 decimal places, is my point.

also lets say it was
z <= 3.13
here we include 3.13
z < 3.13
our tutor taught us to also use the same number for this
apparantly it doesn't matter for the hsc

carrotsss · Oct 2, 2023

synthesisFR said:
also lets say it was
z <= 3.13
here we include 3.13
z < 3.13
our tutor taught us to also use the same number for this
apparantly it doesn't matter for the hsc

yes because the amount that exactly equals 3.1300000 is extremely negligible

synthesisFR · Oct 2, 2023

carrotsss said:
yes because the amount that exactly equals 3.1300000 is extremely negligible

help me wtf is continuity correction, we dont need to know it right? but what is it tho so i know what its about

synthesisFR · Oct 2, 2023

this is using sample sum ^^ could someone explain this a bit to me cuz i can do it but i dont fully get it. like i can do the formulas but not the S = 5x60 and why we do that and set up Pr < 300 etc using the sample sum.
I haven't done any sample sum questions.

breadcrumbs · Oct 2, 2023

synthesisFR said:
help me wtf is continuity correction, we dont need to know it right? but what is it tho so i know what its about

my teacher taught it to us bc they thought it was kinda fascinating but its like a method where u plus or minus 0.5 to the inequalities to give u a more accurate answer in the end when u use normal approx
someone correct me if im wrong

cossine · Oct 2, 2023

synthesisFR said:
View attachment 40123
this is using sample sum ^^ could someone explain this a bit to me cuz i can do it but i dont fully get it. like i can do the formulas but not the S = 5x60 and why we do that and set up Pr < 300 etc using the sample sum.
I haven't done any sample sum questions.

I will need to double check to make sure but it seems like they are using central limit theorem. The issue is there 5 lines. So to make the calculation simpler it is all converted to one line.

synthesisFR · Oct 2, 2023

synthesisFR said:
View attachment 40123
this is using sample sum ^^ could someone explain this a bit to me cuz i can do it but i dont fully get it. like i can do the formulas but not the S = 5x60 and why we do that and set up Pr < 300 etc using the sample sum.
I haven't done any sample sum questions.

OH is it like for this probability since it involves sample sum
u need cumulative time taken for all individuals
so like 60 minutes overall is equivalent to the cumulative sum of 5x60 = 300 minutes because if u can serve these 5 sets of people in 60 minutes each at the same time its 300?
brain is boggled but i kinda get it??

carrotsss · Oct 2, 2023

synthesisFR said:
help me wtf is continuity correction, we dont need to know it right? but what is it tho so i know what its about

i don’t think it’s required for the hsc, you can use it if you want though

synthesisFR · Oct 2, 2023

yoo this topic is not that bad i did some experimenting we can use multiple methods for one question this is kinda nice
so we can use sample mean and sample sum depending on how u approach it

synthesisFR · Oct 2, 2023

synthesisFR said:
View attachment 40123
this is using sample sum ^^ could someone explain this a bit to me cuz i can do it but i dont fully get it. like i can do the formulas but not the S = 5x60 and why we do that and set up Pr < 300 etc using the sample sum.
I haven't done any sample sum questions.

still dont fully get this tho

synthesisFR · Oct 2, 2023

pls

synthesisFR said:
still dont fully get this tho

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