Author Topic: VCE Methods Question Thread! (Read 6060525 times) Tweet Share

knightrider · « **Reply #11955 on:** September 01, 2015, 10:46:57 pm »

Quote from: Splash-Tackle-Flail on September 01, 2015, 08:31:57 pm

Ooooh This one looks really cool!
First, showing $E(z)=0$
$E(z)=E(\frac{x-\mu}{\sigma})=E(\frac{x}{\sigma}-\frac{\mu}{\sigma})=E(\frac{x}{\sigma})-E(\frac{\mu}{\sigma})=\frac{E(x)}{\sigma}-\frac{\mu}{\sigma}=\frac{\mu}{\sigma}-\frac{\mu}{\sigma}=0$

Now, $Var(z)=Var(\frac{x-\mu}{\sigma})=Var(\frac{x}{\sigma}-\frac{\mu}{\sigma})=\frac{1}{\sigma^{2}}*Var(x)=\frac{1}{\sigma^{2}}*\sigma^{2}=1$
Edit: epiphany for second part;easier way to prove

for this working out how did you get from ?

$Var(\frac{x}{\sigma}-\frac{\mu}{\sigma})=\frac{1}{\sigma^{2}}*Var(x)$

StupidProdigy · « **Reply #11956 on:** September 01, 2015, 10:52:51 pm »

Quote from: knightrider on September 01, 2015, 10:46:57 pm

for this working out how did you get from ?

$Var(\frac{x}{\sigma}-\frac{\mu}{\sigma})=\frac{1}{\sigma^{2}}*Var(x)$

using the rule var(ax+b)=a^2*var(x)

Sundal · « **Reply #11957 on:** September 01, 2015, 11:28:57 pm »

For binomial theorem, must you use the pronumerals of X and x or can you use any others?

Cheers.

knightrider · « **Reply #11958 on:** September 01, 2015, 11:37:17 pm »

Quote from: StupidProdigy on September 01, 2015, 10:52:51 pm

using the rule var(ax+b)=a^2*var(x)

yea but using that wont you get

$Var(\frac{x}{\sigma}-\frac{\mu}{\sigma})=\frac{1}{\sigma^{2}}*Var(x)$

$(\frac{x}{\sigma })^{2}\times VAR(X)=\frac{x^{2}}{\sigma ^{2}}\times VAR(X)$

WHERE DOES THE $\frac{1}{\sigma^{2}}*Var(x)$ COME FROM?

knightrider · « **Reply #11959 on:** September 02, 2015, 01:15:44 am »

Would we be expected to do $\sqrt{1.5256}$ by hand on an exam?

knightrider · « **Reply #11960 on:** September 02, 2015, 01:40:34 am »

For this question attached.

If we used $\Pr (-1.3504\leq X\leq 3.5904)$ IS THIS EQUAL TO $\Pr (0\leq X\leq 3)$ ?

knightrider · « **Reply #11961 on:** September 02, 2015, 01:55:48 am »

How would you do this question attached by hand ?

This is what i wrote would this be acceptable.

$\Pr (X> 190)=\Pr (X> 174+2(8))$

$\mu \pm 2\sigma \approx 95 \%$

$\therefore 2.5\% over$

$\implies \Pr (x> 190)=0.025$

knightrider · « **Reply #11962 on:** September 02, 2015, 02:41:25 am »

For this question attached.

Find the inverse function of f

is my working out fine to obtain full marks (3 marks)

let y=f(x) for inverse swap x and y

$\Rightarrow x=(y-2)^{2}+1\Rightarrow (y-2)^{2}=x-1$

$y-2=\pm \sqrt{x-1}=\sqrt{x-1} as ran f^{-1}(x)=\left [ 2,\infty \right )$

$\therefore y=\sqrt{x-1}+2$

$\therefore f^{-1}:[1,\infty)\rightarrow \mathbb{R}, f^{-1}(x)=\sqrt{x-1}+2$

keltingmeith · « **Reply #11963 on:** September 02, 2015, 09:17:23 am »

Quote from: Sundal on September 01, 2015, 11:28:57 pm

For binomial theorem, must you use the pronumerals of X and x or can you use any others?

Cheers.

Ermmm, I'm guessing you mean the distribution from year 12 methods? In which case, yes.

Quote from: knightrider on September 01, 2015, 11:37:17 pm

yea but using that wont you get

$Var(\frac{x}{\sigma}-\frac{\mu}{\sigma})=\frac{1}{\sigma^{2}}*Var(x)$

$(\frac{x}{\sigma })^{2}\times VAR(X)=\frac{x^{2}}{\sigma ^{2}}\times VAR(X)$

WHERE DOES THE $\frac{1}{\sigma^{2}}*Var(x)$ COME FROM?

Woah, calm down. Your working out is off - you don't take the X out.

