 June 07, 2020, 03:02:39 pm

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#### RuiAce

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• Respect: +2518 « Reply #4395 on: October 23, 2019, 06:30:29 pm »
+2
So pretty much the variance and standard deviation are the exact same thing, except that variance is used to make the formulas look nicer and
working out more cleaner.
Yep.
Rui one more thing, for the Expected value of binomial distribution E(X) = np and Variance Var(X) = np(1-p), do we learn how to derive these formulas in school, cause I havent really seen anything in the books and if not, can u please show me how to derive these formulas???

Thanks These proofs are actually heavily involved. I had trouble doing them in first year; wasn't until second year university when I started finding it easy. Here's the proof for the expected value at least just for viewing pleasure.

(Note: So therefore you definitely won't need it at the high school level.)
\begin{align*}
E(X) &= \sum_x x P(X=x)\\
&= \sum_{x=0}^n x \binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=0}^n x \frac{n!}{x!(n-x)!}p^x (1-p)^{n-x}.
\end{align*}
Now, we first evaluate the sum at $x=0$. We pull the value when $x=0$ outside of the sum.
\begin{align*}
E(X) &= 0 \binom{n}{0}p^0 (1-p)^{n-0} + \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}
\end{align*}
because that first term actually evaluates to 0. Now we note that the terms in the sum are indexed by $x=1, 2, 3, \dots, n$. So every term in the sum now always has the property $x=0$, and hence we can cancel any $x$'s in the numerator and denominator.

In addition to this, we need the trick $N! = N(N-1)!$. (You can try to convince yourself that this formula is true by simply expanding both of the factorials.) Here, I first use that $x! = x(x-1)!$. (Note: The assumption for this question is that $x$ is an integer; not any real number.)
\begin{align*}
E(X)& = \sum_{x=1}^n x\binom{n}{x}p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n x \frac{n!}{x!(n-x)!} p^x (1-p)^{n-x}\\
&= \sum_{x=1}^n \frac{n!}{(x-1)(n-x)!} p^x (1-p)^{n-x}
\end{align*}
Observe that we had to tear apart the binomial coefficient (i.e. convert it to its factorial notation) to cancel out the $x$ in front. We now have to rebuild a new binomial coefficient using the remaining factorials. It turns out that we now need to manipulate it to obtain $\binom{n-1}{x-1}$.

This requires that we use $n! = n(n-1)!$ in the numerator. Once we do that, observe that the sum is in terms of $x$, and not in terms of $n$. Hence the extra $n$ can now be moved in front.
\begin{align*}
E(X) &= \sum_{x=1}^n \frac{n(n-1)!}{(x-1)!(n-x)!}p^x (1-p)^{n-x}\\
&= n \sum_{x=1}^n \frac{(n-1)!}{(x-1)!(n-x)!} p^x (1-p)^{n-x}\\
&= n \sum_{x=1}^n \frac{(n-1)!}{(x-1)![(n-1)-(x-1)!]} p^x (1-p)^{n-x}\\
&=  n \sum_{x=1}^n \binom{n-1}{x-1}p^x (1-p)^{n-x}
\end{align*}
Now I'll decompose $p^x$ into $p\times p^{x-1}$, which allows me to pull out a factor of $p$ in front as well.
\begin{align*}
E(X) &=  n \sum_{x=1}^n\binom{n-1}{x-1} p\, p^{x-1}(1-p)^{n-x}\\
&= np \sum_{x=1}^n \binom{n-1}{x-1} p^{x-1}(1-p)^{n-x}
\end{align*}
__________________________________________________________________

Observe that our end goal $np$ has now showed up. It remains to prove that the ugly sum actually equals to 1. To do this, I will now tear apart the sum, by subbing every value of $x$ from $1$ to $n$ in. I'll focus only on the sum here.
\begin{align*}
&= \binom{n-1}{1-1} p^{1-1} (1-p)^{n-1} + \binom{n-1}{2-1} p^{2-1}(1-p)^{n-2} + \binom{n-1}{3-1} p^{3-1}(1-p)^{n-3} + \cdots + \binom{n-1}{n-1} p^{n-1} (1-p)^{n-n}\\
&= \binom{n-1}{0} (1-p)^{n-1} + \binom{n-1}{1} p(1-p)^{(n-1)-1} + \binom{n-1}{2} p^2 (1-p)^{(n-1)-2} + \cdots + \binom{n-1}{n-1} p^{n-1}
\end{align*}
It may or may not be obvious here, but this now looks disturbing like the statement of the binomial theorem!
Recall: Statement of the binomial theorem
$(x+y)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1}y + \binom{n}{2} x^{n-2} y^2 + \cdots + \binom{n}{n}y^n.$
Here, the power should be to $n-1$ instead. Observe how $p$ replaces the role of $x$, and $1-p$ replaces the role of $y$.

