Author Topic: VCE Methods Question Thread! (Read 5740939 times) Tweet Share

lzxnl · « **Reply #2220 on:** July 12, 2013, 11:31:28 am »

DUDE...is that 10 subjects I see you doing in your sig? That's ridiculous.

Sanguinne · « **Reply #2221 on:** July 12, 2013, 02:02:38 pm »

The life (in hours) of a particular brand of batteries is a random variable with probability density function given by
f(x)=(e^(-x/1000))/1000, x≥0 sry i dont know how to use latex yet.. $:-\$
0,elsewhere

a) no problem
b) if after 180 hours of operation a battery is still working, what is the probability correct to 4 decimal places that it will last at least another 50 hours
c) A manufacturer of the batteries claim that 90% of their batteries will work for at least n hours.
Find the largest possible value of n.

Professor Polonsky · « **Reply #2222 on:** July 12, 2013, 02:23:59 pm »

Quote from: fleet street on July 12, 2013, 09:30:42 am

I think that angle between two curves isn't on the study design. From the Essentials textbook, "This topic is not listed in the study design, but can be included for the understanding of other topics that are included."But yeah, it's good to know anyway. Also, if your teacher does decide to put it on SACs, then I guess you'd have to know it.

They shouldn't do that.

Homer · « **Reply #2223 on:** July 12, 2013, 02:30:59 pm »

Quote from: Sanguinne on July 12, 2013, 02:02:38 pm

The life (in hours) of a particular brand of batteries is a random variable with probability density function given by
f(x)=(e^(-x/1000))/1000, x≥0 sry i dont know how to use latex yet.. $:-\$
0,elsewhere

a) no problem
b) if after 180 hours of operation a battery is still working, what is the probability correct to 4 decimal places that it will last at least another 50 hours
c) A manufacturer of the batteries claim that 90% of their batteries will work for at least n hours.
Find the largest possible value of n.

b) $\frac{\int_{230}^{\infty }f(x)dx}{\int_{180}^{\infty }f(x)dx}$

c)upper limit =unknown lower limit 0 integrate f(x)=0.1, cas should solve for unknown giving 105.3605157=105.36hours

pi · « **Reply #2224 on:** July 12, 2013, 02:42:35 pm »

Quote from: nliu1995 on July 12, 2013, 11:31:28 am

DUDE...is that 10 subjects I see you doing in your sig? That's ridiculous.

# y o l o

Also Homer, from memory better notation of improper integrals is: $\lim_{a \rightarrow \infty} \int^a_{180}f(x) dx$ . Apparently VCAA likes it that way (teacher hearsay).

Regards.

lzxnl · « **Reply #2225 on:** July 12, 2013, 03:22:01 pm »

For an explanation of what Homer is doing, this is a conditional problem; the probability that the battery works for 230 hours at least given that it works for 180 hours. That's what the division is for.
As for the third one, integrate your function from 0 to n. This probability should be 0.1 as this is the probability the battery lasts for less than n hours. Then just solve for n.

asian5 · « **Reply #2226 on:** July 12, 2013, 07:48:56 pm »

The probability that a person chooses to drink coffee is dependent on what they had for their previous drink. If they drink coffee the probability they will drink coffee for their next drink is 0.4. If they drink something else the probability they drink coffee the next time is 0.75. The probability the person drinks coffee for their first drink of the day is 0.9

a)What is the probability they will drink coffee for the fourth drink of the day?
b)What is the steady state probability the person will drink coffee?

Professor Polonsky · « **Reply #2227 on:** July 12, 2013, 08:32:08 pm »

You would do these with transition matrices.

a) You would use $S_{n}=T^n \times S_{0}$ , where $T$ is your tansition matrix, $S_{0}$ is the probability of the first step. It's important to note that as we're starting with 0, $n$ is equal to 3.

