binomial dist : ( (3 Viewers)

synthesisFR · Oct 1, 2023

A company produces fuses that are 95% non defective. A batch of 1000 is produced. 5 are randomly picked and tested. What is the probability at least 1 defective is found.

the solutions have
Pr at least 1 = 1 - pr none
pr (none) = 5C0 (0.95)^5(0.05)^0

But like for my working i did
pr (none) = 5C5 (0.95)^5(0.05)^0

sorry if this is a stupid question but does this matter im not too sure? Ik the answer is obv the same.
But like understanding this notation. Pr (none) is so that none of them are defective yeah? and so u would do 5C5 bc u want none of them to be defective right!! because we defined 0.95 as the probability of a success since thats what we want?

wfj · Oct 1, 2023

5C0 and 5C5 are equal so it doesnt really matter. It depends as what you define as a success in this trial, but the answer you'll get is the same either way.

synthesisFR · Oct 1, 2023

so here if u defined 0.05 i.e faulty as the success
so u do:
pr (none) = 5C0 (0.05)^0(0.95)^5 because u want to get none of them as faulty and so its just 5 of them as not faulty?

synthesisFR · Oct 1, 2023

@carrotsss u lied to tf is this
its just applying the formula on the data sheet right

scaryshark09 · Oct 1, 2023

synthesisFR said:
View attachment 40112
@carrotsss u lied to tf is this
its just applying the formula on the data sheet right

ummmm how do you do c to e @carrotsss

synthesisFR · Oct 1, 2023

theres these formula u need to know

E(X^2) = Var(X) + E^2(X)

E(3-2X) = E(3) - E(2X)
= E(3) - 2E(X)

Var(2/3X-900)
= (2/3)^2VarX - Var (900)
= (2/3)^2VarX - 0

scaryshark09 · Oct 1, 2023

synthesisFR said:
theres these formula u need to know

E(X^2) = Var(X) + E^2(X)

E(3-2X) = E(3) - E(2X)
= E(3) - 2E(X)

Var(2/3X-900)
= (2/3)^2VarX - Var (900)
= (2/3)^2VarX - 0

bruh are you joking??
ive never seen those? are they in the syllubus

synthesisFR · Oct 1, 2023

scaryshark09 said:
bruh are you joking??
ive never seen those? are they in the syllubus

isdk

carrotsss · Oct 1, 2023

synthesisFR said:
View attachment 40112
@carrotsss u lied to tf is this
its just applying the formula on the data sheet right

is that a hsc question though

synthesisFR · Oct 1, 2023

NESA YOU HAVE DONE IT FOR THE PAST 2 HSC EXAMS SO PLEASE THIS YEAR DONT PUT A Q13/14 ON THIS TOPIC PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE PLEASE

Luukas.2 · Oct 1, 2023

scaryshark09 said:
bruh are you joking??
ive never seen those? are they in the syllubus

The first one is the standard formula used to find the variance:

Var(X) = E(X²) - [E(X)]²

...but rearranged to make E(X²) the subject.

The others are results you could be asked to prove, but should still know...

E(aX + b) = aE(X) + b

which means if I take a set of data, multiple each data point by a and then add b (that is, I do a linear transformation), the new average will be the old average transformed in the same way - multiplied by a and then b is added.

Var(aX + b) = a²Var(X)

which means if I take a set of data and do a linear transformation (X -----> aX + b), the adding of b part will shift the position of the distribution but will not change its spread or standard deviation. The spread will increase by multiplying by a, though, so the standard deviation of the new distribution will be a times the standard deviation of the original distribution... and, since variance is the square of standard deviation, the new variance will be a² times the old variance.

scaryshark09 · Oct 1, 2023

Luukas.2 said:
The first one is the standard formula used to find the variance:

Var(X) = E(X²) - [E(X)]²

explain this please

Luukas.2 · Oct 2, 2023

scaryshark09 said:
explain this please

Suppose you toss a fair coin three times.

