Author Topic: Maths 3A/3B (Read 83812 times) Tweet Share

jamonwindeyer · « **Reply #105 on:** August 23, 2017, 11:08:41 am »

Quote from: anotherworld2b on August 23, 2017, 01:12:23 am

Can I have help with q 21 and q22c please?

For Q21, you are given the mean and standard deviation of the normal distribution of the fruit juice. You can standardise it using the formula:

$Y=\frac{X-\mu}{\sigma}$

And then use a table to find the normalised probability that the juice contains less than the daily recommended intake. 110% is 44mg, so we are looking for $X<44$ (use the formula above to convert to a Y-value, that's what you use in the table)

Q22, you should already have your distribution normalised to use with a table. Look for the values of the normalised variable that give:

$P \left( X < X_1 \right)=0.3\\P \left( X < X_2 \right) =0.7$

Those values, $X_1,X_2$ correspond to the lower and upper limits of the middle 40%. Then just use the formula you used to be able to use the table, in reverse, to get back to original scores

anotherworld2b · « **Reply #106 on:** August 26, 2017, 07:00:03 pm »

Can I have help with q32 b?

Syndicate · « **Reply #107 on:** August 26, 2017, 07:23:26 pm »

Quote from: anotherworld2b on August 26, 2017, 07:00:03 pm

Can I have help with q32 b?

Using a CAS calculator, I got that a (where a is the point at which Pr(Z<a) = 0.015) is equal to -2.17009

a = (x - u)/(sd(x) (Note: u is the population mean, and x is the value being tested)
sd(x) = (x-u)/a
= (490-500)/-2.17009

therefore sd(x) = 4.608 (as per Shadow's correction)

anotherworld2b · « **Reply #108 on:** August 27, 2017, 09:56:11 am »

the answer for a is 2.3% while the answer for b is 4.6?

Quote from: Syndicate on August 26, 2017, 07:23:26 pm

Using a CAS calculator, I got that a (where a is the point at which Pr(Z<a) = 0.015) is equal to -2.17009

a = (x - u)/(sd(x) (Note: u is the population mean, and x is the value being tested)
sd(x) = (x-u)/a
= (490-500)/-2.71009

therefore sd(x) = 3.69

Shadowxo · « **Reply #109 on:** August 27, 2017, 10:05:50 am »

Quote from: anotherworld2b on August 27, 2017, 09:56:11 am

the answer for a is 2.3% while the answer for b is 4.6?

His method is right, he just used a as -2.71 instead of -2.17
sd(x) = (x-u)/a
= (490-500)/-2.17009
=4.608 = 4.6

Syndicate · « **Reply #110 on:** August 27, 2017, 10:52:09 am »

Quote from: Shadowxo on August 27, 2017, 10:05:50 am

His method is right, he just used a as -2.71 instead of -2.17
sd(x) = (x-u)/a
= (490-500)/-2.17009
=4.608 = 4.6

Haha thanks for the correction

anotherworld2b · « **Reply #111 on:** August 29, 2017, 10:46:10 pm »

I was wondering how do you these questions. How do you find b for q18? How do you find a and k for q16?

RuiAce · « **Reply #112 on:** August 29, 2017, 10:48:53 pm »

Quote from: anotherworld2b on August 29, 2017, 10:46:10 pm

I was wondering how do you these questions. How do you find b for q18? How do you find a and k for q16?

$\text{Recall that the integral over the entire density must aggregate back to 1}\\ \text{In 18. this would be }\int_0^\infty ae^{-bx}dx = 1$
_______________________________
$\text{In 16, we would have }\int_0^2 (ax^2+k) dx = 1$
$\text{Also in 16, we're told }P(X\le 1) = 0.2\\ \text{Hence }\int_0^1 (ax^2+k) dx = 0.2$
These are the ingredients set up for you. Have a go at the actual computations and come back if you need further help

anotherworld2b · « **Reply #113 on:** August 30, 2017, 12:03:52 am »

I manually integrated q18 and substituted he upper and lower boundaries but I'm not sure what to do from there.
I got ( x^2 x e^ - bc) divided by 8

I was wondering where do u get the second equation from in q16?
So I far I have 8a/ 3 + 2k = 0.4