Spoiler

$S_{3}= \begin{bmatrix} 0.4 & 0.75 \\ 0.6 & 0.25 \\\end{bmatrix}^3 \times \begin{bmatrix} 0.9\\0.1\\ \end{bmatrix} = \begin{bmatrix} 0.5408\\0.4592\\ \end{bmatrix}$ (note that this can be done with fractions on your calculator to get an exact answer).

Therefore the probability they will drink coffee for the fourth drink of the day is 0.5408

b) With the steady-state, you no longer consider the initial probability as this is the long-term trend - essentially, what the probability converges towards as $n$ grows (reaches infinity).

If your transition matrix is $\begin{bmatrix} 1-a & b \\ a & 1-b \\\end{bmatrix}$ , then the steady state for the event in the first row is $\frac{b}{a+b}$ , and for the one in the second row $\frac{a}{a+b}$

Spoiler

$\frac{0.75}{0.75+0.6} = 0.55556 = \frac{5}{9}$

shadows · « **Reply #2228 on:** July 12, 2013, 10:00:57 pm »

Just stumbled upon a question I can't seem to do by hand

how does the derivative of loge(x+ sq(1+x^2))

equal 1/(sq(1+x^2))

Phy124 · « **Reply #2229 on:** July 12, 2013, 10:19:25 pm »

Quote from: shadows on July 12, 2013, 10:00:57 pm

Just stumbled upon a question I can't seem to do by hand

how does the derivative of loge(x+ sq(1+x^2))

equal 1/(sq(1+x^2))

$\frac{d}{dx}\left(\log_e\left(x+\sqrt{1+x^2}\right)\right) = \frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}}$

$=\frac{1+\frac{x}{\sqrt{1+x^2}}}{x+\sqrt{1+x^2}} \times \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}$

$= \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}(x+\sqrt{1+x^2})}$

$=\frac{1}{\sqrt{1+x^2}}$

b^3 · « **Reply #2230 on:** July 12, 2013, 10:21:27 pm »

$\begin{alignedat}{1}\frac{d}{dx}\left(\log_{e}\left(x+\sqrt{1+x^{2}}\right)\right) & =\frac{1}{x+\sqrt{1+x^{2}}}\times\left(1+\frac{1}{2}\left(1+x^{2}\right)^{-\frac{1}{2}}\left(2x\right)\right) \\ & =\frac{1}{x+\sqrt{1+x^{2}}}\times\left(1+\frac{x}{\sqrt{1+x^{2}}}\right) \\ & =\frac{1}{x+\sqrt{1+x^{2}}}\times\left(\frac{\sqrt{1+x^{2}}+x}{\sqrt{1+x^{2}}}\right) \\ & =\frac{1}{\sqrt{1+x^{2}}} \end{alignedat}$

EDIT: Beaten.

Probably should add an explanation too.
If we let $u=x+\sqrt{1+x^{2}}$ then we have $y=\log_{e}(u)$ , thus $\frac{dy}{du}=\frac{1}{u}$ and $\frac{du}{dx}=1+\frac{1}{2}\left(1+x^{2}\right)^{-\frac{1}{2}}\left(2x\right)$ (remembering to apply the chain rule for this derivative).
Then finally we have $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ by applying the chain rule again.

shadows · « **Reply #2231 on:** July 12, 2013, 10:28:15 pm »

i got up to the second last step?

my algebras not that good. can you elaborate on how to got to the answer from the second last step D:?

Phy124 · « **Reply #2232 on:** July 12, 2013, 10:31:06 pm »

The $\sqrt{1+x^2}+x$ in the numerator "cancels out" with the $x+\sqrt{1+x^2}$ in the denominator, as if you divide something by itself, it is equal to 1.

shadows · « **Reply #2233 on:** July 12, 2013, 10:43:48 pm »

oh! i approached it differently, and totally confused myself.
facepalm.

Nato · « **Reply #2234 on:** July 13, 2013, 12:42:45 pm »

hey guys, does anybody have any neat ways to identify when events (in probability) and independent or not, it's getting quite confusing $:-\$

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