You get a probability distribution where

P(0 heads) = 1 / 8

P(1 head) = 3 / 8

P(2 heads) = 3 / 8

P(3 heads) = 1 / 8

E(X) = add up X values times their probabilities:

$E(X) = 0 \times \cfrac{1}{8} + 1 \times \cfrac{3}{8} + 2 \times \cfrac{3}{8} + 3 \times \cfrac{1}{8} = \cfrac{0 + 3 + 6 + 3}{8} = \cfrac{3}{2}$

E(X²) = add up X² values times their probabilities:

$E(X^2) = 0^2 \times \cfrac{1}{8} + 1^2 \times \cfrac{3}{8} + 2^2 \times \cfrac{3}{8} + 3^2 \times \cfrac{1}{8} = \cfrac{0 + 3 + 12 + 9}{8} = 3$

The variance (and standard deviation) is then determinable:

$Var(X) = E\left(X^2\right) - [E(X)]^2 = 3 - \left(\cfrac{3}{2}\right)^2 = \cfrac{3}{4}$

$Var(X) = \sigma^2 \quad \implies \quad \sigma = \sqrt{Var(X)} = \sqrt{\cfrac{3}{4}} = \cfrac{\sqrt{3}}{2} \approx 0.866...$

So, tossing a coin 3 times, you expect $\mu \pm \sigma = 1.5 \pm 0.866...$ heads on average.

scaryshark09 · Oct 2, 2023

temporarylol said:
I hated the question they gave last year, I did two questions similar to it and completely forgot how to do it on the day, truly the best time to have a mind-blank

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

scaryshark09 · Oct 2, 2023

Luukas.2 said:
Suppose you toss a fair coin three times.

You get a probability distribution where

P(0 heads) = 1 / 8

P(1 head) = 3 / 8

P(2 heads) = 3 / 8

P(3 heads) = 1 / 8

E(X) = add up X values times their probabilities:

$E(X) = 0 \times \cfrac{1}{8} + 1 \times \cfrac{3}{8} + 2 \times \cfrac{3}{8} + 3 \times \cfrac{1}{8} = \cfrac{0 + 3 + 6 + 3}{8} = \cfrac{3}{2}$

E(X²) = add up X² values times their probabilities:

$E(X^2) = 0^2 \times \cfrac{1}{8} + 1^2 \times \cfrac{3}{8} + 2^2 \times \cfrac{3}{8} + 3^2 \times \cfrac{1}{8} = \cfrac{0 + 3 + 12 + 9}{8} = 3$

The variance (and standard deviation) is then determinable:

$Var(X) = E\left(X^2\right) - [E(X)]^2 = 3 - \left(\cfrac{3}{2}\right)^2 = \cfrac{3}{4}$

$Var(X) = \sigma^2 \quad \implies \quad \sigma = \sqrt{Var(X)} = \sqrt{\cfrac{3}{4}} = \cfrac{\sqrt{3}}{2} \approx 0.866...$

So, tossing a coin 3 times, you expect $\mu \pm \sigma = 1.5 \pm 0.866...$ heads on average.

Oh ok thanks!
I use the formula var(x) = npq where q is 1-p so over never come across this before
And also I use E(X) = np

Luukas.2 · Oct 2, 2023

scaryshark09 said:
Oh ok thanks!
I use the formula var(x) = npq where q is 1-p so over never come across this before
And also I use E(X) = np

Those apply to a binomial distribution, and you are correct that they will work for the example that I gave (as it is binomial).

However, I am talking about a method applicable to any discrete probability distribution, and all of this was covered in Year 11 Advanced content, and is extended to continuous probability distributions in year 12 Advanced.

If this is new to you then you have a significant gap in advanced content.

synthesisFR · Oct 2, 2023

do we need to know these because in tutoring we used these formulas at the bottom to solve the questions

synthesisFR · Oct 2, 2023

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

synthesisFR · Oct 2, 2023

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????

scaryshark09 · Oct 2, 2023

Luukas.2 said:
However, I am talking about a method applicable to any discrete probability distribution, and all of this was covered in Year 11 Advanced content, and is extended to continuous probability distributions in year 12 Advanced.

If this is new to you then you have a significant gap in advanced content.

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

binomial dist : ( (3 Viewers)

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