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#### ArtyDreams

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##### Re: VCE Methods Question Thread!
« Reply #18120 on: September 12, 2019, 04:35:54 pm »
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Hey! I just had a question related to probability.

So lets say that E(X) (the mean) is 2 for a function, and The variance, is 0.84.

Then, to find E(2/3X), I simply times 2 by 2/3, which will give 4/3.

So, in order to find the variance that corresponds with E(2/3X), do I just simply multiply 0.84 by 2/3?

Sorry if this question doesnt make sense!

Would really appreciate any help though
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#### RuiAce

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##### Re: VCE Methods Question Thread!
« Reply #18121 on: September 12, 2019, 04:50:31 pm »
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Hey! I just had a question related to probability.

So lets say that E(X) (the mean) is 2 for a function, and The variance, is 0.84.

Then, to find E(2/3X), I simply times 2 by 2/3, which will give 4/3.

So, in order to find the variance that corresponds with E(2/3X), do I just simply multiply 0.84 by 2/3?

Sorry if this question doesnt make sense!

Would really appreciate any help though
The formulas are different. Your first answer is correct, because $\operatorname{E}(aX)=a\operatorname{E}(X)$.

But for the variance, we instead have $\boxed{\operatorname{Var}(aX) = a^2\operatorname{Var}(X)}$. You need to multiply by the square of that extra constant factor. (In your case, it would be multiplication by 4/9 instead of 2/3.)

#### ArtyDreams

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##### Re: VCE Methods Question Thread!
« Reply #18122 on: September 12, 2019, 04:56:29 pm »
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The formulas are different. Your first answer is correct, because $\operatorname{E}(aX)=a\operatorname{E}(X)$.

But for the variance, we instead have $\boxed{\operatorname{Var}(aX) = a^2\operatorname{Var}(X)}$. You need to multiply by the square of that extra constant factor. (In your case, it would be multiplication by 4/9 instead of 2/3.)

Right I understand now. Thank you so much!!!
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#### persistent_insomniac

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##### Re: VCE Methods Question Thread!
« Reply #18123 on: September 12, 2019, 07:30:39 pm »
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When we are approximating the distribution of the sample proportion to find probabilities, do we just assume that p hat = x/n is the same as the probability of success of a particular event?
e.g.) "Find the approx probability that in the next 50 tosses of a fair coin, the proportion of heads observed will be less than or equal to 0.46).
So to find the mean and sd of p hat by using the formula, do we assume pr(sucess) = 1/2 is the same as x/n??

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #18124 on: September 12, 2019, 07:58:11 pm »
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When we are approximating the distribution of the sample proportion to find probabilities, do we just assume that p hat = x/n is the same as the probability of success of a particular event?
e.g.) "Find the approx probability that in the next 50 tosses of a fair coin, the proportion of heads observed will be less than or equal to 0.46).
So to find the mean and sd of p hat by using the formula, do we assume pr(sucess) = 1/2 is the same as x/n??

The definition of the distribution of the sample proportion is $\hat{P}=\dfrac{X}{n}$, where $X$ denotes the random variable for the number of successes observed from a number of items.

In the coins example, $X$ happens to have the following binomial distribution: $X\sim\text{Bi}(50,\ 1/2)$. The parameter $p=1/2$ comes from an inherent property of fair coins, not from sample proportions. The sample proportion varies from sample to sample, and won't always be $\hat{p}=1/2$, if that's what you're asking.

Using these formulas which you should have derived in class: $\text{E}(\hat{P})=p\quad\text{and}\quad \text{sd}(\hat{P})=\sqrt{\frac{p(1-p)}{n}},$ it shouldn't be too difficult to show that the distribution of $\hat{P}$ can be approximated by a normal distribution with mean $\dfrac12$ and standard deviation $\dfrac{\sqrt{2}}{20}$.

Use your CAS to then find $\Pr(\hat{P}\leq 0.46)$.
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#### persistent_insomniac

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##### Re: VCE Methods Question Thread!
« Reply #18125 on: September 13, 2019, 06:35:12 am »
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The definition of the distribution of the sample proportion is $\hat{P}=\dfrac{X}{n}$, where $X$ denotes the random variable for the number of successes observed from a number of items.

In the coins example, $X$ happens to have the following binomial distribution: $X\sim\text{Bi}(50,\ 1/2)$. The parameter $p=1/2$ comes from an inherent property of fair coins, not from sample proportions. The sample proportion varies from sample to sample, and won't always be $\hat{p}=1/2$, if that's what you're asking.

