Author Topic: BEC'S methods questions (Read 117882 times) Tweet Share

bec · « **Reply #45 on:** January 03, 2008, 11:14:10 pm »

hahaha that's so obvious! you know why i couldn't get it? i didn't think of y=3 as being a curve...righhhhht

AppleXY · « **Reply #46 on:** January 03, 2008, 11:20:01 pm »

Darn i knew you used MathsWorld Bec. I knew it

Cause I used the same book last year when I completed methods.

(the last question about the council sounded so familiar)

bec · « **Reply #47 on:** January 04, 2008, 01:57:18 pm »

how did you complete methods last year if you did 2 other 3/4s as well??

bec · « **Reply #48 on:** January 04, 2008, 02:17:29 pm »

but wait, there's more!

The functions f and g are defined by $f(t)=e^{-0.2t}$ and $g(t)=sin2x\pi$ where $0\le t\le 4$
h(t) = f(t)g(t)
Find h'(t)

The book says: $e^{-0.2t}(2 \pi cos2 \pi t-0.2sin2 \pi t)$
My calc says: $2 \pi e^{-2x}cos(2 \pi x)-2e^{-2x}sin(2 \pi x)$
and when i did it by hand i got: $2e^{-2x}(-sin2 \pi x + \pi cos2 \pi x)$

which is the right answer? i'm guessing i'm wrong, but then i don't know if it's the book or calc that's correct - the book has so many errors in its solutions, and the calc is pretty likely to be wrong since it's me keying things into it haha

humph · « **Reply #49 on:** January 04, 2008, 02:27:31 pm »

Quote from: bec on January 04, 2008, 02:17:29 pm

but wait, there's more!

The functions f and g are defined by $f(t)=e^{-0.2t}$ and $g(t)=sin2x\pi$ where $0\le t\le 4$
h(t) = f(t)g(t)
Find h'(t)

The book says: $e^{-0.2t}(2 \pi cos2 \pi t-0.2sin2 \pi t)$
My calc says: $2 \pi e^{-2x}cos(2 \pi x)-2e^{-2x}sin(2 \pi x)$
and when i did it by hand i got: $2e^{-2x}(-sin2 \pi x + \pi cos2 \pi x)$

which is the right answer? i'm guessing i'm wrong, but then i don't know if it's the book or calc that's correct - the book has so many errors in its solutions, and the calc is pretty likely to be wrong since it's me keying things into it haha

$h(t) = f(t)g(t)$
$\Rightarrow h'(t) = f'(t)g(t)+f(t)g'(t)$

$f(t) = e^{-0.2t}$
$\Rightarrow f'(t) = -0.2e^{-0.2t}$

$g(t)=sin2\pi t$
$\Rightarrow g'(t)=2\pi cos2\pi t$

$\Rightarrow h'(t) = (-0.2e^{-0.2t})sin2\pi t + e^{-0.2t}(2\pi cos2\pi t)$
$\Rightarrow h'(t) = e^{-0.2t}(2 \pi cos2 \pi t-0.2sin2 \pi t)$

so the book is right.

bec · « **Reply #50 on:** January 04, 2008, 04:33:29 pm »

ok thanks!

anyone any good at probability? sorry this is such a long post, it's all spaces though with that mathtype thing

A card is chosen randomly from a standard deck of 52 playing cards. A second card is then chosen randomly Construct a table of the probability distribution for the number of clubs chosen, if at least one club is chosen and

a) Each card is replaced before the next card is chosen
b) Each card is chosen without the previous card being replaced

This is actually a worked example in my book but i still don't understand question b! (or for people with the mathsworld book, it's actually question (c)ii on p. 422)

The solution to (a) is:
$Pr(X=0|X \ge 1)=0$
$Pr(X=1|X \ge 1)= \frac{Pr((X=1) \cap (X \ge 1))}{Pr(X \ge 1)}$

$=\frac {Pr(X=1)}{Pr(X \ge 1)}$

$=\frac {3}{8} \div \frac {7}{16}$

$=\frac {3}{8} \times \frac {16}{7}$

$= \frac {6}{7}$

$Pr(X=2|X \ge 1)= \frac{Pr((X=2) \cap (X \ge 1))}{Pr(X \ge 1)}$

$=\frac {Pr(X=2)}{Pr(X \ge 1)}$

$=\frac {1}{16} \div \frac {7}{16}$

$=\frac {1}{16} \times \frac {16}{7}$

$= \frac {1}{7}$

So the probability distribution can be represented as (pretend it's a table):

x 1 2
Pr(X=2|X>1) 6/7 1/7

I understand that (sort of)... but then, in the working in (b), they say:

Again, $Pr(X=0|X \ge 1)=0$
$Pr(X=1|X \ge 1)= \frac{Pr((X=1) \cap (X \ge 1))}{Pr(X \ge 1)}$

$=\frac {Pr(X=1)}{Pr(X \ge 1)}$

$=\frac {13}{34} \div \frac {15}{34}$ <----why?? in (a), we did exactly the same sum and got 3/8 divided by 7/16...why does it change?