So from using the binomial theorem, that expression now simplifies to
$\left[ p + (1-p)\right]^{n-1}.$
Hopefully it is clear that this expression indeed evaluates to 1. Thus upon substituting into what we had earlier,
$E(X) = np \times 1 = np$
as required.

For the variance, one uses a similar strategy to this, but they start by computing $E(X(X-1))$ instead. Since $E(X(X-1)) = E(X^2 - X) = E(X^2) - E(X)$, this then allows us to find the second moment $E(X^2)$. And as usual, conclude with $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$.
« Last Edit: October 23, 2019, 06:33:56 pm by RuiAce »  #### Hawraa « Reply #4396 on: October 24, 2019, 08:50:48 pm »
0
Hi there,
Just a question from 2011 Q5-iii
After they find the number of members using Sn, shouldn't they multiply the answer by (0.5)? They multiplied it by (0.005) and I don't understand why?

#### fun_jirachi

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• Respect: +355 « Reply #4397 on: October 24, 2019, 09:03:53 pm »
+2
Hey there!

They earn half a cent per member per day, not half a dollar. This means they are correct in multiplying by $0.005 as opposed to the$0.50 that half a dollar would get you.

Hope this helps Failing everything, but I'm still Flareon up.

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#### Hawraa « Reply #4398 on: October 24, 2019, 09:04:48 pm »
0
2011-Q9-b-ii
Can someone please explain how did they get the answer as 18 litres? And like how do you know that you have to integrate (t)? Because I tried to sub 4 into both equations of liquid A and B and then subtract to get the answer as 4 litres? Help please!

#### Hawraa « Reply #4399 on: October 24, 2019, 09:08:02 pm »
0
Hey there!

They earn half a cent per member per day, not half a dollar. This means they are correct in multiplying by $0.005 as opposed to the$0.50 that half a dollar would get you.

Hope this helps But then would you do 0.5x1/2? But that would equal 0.25? Sorry I know it sounds stupid but I'm actually confused how to get half a cent?

#### fun_jirachi

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• Respect: +355 « Reply #4400 on: October 24, 2019, 09:46:43 pm »
+2
But then would you do 0.5x1/2? But that would equal 0.25? Sorry I know it sounds stupid but I'm actually confused how to get half a cent?

There are never stupid questions, ever. You can't know what you don't know, so don't be afraid to ask!

Recall that 100 cents make a dollar; this means that 1 cent is $0.01. Then, we halve it to get half a cent ie.$0.005 Does this make more sense?

2011-Q9-b-ii
Can someone please explain how did they get the answer as 18 litres? And like how do you know that you have to integrate (t)? Because I tried to sub 4 into both equations of liquid A and B and then subtract to get the answer as 4 litres? Help please!

You have to integrate here to find the volumes; you can't subtract the rates of change then multiply by the time taken. The rate of change given in the question is essentially just telling you how fast the water is coming out; it doesn't tell us anything about how much water is in the actual tank. And the answer is wrong; it's actually supposed to be 8 litres, not 18. There's a load of typos in the various sample answers, so just be wary of that Failing everything, but I'm still Flareon up.

HSC 2018: Modern History  | 2U Maths 
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ATAR: 99.05

UCAT: 3310 - Verbal Reasoning  | Decision Making  | Quantitative Reasoning  | Abstract Reasoning 

#### Hawraa « Reply #4401 on: October 24, 2019, 09:55:03 pm »
0
There are never stupid questions, ever. You can't know what you don't know, so don't be afraid to ask!

Recall that 100 cents make a dollar; this means that 1 cent is $0.01. Then, we halve it to get half a cent ie.$0.005 Does this make more sense?