Quote from: RuiAce on August 29, 2017, 10:48:53 pm

$\text{Recall that the integral over the entire density must aggregate back to 1}\\ \text{In 18. this would be }\int_0^\infty ae^{-bx}dx = 1$
_______________________________
$\text{In 16, we would have }\int_0^2 (ax^2+k) dx = 1$
$\text{Also in 16, we're told }P(X\le 1) = 0.2\\ \text{Hence }\int_0^1 (ax^2+k) dx = 0.2$
These are the ingredients set up for you. Have a go at the actual computations and come back if you need further help

jamonwindeyer · « **Reply #114 on:** August 30, 2017, 09:13:42 pm »

Quote from: anotherworld2b on August 30, 2017, 12:03:52 am

I manually integrated q18 and substituted he upper and lower boundaries but I'm not sure what to do from there.
I got ( x^2 x e^ - bc) divided by 8

I was wondering where do u get the second equation from in q16?
So I far I have 8a/ 3 + 2k = 0.4

For Question 18, you shouldn't have any x'es left after the substitution:

$\left[\frac{-a}{b}e^{-bx}\right]^\infty_0=0\\\frac{a}{b}=1\\b=a$

In 16, we know we can integrate from 0 to 1 to find the probability, $P\left(X\le1\right)$. And we know it is equal to 0.2 from the question - That is where the second equation comes from

anotherworld2b · « **Reply #115 on:** September 08, 2017, 11:19:15 pm »

I was wondering if I could have help with this question please. We just started to learn about sample proportions but I'm really confused about this topic in general

RuiAce · « **Reply #116 on:** September 08, 2017, 11:45:05 pm »

Quote from: anotherworld2b on September 08, 2017, 11:19:15 pm

I was wondering if I could have help with this question please. We just started to learn about sample proportions but I'm really confused about this topic in general

$\text{Because we are rolling two fair dice}\\ \text{We can just list all the outcomes that give a total less than 9}\\ \text{to determine the theoretical probability }p\\ \text{(for one throw of two dice)}$
(1,1) (1,2) (2,1) (1,3) (3,1) (2,2) (1,4) (2,3) (3,2) (4,1) (1,5) (2,4) (3,3) (4,2) (5,1) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) (2,6) (3,5) (4,4) (5,3) (6,2) - 26 outcomes
$\text{Hence }p = \frac{26}{36} = \frac{13}{18}$
(Of course, a frequency histogram or something would've made this a lot faster)
_____________________________
$\hat{p}\text{ is just the 'probability' we obtained experimentally}\\ \text{This is just }\hat{p}=\frac{67}{100}$
In practice, such a thing is an "estimator" for the theoretical probability.
_____________________________
$\text{Note that the probability above is the }\textit{observed value}\text{ of }\hat{p}\\ \text{Prior to that, }\hat{p}\text{ was unknown}$
Admittedly, at this point I'm not too confident (no pun intended) anymore. I might get things wrong from here, so please comment the correct answers if that does happen.
$\text{One possible approach (albeit unsure if intended approach)}\\ \text{is to now identify the distribution of just one dice roll}\\ \text{The distribution of just any one dice roll, say, }p_k\text{ will be}\\ p_k \sim \text{Bin}\left(1, \frac{13}{18}\right)$
$\text{As the sum of independent binomial random variables is still a binomial r.v.}\\ \text{The distribution of 100 dice rolls, denoted }P,\text{ will be }\\ P = \sum_{k=1}^{100}p_k \sim \text{Bin}\left(100, \frac{13}{18}\right)$
$\text{And the trick is that }\hat{p} = \frac{P}{100}\\ \text{This is because our sample proportion is really effective of the ratio between}\\ \text{the number of times we get less than 9, against the TOTAL number of rolls.}$
$\text{We may now use the properties of the mean and variance to help us}\\ \begin{align*}\mathbb{E}[\hat{p}] &= \mathbb{E}\left[ \frac{P}{100}\right]\\ &= \frac{1}{100}\mathbb{E}[P]\tag{properties of the expected value}\\ &= \frac1{100} \times \left(100 \times \frac{13}{18}\right)\tag{formula for expected value of binomial}\\&= \frac{13}{18}\end{align*}$
$\begin{align*}\text{Var}(\hat{p}) &= \text{Var}\left(\frac{P}{100}\right)\\ &= \frac{1}{10000}\text{Var}(P)\tag{properties of the variance}\\ &= \frac{1}{10000}\times \left(100\times \frac{13}{18}\times \frac{5}{18}\right)\tag{formula for variance of binomial}\\ &= \frac{13}{6480}\\ \implies \sigma_{\hat{p}}&= \sqrt{\frac{13}{6480}}\end{align*}$
_____________________________
$\text{Assuming everything else is right, at this point all we need to do}\\ \text{is basically compute a }z\text{-score}\\ \text{for our observed value of }\hat{p}$
$i.e.\text{ plug into the formula }\frac{x - \mu}{\sigma}\\ \text{where }\mu\text{ is the mean}$
$\text{No. of standard deviations away} = \frac{67/100 - 13/18}{\sqrt{13/6480}}$

