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June 15, 2021, 01:05:51 pm

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#### ethan.lozevski

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« Reply #4530 on: June 03, 2021, 03:53:50 pm »
+1
Hey guys,

I have attached a question that has confused me a little. I used Ali's model and put all my solutions over 19, to sum up to 1 as a pdf requires. Please let me know if I am doing this question correctly. Thanks, guys.

#### fun_jirachi

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« Reply #4531 on: June 03, 2021, 06:44:43 pm »
+3
Welcome to the forums!

A few questions in return:
- What are the chances you roll a 1 (and subsequently score 4 points) on the new distribution? It is still a fair die.
- Do some values on the original die yield duplicate numbers of points when rolled? What does this do for the probability of scoring a particular y value?

Hope this helps
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#### RuiAce

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« Reply #4532 on: June 03, 2021, 09:47:11 pm »
+3
Hey guys,

I have attached a question that has confused me a little. I used Ali's model and put all my solutions over 19, to sum up to 1 as a pdf requires. Please let me know if I am doing this question correctly. Thanks, guys.
Building onto fun_jirachi's answer here. I would strongly advise considering the second point that he mentioned.

Your table of values is good up until the second row. Then I believe you are confusing yourself with the third row. You want to find the probability distribution for $Y$. Just using the actual values that $Y$ takes on, is not the correct approach here.

Hint: Firstly, what are the values that $Y$ can actually be? (Refer to the second row of your table of values. There are 3 distinct values that $Y$ can take on.) Secondly, for each of those three values of $y$, what is $P(Y=y)$?

#### ethan.lozevski

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« Reply #4533 on: June 04, 2021, 09:17:11 am »
0
Thank you for your help. I think I may have read the question incorrectly. Is the following attached image correct?
Thank you.

#### fun_jirachi

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« Reply #4534 on: June 04, 2021, 12:34:29 pm »
+3
You've addressed my first question - well done!

The second question still stands however.

Hint: Firstly, what are the values that $Y$ can actually be? (Refer to the second row of your table of values. There are 3 distinct values that $Y$ can take on.) Secondly, for each of those three values of $y$, what is $P(Y=y)$?

Should such a distribution have duplicate values for $Y$? If so, how would we amend our probability $P(Y = y)$? Definitely getting closer though, give it another shot
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HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

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#### Maroon and Gold Never Fold

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« Reply #4535 on: June 06, 2021, 08:45:07 pm »
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Is anyone able to look over this answer as I am not sure if I have proved it properly or used correct notation. It is the second question of the phtoto (5.ii)

Thanks

#### fun_jirachi

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« Reply #4536 on: June 06, 2021, 08:52:03 pm »
+2
Seems about right - just be careful with the first line, you haven't subscripted the i, it looks like it's being multiplied onto the numerator as opposed to indexing the score
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HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

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#### Maroon and Gold Never Fold

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« Reply #4537 on: June 07, 2021, 06:13:01 pm »
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Seems about right - just be careful with the first line, you haven't subscripted the i, it looks like it's being multiplied onto the numerator as opposed to indexing the score

What do you mean by I haven't subscripted the 'i'. In my picture it is being multiplied on my numerator, is that wrong?

#### fun_jirachi

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« Reply #4538 on: June 07, 2021, 06:24:52 pm »
+2
Yes - $\sum_{i=1}^{n} \frac{(a+x)i}{n} = \frac{(a+x)(n - 1)}{2}$, which will lead you to an incorrect conclusion. If you notice in the formula that is given for the mean, i is a subscript denoting index (ie. $x_i$ denotes the $i^{\text{th}}$ score). Your first line should be something like $\sum_{i=1}^{n} \frac{(a+x_i)}{n}$.
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HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
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#### Maroon and Gold Never Fold

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« Reply #4539 on: June 07, 2021, 09:50:32 pm »
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Yes - $\sum_{i=1}^{n} \frac{(a+x)i}{n} = \frac{(a+x)(n - 1)}{2}$, which will lead you to an incorrect conclusion. If you notice in the formula that is given for the mean, i is a subscript denoting index (ie. $x_i$ denotes the $i^{\text{th}}$ score). Your first line should be something like $\sum_{i=1}^{n} \frac{(a+x_i)}{n}$.

thanks, should I leave the subscript 'i' on 'a' in my second line of working out then.
« Last Edit: June 07, 2021, 09:57:29 pm by Maroon and Gold Never Fold »

#### fun_jirachi

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