Using these formulas which you should have derived in class: $\text{E}(\hat{P})=p\quad\text{and}\quad \text{sd}(\hat{P})=\sqrt{\frac{p(1-p)}{n}},$ it shouldn't be too difficult to show that the distribution of $\hat{P}$ can be approximated by a normal distribution with mean $\dfrac12$ and standard deviation $\dfrac{\sqrt{2}}{20}$.

Use your CAS to then find $\Pr(\hat{P}\leq 0.46)$.
The definition of the distribution of the sample proportion is $\hat{P}=\dfrac{X}{n}$, where $X$ denotes the random variable for the number of successes observed from a number of items.

In the coins example, $X$ happens to have the following binomial distribution: $X\sim\text{Bi}(50,\ 1/2)$. The parameter $p=1/2$ comes from an inherent property of fair coins, not from sample proportions. The sample proportion varies from sample to sample, and won't always be $\hat{p}=1/2$, if that's what you're asking.

Using these formulas which you should have derived in class: $\text{E}(\hat{P})=p\quad\text{and}\quad \text{sd}(\hat{P})=\sqrt{\frac{p(1-p)}{n}},$ it shouldn't be too difficult to show that the distribution of $\hat{P}$ can be approximated by a normal distribution with mean $\dfrac12$ and standard deviation $\dfrac{\sqrt{2}}{20}$.

Use your CAS to then find $\Pr(\hat{P}\leq 0.46)$.
The definition of the distribution of the sample proportion is $\hat{P}=\dfrac{X}{n}$, where $X$ denotes the random variable for the number of successes observed from a number of items.

In the coins example, $X$ happens to have the following binomial distribution: $X\sim\text{Bi}(50,\ 1/2)$. The parameter $p=1/2$ comes from an inherent property of fair coins, not from sample proportions. The sample proportion varies from sample to sample, and won't always be $\hat{p}=1/2$, if that's what you're asking.

Using these formulas which you should have derived in class: $\text{E}(\hat{P})=p\quad\text{and}\quad \text{sd}(\hat{P})=\sqrt{\frac{p(1-p)}{n}},$ it shouldn't be too difficult to show that the distribution of $\hat{P}$ can be approximated by a normal distribution with mean $\dfrac12$ and standard deviation $\dfrac{\sqrt{2}}{20}$.

Use your CAS to then find $\Pr(\hat{P}\leq 0.46)$.

Thanks for the reply. I understand what you're saying but what I don't understand is that in another question ("suppose a fair coin is tossed 4 times and the no. of heads is observed. Find the mean and sd of p hat") - for this qs although we use the binomial distribution to find the probabilties, how come they don't use the formula to find the mean and sd and instead do it using E(P hat) = E(P hat^2) - E(P hat)^2 ?

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #18126 on: September 14, 2019, 03:06:07 pm »
+1
Thanks for the reply. I understand what you're saying but what I don't understand is that in another question ("suppose a fair coin is tossed 4 times and the no. of heads is observed. Find the mean and sd of p hat") - for this qs although we use the binomial distribution to find the probabilties, how come they don't use the formula to find the mean and sd and instead do it using E(P hat) = E(P hat^2) - E(P hat)^2 ?

I'm not sure why you would want to evaluate $\text{Var}(\hat{P})$ by using $\text{E}\Big(\hat{P}^2\Big)-\Big[\text{E}(\hat{P})\Big]^2$ for this particular question. While it will give you the correct answer, since $X$ (the number of heads observed) is clearly a binomial variable (with $n=4$ and $p=1/2$), it simplifies down to the formula given in my previous post.

For a proof, check the textbook.
« Last Edit: September 14, 2019, 03:12:59 pm by AlphaZero »
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#### maths_genius246

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##### Re: VCE Methods Question Thread!
« Reply #18127 on: September 15, 2019, 04:28:21 pm »
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Hey guys,
wondering what would a sac avg of 85 (B+) and a A on exam 1 and a A+ on exam 2 get me?
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#### colline

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##### Re: VCE Methods Question Thread!
« Reply #18128 on: September 15, 2019, 04:55:14 pm »
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Hey guys,
wondering what would a sac avg of 85 (B+) and a A on exam 1 and a A+ on exam 2 get me?

I would assume very high 30s based on last year’s distribution. I know a friend who had B+, A+, A+, 40.

Though don’t you already have a 46 in methods?
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#### forsande

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##### Re: VCE Methods Question Thread!
« Reply #18129 on: September 15, 2019, 05:57:27 pm »
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Can anyone help?

A bag contains five red marbles and four blue marbles. Two marbles are drawn from the bag, without replacement, and the results are recorded.
What is the probability that the marbles are different?