Quote from: knightrider on September 02, 2015, 01:15:44 am

Would we be expected to do $\sqrt{1.5256}$ by hand on an exam?

No

Quote from: knightrider on September 02, 2015, 01:40:34 am

For this question attached.

If we used $\Pr (-1.3504\leq X\leq 3.5904)$ IS THIS EQUAL TO $\Pr (0\leq X\leq 3)$ ?

Yes.

Quote from: knightrider on September 02, 2015, 01:55:48 am

How would you do this question attached by hand ?

This is what i wrote would this be acceptable.

$\Pr (X> 190)=\Pr (X> 174+2(8))$

$\mu +2\sigma \approx 95 \%$

$\therefore 2.5\% over$

$\implies \Pr (x> 190)=0.025$

Change that plus to a plus/minus and you should be good.

Quote from: knightrider on September 02, 2015, 02:41:25 am

For this question attached.

Find the inverse function of f

is my working out fine to obtain full marks (3 marks)

let y=f(x) for inverse swap x and y

$\Rightarrow x=(y-2)^{2}+1\Rightarrow (y-2)^{2}=x-1$

$y-2=\pm \sqrt{x-1}=\sqrt{x-1} as ran f^{-1}(x)=\left [ 2,\infty \right )$

$\therefore y=\sqrt{x-1}+2$

$\therefore f^{-1}:[1,\infty)\rightarrow \mathbb{R}, f^{-1}(x)=\sqrt{x-1}+2$

That's fine.

knightrider · « **Reply #11964 on:** September 02, 2015, 09:22:47 am »

Quote from: EulerFan101 on September 02, 2015, 09:17:23 am

Ermmm, I'm guessing you mean the distribution from year 12 methods? In which case, yes.

Woah, calm down. Your working out is off - you don't take the X out.

No

Yes.

Change that plus to a plus/minus and you should be good.

That's fine.

Thanks EulerFan101

EDIT:

is my working out fine now i think i found out?

$Var(\frac{x}{\sigma}-\frac{\mu}{\sigma})=\frac{1}{\sigma^{2}}*Var(x)$

$(\frac{1}{\sigma })^{2}\times VAR(X)=\frac{1^{2}}{\sigma ^{2}}\times VAR(X)=\frac{1}{\sigma ^{2}}\times\sigma ^{2}=1$

keltingmeith · « **Reply #11965 on:** September 02, 2015, 09:30:19 am »

Quote from: knightrider on September 02, 2015, 09:22:47 am

Thanks EulerFan101

could you show me how $Var(\frac{x}{\sigma}-\frac{\mu}{\sigma})=\frac{1}{\sigma^{2}}*Var(x)$ ?

I would, but I'd basically be copying what StupidProdigy wrote. Maybe you should check your textbook, become a bit more comfortable with that rule, and then have another look?

knightrider · « **Reply #11966 on:** September 02, 2015, 09:34:44 am »

Quote from: EulerFan101 on September 02, 2015, 09:30:19 am

I would, but I'd basically be copying what StupidProdigy wrote. Maybe you should check your textbook, become a bit more comfortable with that rule, and then have another look?

i have edited my post could you please check my working out?

knightrider · « **Reply #11967 on:** September 02, 2015, 10:03:40 am »

For this info attached.

for this question -If two individuals are selected at random, find the probability that both of their heights will exceed 190 cm.

is my working out fine ?

let A and B represent probaility that heights exceed 190 cm.

$\Pr (A)=\Pr (X> X_{A})=0.025 AND \Pr (B)=\Pr (X> X_{B})=0.025$

$\Rightarrow \Pr (A\cap B) =\Pr (A)\times \Pr (B)=0.025\times 0.025=0.000625$

Sundal · « **Reply #11968 on:** September 03, 2015, 02:30:42 pm »

Is the reason that it is said, that " the probability of a continuous random variable for an exact value is 0 " is because since a continuous random variable is something which is measured, you will never actually have an exact value?

For e.g. with height, since it is measured you wont ever really have someone who is exactly 150cm?

Bit confused, here.

Hope someone can explain.
Cheers.

knightrider · « **Reply #11969 on:** September 03, 2015, 04:07:54 pm »

Quote from: knightrider on September 02, 2015, 10:03:40 am

For this info attached.

for this question -If two individuals are selected at random, find the probability that both of their heights will exceed 190 cm.

is my working out fine ?

let A and B represent probaility that heights exceed 190 cm.

$\Pr (A)=\Pr (X> X_{A})=0.025 AND \Pr (B)=\Pr (X> X_{B})=0.025$

$\Rightarrow \Pr (A\cap B) =\Pr (A)\times \Pr (B)=0.025\times 0.025=0.000625$

Can anyone please check my working out and answer for this question?

Thanks

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