$=\frac {13}{38} \times \frac {34}{15}$

$= \frac {13}{15}$

(and it goes on to find [tex] Pr(X=2|X \ge 1) and the probability distribution. )

so basically what i'm saying i don't understand is why the same expression can have two totally different answers.

thanks so much to anyone who can be bothered reading and replying to this!

Collin Li · « **Reply #51 on:** January 04, 2008, 04:47:52 pm »

You shouldn't really be using $X$ for parts a and b, because they are different probability distributions. One is talking about the case where you are replacing cards, the other is talking about the case where you are not replacing cards. It doesn't matter if you use $X$ twice, but just don't be misled into thinking they are the same. (i.e.: $x+2 = 3$ has nothing to do with $x-5 = 7$ , it's just a pronumeral)

You are essentially asking for the same expression, but with a different game (or a different set of rules).

Mao · « **Reply #52 on:** January 04, 2008, 04:50:53 pm »

in one case the card is replaced

$Pr(x \ge 1) = pr(x=1)+pr(x=2)$

$Pr(x \ge 1)=^1C_2 \frac{1}{4}\cdot \frac{3}{4} + ^2C_2 \frac{1}{4}\cdot \frac{1}{4}$

$Pr(x \ge 1)=\frac{7}{16}$

in one case it is not

$Pr(x \ge 1) = pr(x=1)+pr(x=2)$

$Pr(x \ge 1) =^1C_2 \frac{1}{4} \cdot \frac{38}{52} + ^2C_2 \frac{1}{4} \cdot \frac{12}{13}$

$Pr(x \ge 1)=\frac{15}{34}$

this will become clearer as soon as you have done binomial distributions

hope that helps

bec · « **Reply #53 on:** January 04, 2008, 04:57:13 pm »

thanks, yeah that does help. i didn't realise the X wasn't a constant. actually i don't think i really understand probability full stop haha. you'll here from me again soon!

doboman · « **Reply #54 on:** January 04, 2008, 09:48:56 pm »

Bec are you planning to finish the whole course before going into year 12?

AppleXY · « **Reply #55 on:** January 05, 2008, 09:38:08 am »

lol. We should rename this thread as "Bec's Methods Questions" :p

bec · « **Reply #56 on:** January 05, 2008, 01:31:25 pm »

lol yes i think i'll have to!

umm doing discrete random variables still and i don't really understand it.
Given $\sigma =3$ , find:
a) var(6X)
b) var(X-5)
c) var(2X+7)

If $\sigma = \sqrt {var(X)}$ , does that mean that $\sigma^{2} =var(x)$ ?
And does var(6X)=6var(X) or does that rule only work with E(6X) etc?

Obviously it doesn't work because what i did was:
$var(X) = \sigma ^{2} = 9$
$var(6X) = 6var(X) =6(9) = 54$ .......and the answer is supposed to be 324
can anyone explain it because i know i'm very off track.
thanks!

(and rooboy, i'm not doing any more school work after tomorrow so nope i won't finish unit 4)

brendan · « **Reply #57 on:** January 05, 2008, 01:38:39 pm »

that's cos you are using the wrong formula
the correct one is
var(aX + b) = (a^2)var(X)

bec · « **Reply #58 on:** January 05, 2008, 01:41:48 pm »

ohhhh ok

so...
$var(x) = \sigma ^{2} = 9$
$var(6X) = 6^{2}var(x)$
$= 36 \times 9$
$= 324$

wooo thanks

Collin Li · « **Reply #59 on:** January 05, 2008, 03:38:43 pm »

Let me provide a little bit of justification of those formulae:

$\mbox{E}(aX + b) = a\cdot\mbox{E}(X) + b$

This is saying: if you take a random variable $X$ , and you multiply all its values by $a$ , then the mean will be multiplied by $a$ . This makes sense. If you add $b$ to every value, you will simply add $b$ to the mean. This also makes sense!

Now, for $\mbox{Var}(aX+b) = a^2\cdot\mbox{Var}(X)$

First, let's deal with $b$ . If you add $b$ to every value in the distribution, the variance will not change at all. Why? The variance is a measure of spread, a measure of deviation from the mean. Adding $b$ will not change the spread of the distribution, nor will it change the deviation from the mean.

Now, if you multiply every value by $a$ , the variance will change by a factor of $a^2$ . This happens simply by virtue of the fact that $\mbox{Var}(X) = \sigma ^2$ .

So, that's a semi-flip-floppy explanation of the formulae above, but hopefully it will help you remember them (not sure if they're on the formula sheet).

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Author Topic: BEC'S methods questions (Read 117882 times) Tweet Share

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