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svnflower

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« Reply #4515 on: October 24, 2020, 10:25:33 pm »
+5
Hi Everyone!

Could I please have help with working out the correlation coefficient (r) and part ii, in the attached question?

Coolmate

Hello

To find r, input your values into a table. To do this, click MODE > 2 > 2 then plug in your values. After that, hit the AC button > SHIFT > 1 > 5 > 3. And there's your correlation coefficient number!

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p.s. i got an answer of 0.959 for that question
« Last Edit: October 24, 2020, 10:27:57 pm by svnflower »

Coolmate

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« Reply #4516 on: October 24, 2020, 10:47:07 pm »
0
Hello

To find r, input your values into a table. To do this, click MODE > 2 > 2 then plug in your values. After that, hit the AC button > SHIFT > 1 > 5 > 3. And there's your correlation coefficient number!

Spoiler
p.s. i got an answer of 0.959 for that question

Wow! Thanks, svnflower This really helped me, I was so confused, but now it is so clear . Also, I got the same answer as you

Coolmate
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svnflower

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« Reply #4517 on: October 24, 2020, 11:00:21 pm »
+1
Wow! Thanks, svnflower This really helped me, I was so confused, but now it is so clear . Also, I got the same answer as you

Coolmate

YAY happy to help, all the best for Monday!

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« Reply #4518 on: October 25, 2020, 11:14:50 am »
0
When using trapezoidal rule, how do you figure out whether the answer is less than or greater than the actual, integrated value?

fun_jirachi

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« Reply #4519 on: October 25, 2020, 01:38:37 pm »
+6
Hi

If the graph is concave down over an interval [a, b], then the trapezoidal rule will evaluate a value less than the actual integrated value over that same interval, and vice versa - you can think about this by drawing it out and seeing that the trapezoidal rule effectively truncates area/adds area on. If the slope of the curve is constant, then the trapezoidal rule will evaluate a value equal to the actual integrated value.
« Last Edit: October 25, 2020, 02:35:25 pm by fun_jirachi »
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« Reply #4520 on: October 25, 2020, 02:11:54 pm »
0
Thanks for the response! I thought it would've been the other way around! Also, what does 2 applications of the trapezoidal rule mean? Does it mean use two intervals, or does it mean do the trapezoidal rule process twice, with one interval each?
« Last Edit: October 25, 2020, 02:13:37 pm by BakerDad12 »

fun_jirachi

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« Reply #4521 on: October 25, 2020, 02:51:50 pm »
+4
Those two things are effectively the same thing - but that is what it means

Consider what happens if we have a function $f(x)$ bounded by the interval $[a, c]$ and $b = \frac{a+c}{2}$. By the trapezoidal rule, $\int_a^c f(x) \ dx \approx \frac{b-a}{2}(f(a) + 2f(b) + f(c))$, if we use two intervals. However, we can rewrite the right hand side as $\frac{b-a}{2}(f(a)+f(b)) + \frac{c-b}{2}(f(b)+f(c))$ since b is the midpoint of a and c. Note that this also approximates the integral and is also the application of the trapezoidal rule on one interval twice!

Hope this helps
« Last Edit: October 25, 2020, 02:53:42 pm by fun_jirachi »
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Coolmate

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« Reply #4522 on: October 25, 2020, 05:20:15 pm »
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YAY happy to help, all the best for Monday!

Thanks, you too!
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Coolmate

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« Reply #4523 on: October 25, 2020, 08:19:52 pm »
0
Hi Everyone!

Just a quick question before tomorrows exam:
Using the attached question as an example, would someone please show me how to determine the equation of the least-squares regression line?

Coolmate
« Last Edit: October 25, 2020, 09:06:38 pm by Coolmate »
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needhelp101

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« Reply #4524 on: February 03, 2021, 10:17:03 pm »
0
Hey everyone,

I'm in Year 11 and I need help on this Question:

The questions asks me to prove the following inequality is true by considering the reciprocals of both sides:

fun_jirachi

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« Reply #4525 on: February 03, 2021, 10:27:18 pm »
+3
Welcome to the forums!

They're basically asking you to take the reciprocals then rationalise the denominator without explicitly telling you the second step.
$\frac{1}{\sqrt{6}-\sqrt{5}} > \frac{1}{\sqrt{5}-2}
\\ \sqrt{6}+\sqrt{5} > \sqrt{5} + 2
\\ \sqrt{6} > 2$

This is the true statement you're supposed to find. Note that you can't actually present this as a proof because you're starting with what you're trying to prove, but working backwards from the last line of working is an acceptable proof of the inequality in the question (and this method of 'discovering a proof' is very handy in a lot of HS maths, especially compared to blindly using brute force). Working backwards is fine because you're starting with a statement of fact, then proving (or maybe in some other case disproving) the statement presented in the question.

Hope this helps
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Maroon and Gold Never Fold

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« Reply #4526 on: May 31, 2021, 11:18:56 pm »
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Attached a question I am unable to do.

RuiAce

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« Reply #4527 on: June 01, 2021, 12:17:31 am »
+2
Attached a question I am unable to do.

For $y>0$,
\begin{align*}
F_Y(y) &= P(Y\leq y)\\
&= P(\sqrt{X} \leq y)\\
&= P(X \leq y^2) \tag{square both sides}\\
&= F_X(y^2) \tag{because we now have $X$}\\
&= 1-e^{-y^2}.
\end{align*}
You can then find the density function through the usual way (taking derivative).

For any future questions, you should post your attempts at solving them, no matter how right or wrong they are. That way we can be sure you've given them a go first, and can also understand your thought process to give you actual valuable help.
« Last Edit: June 01, 2021, 12:19:07 am by RuiAce »

Maroon and Gold Never Fold

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« Reply #4528 on: June 01, 2021, 08:10:45 am »
0

For any future questions, you should post your attempts at solving them, no matter how right or wrong they are. That way we can be sure you've given them a go first, and can also understand your thought process to give you actual valuable help.

Thanks for your help. I was really stuck on the wording of the question and I'm still a bit lost on what I'm sort of meant to do.

RuiAce

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« Reply #4529 on: June 01, 2021, 12:32:36 pm »
+4
Thanks for your help. I was really stuck on the wording of the question and I'm still a bit lost on what I'm sort of meant to do.
You started with a random variable $X$. You were given the CDF of $X$.

You then pull out a new random variable $Y$. You are told that this new random variable is related to the old one, by the equation $Y = \sqrt{X}$. (So you see that $Y$ is in fact a function of another random variable, and hence also a random variable.)

Since $Y$ is a new random variable, it should also have a CDF. You are asked to find this CDF in the first part. (And then using it, you can find the PDF in the second part.)

(But how would you be able to find the CDF of $Y$ to begin with? Well, you're given the relationship $Y = \sqrt{X}$, and the CDF of $X$. So somehow you had to puzzle the two together.)

The actual technique used is understandably one you might've not seen before. But as for what the question was asking, you needed to realise most of the above.
« Last Edit: June 01, 2021, 12:34:25 pm by RuiAce »