You have to integrate here to find the volumes; you can't subtract the rates of change then multiply by the time taken. The rate of change given in the question is essentially just telling you how fast the water is coming out; it doesn't tell us anything about how much water is in the actual tank. And the answer is wrong; it's actually supposed to be 8 litres, not 18. There's a load of typos in the various sample answers, so just be wary of that Yup, got it now. Thanks a lot 🌷

#### Coolmate « Reply #4402 on: November 05, 2019, 07:09:34 pm »
0
Hi Everyone! I am currently studying the topic: "Series and Sequences" and have come across both of these questions (questions are attached) and have no idea about how to go about solving them. Could someone please help me with them? --> btw, the question states:

"Find the value of the pronumeral in each arithmetic sequence"

Coolmate « Last Edit: November 05, 2019, 07:11:05 pm by Coolmate »
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#### r1ckworthy

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• Respect: +294 « Reply #4403 on: November 05, 2019, 07:30:04 pm »
+3
Hi Everyone! I am currently studying the topic: "Series and Sequences" and have come across both of these questions (questions are attached) and have no idea about how to go about solving them. Could someone please help me with them? --> btw, the question states:

"Find the value of the pronumeral in each arithmetic sequence"

Coolmate Hey!
Arithmetic series have the following property (due to the common difference):
$T_2 - T_1 = T_3 - T_2$
We can use this to find the pro-numeral:
$(5k + 2) - 3 = 21 - (5k +2) \\ 5k + 2 -3 = 21 -5k -2 \\ 5k -1 = 19 - 5k \\ 10k = 20 \\ k = 2$
I'll let you have a go with the other question Let me know if you need any more help!
« Last Edit: November 05, 2019, 07:37:19 pm by r1ckworthy »
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#### Coolmate « Reply #4404 on: November 05, 2019, 07:46:52 pm »
0
Hey!
Arithmetic series have the following property (due to the common difference):
$T_2 - T_1 = T_3 - T_2$
We can use this to find the pro-numeral:
$(5k + 2) - 3 = 21 - (5k +2) \\ 5k + 2 -3 = 21 -5k -2 \\ 5k -1 = 19 - 5k \\ 10k = 20 \\ k = 2$
I'll let you have a go with the other question Let me know if you need any more help!

Hi r1ckworthy,

Thankyou for your reply, I do have a couple of questions though. For the first one why didn't you times (5k+2) by -3? and for the second question (h) I wrote this as my working out and got a wrong answer of x = 11:
(h):
$(x+3) - x = (2x +5) - (x+3) \\ 3 = 2x + 5 - x - 3 \\ 6 = x + 5 \\ x = 11$

Could you please suggest what I have done wrong with my working out please?

Coolmate 🤯🤯 HSC 2020:

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#### r1ckworthy

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• Respect: +294 « Reply #4405 on: November 05, 2019, 07:58:56 pm »
+3
Hi r1ckworthy,

Thankyou for your reply, I do have a couple of questions though. For the first one why didn't you times (5k+2) by -3? and for the second question (h) I wrote this as my working out and got a wrong answer of x = 11:
(h):
$(x+3) - x = (2x +5) - (x+3) \\ 3 = 2x + 5 - x - 3 \\ 6 = x + 5 \\ x = 11$

Could you please suggest what I have done wrong with my working out please?

Coolmate Hey Coolmate!

As for your first question, I don't really need to multiply by -3, just subtract 3 from (5k + 2) as per the formula. As for your second question, you made an error on the third line of your working out. You should subtract 5 from 6 instead of add , so it will look like this:
$(x+3) - x = (2x +5) - (x+3) \\ 3 = 2x + 5 - x - 3 \\ 6 = x + 5 \\ 6 \color{red}{- 5} = x \\ x = 1$

Hope that helps! Let me know if you need further help HSC 2019: English Advanced || Mathematics || Mathematics Extension 1 || Physics || Chemistry || Science Extension || Ancient History ||

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#### Coolmate « Reply #4406 on: November 05, 2019, 08:04:52 pm »
+2
Hey r1ckworthy!

Thankyou so much, this makes a whole lot more sense and I am getting the questions right now! I like how you previously did the formula as:

T2 - T1 = T3 - T2

And for my second question...... that was such a silly mistake😂😂, thanks for clarifying this for me

Thanks again!
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#### sharlt

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• Respect: 0 « Reply #4407 on: November 28, 2019, 03:44:40 pm »
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Hi there,

I was wondering if I could have help with this question - thank you!

A bag contains 6 white, 7 red and 8 blue balls. Create a probability function table for the number of white balls selected when drawing two balls from the bag:

A) With replacement
B) Without replacement

#### RuiAce

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• Respect: +2518 « Reply #4408 on: November 28, 2019, 10:26:40 pm »
+1
Hi there,

I was wondering if I could have help with this question - thank you!

A bag contains 6 white, 7 red and 8 blue balls. Create a probability function table for the number of white balls selected when drawing two balls from the bag:

A) With replacement
B) Without replacement
Can you please be more specific on where it is you're having trouble? Or alternatively provide some insight on what you tried thus far?  #### sharlt

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