The answer is 5/9, but I'm getting a completely different answer  (I used a tree diagram instead of hypergeometric, don't rly think I need to know how to use hypergeomtric )

#### Evolio

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##### Re: VCE Methods Question Thread!
« Reply #18130 on: September 15, 2019, 06:06:56 pm »
+1
I used a tree diagram as well.
So basically you do the calculation:
(5/9*1/2)+(4/9)*(5/8). This will give you 5/9.
Did you use the values above or did you use different values?

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#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #18131 on: September 15, 2019, 06:08:25 pm »
+2
Can anyone help?

A bag contains five red marbles and four blue marbles. Two marbles are drawn from the bag, without replacement, and the results are recorded.
What is the probability that the marbles are different?

The answer is 5/9, but I'm getting a completely different answer  (I used a tree diagram instead of hypergeometric, don't rly think I need to know how to use hypergeomtric )

Edit: Evolio beat me to it
$\Pr(\text{different})=\Pr(R,B)+\Pr(B,R)=\frac59\!\times\!\frac48+\frac49\!\times\!\frac58=\frac59$

If you want to use the hypergeometric distribution (even though it's not technically required by the course), you would want the following: $\Pr(\text{different})=\frac{\displaystyle \binom{5}{1}\binom{4}{1}}{\displaystyle \binom{9}{2}}=\frac59$
« Last Edit: September 15, 2019, 06:10:51 pm by AlphaZero »
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#### forsande

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##### Re: VCE Methods Question Thread!
« Reply #18132 on: September 15, 2019, 06:51:05 pm »
+1
I used a tree diagram as well.
So basically you do the calculation:
(5/9*1/2)+(4/9)*(5/8). This will give you 5/9.
Did you use the values above or did you use different values?
Ohh thank you, makes sense now

For Pr(R,B), I did 5/9 * 4/7 because I thought two marbles were chosen after Pr(R).

Cheers guys

#### forsande

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##### Re: VCE Methods Question Thread!
« Reply #18133 on: September 15, 2019, 09:00:58 pm »
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Trains from South Yarra station always run late.
The length of time that a train runs late, in minutes, is a random variable with a pdf given by

f(x)={(0.02x                         0≤x≤5
(−1/150x) + 2/15         5<x≤20
0                                 elsewhere
(sorry, supposed to be a piecewise function but dunno how to do it on AN)

b) There is an 80% chance that the train will leave, at most, k minutes late.  Determine k, correct to two decimal places.
c) The length of time that a train runs late is independent of the length of time that any other train runs late.  Of the next five trains that leave South Yarra station, what is the probability that exactly four of them run less than 5 minutes late

Answer for b is 12.25mins and c is 0.0146. I wish I could give show you my working to make sure I'm not leaching off, but I seriously don't know how to do this.
Cheers everyone.

#### AlphaZero

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##### Re: VCE Methods Question Thread!
« Reply #18134 on: September 16, 2019, 01:06:48 pm »
+1
Hey! I worked out the question - I'm not really sure how to explain this properly, as I am only doing 3/4 methods this year, but I tried to take it step by step through the working out Sorry its not in much detail, but I do hope it helps you!

Essentially, the best thing is to sketch a graph of the hybrid function first, and then remember that the sum of the area under the function must always equal to 1.

Well done for having a go at the question. Since you're taking Methods 3&4, I'm going to give you some feedback.

Sketching the graph of the function in this question wouldn't get you any marks. While it might help with your understanding, just be wary that in an exam situation, you're pressed for time, so just be careful spending extra time on things that don't get you any marks.

Please do not write the terminals of integration on the left side of your integral sign. This is an abuse of notation. The question also already defines the critical value as $k$ but you've used $m$. Stick to what the question has already defined for you.

In part b, since $\Pr(X<5)=\int_0^5 \frac{x}{50}\,\text{d}x=\frac14,$ we know that $5\leq k\leq 20$, and so one can just solve $\int_k^{20} \left(\frac{-x}{150}+\frac{2}{15}\right)\text{d}x=\frac15\implies \left[\frac{-x^2}{300}+\frac{2x}{15}\right]_k^{20}=\frac15\implies \cdots$ which is a little nicer. Guessing that $5\leq k\leq 20$ is a bit risky.

For part c, you should define the binomial variable $N\sim\text{Bi}(5,\,1/4)$ for clarity. Then it becomes much clearer that $\Pr(N=4)=\binom{5}{4}\left(\frac14\right)^{\!4}\!\left(\frac34\right)=\frac{15}{1024}.$ The notation $^n\text{C}_k$ is correct, but using the binomial coefficient notation, $\displaystyle \binom{n}{k}$, is preferred.
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