anotherworld2b · « **Reply #117 on:** September 09, 2017, 11:56:16 am »

I was wondering would it be correct to say that
Population proportion: is the experimental probability
sample proportion: is practical probability?

I was also wondering for part d when I typed it into my calculator it gave -1.1978 but when I put 0.0448 instead of the square root of 13/6840 it gave me -1.1656?

Quote from: RuiAce on September 08, 2017, 11:45:05 pm

$\text{Because we are rolling two fair dice}\\ \text{We can just list all the outcomes that give a total less than 9}\\ \text{to determine the theoretical probability }p\\ \text{(for one throw of two dice)}$
(1,1) (1,2) (2,1) (1,3) (3,1) (2,2) (1,4) (2,3) (3,2) (4,1) (1,5) (2,4) (3,3) (4,2) (5,1) (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) (2,6) (3,5) (4,4) (5,3) (6,2) - 26 outcomes
$\text{Hence }p = \frac{26}{36} = \frac{13}{18}$
(Of course, a frequency histogram or something would've made this a lot faster)
_____________________________
$\hat{p}\text{ is just the 'probability' we obtained experimentally}\\ \text{This is just }\hat{p}=\frac{67}{100}$
In practice, such a thing is an "estimator" for the theoretical probability.
_____________________________
$\text{Note that the probability above is the }\textit{observed value}\text{ of }\hat{p}\\ \text{Prior to that, }\hat{p}\text{ was unknown}$
Admittedly, at this point I'm not too confident (no pun intended) anymore. I might get things wrong from here, so please comment the correct answers if that does happen.
$\text{One possible approach (albeit unsure if intended approach)}\\ \text{is to now identify the distribution of just one dice roll}\\ \text{The distribution of just any one dice roll, say, }p_k\text{ will be}\\ p_k \sim \text{Bin}\left(1, \frac{13}{18}\right)$
$\text{As the sum of independent binomial random variables is still a binomial r.v.}\\ \text{The distribution of 100 dice rolls, denoted }P,\text{ will be }\\ P = \sum_{k=1}^{100}p_k \sim \text{Bin}\left(100, \frac{13}{18}\right)$
$\text{And the trick is that }\hat{p} = \frac{P}{100}\\ \text{This is because our sample proportion is really effective of the ratio between}\\ \text{the number of times we get less than 9, against the TOTAL number of rolls.}$
$\text{We may now use the properties of the mean and variance to help us}\\ \begin{align*}\mathbb{E}[\hat{p}] &= \mathbb{E}\left[ \frac{P}{100}\right]\\ &= \frac{1}{100}\mathbb{E}[P]\tag{properties of the expected value}\\ &= \frac1{100} \times \left(100 \times \frac{13}{18}\right)\tag{formula for expected value of binomial}\\&= \frac{13}{18}\end{align*}$
$\begin{align*}\text{Var}(\hat{p}) &= \text{Var}\left(\frac{P}{100}\right)\\ &= \frac{1}{10000}\text{Var}(P)\tag{properties of the variance}\\ &= \frac{1}{10000}\times \left(100\times \frac{13}{18}\times \frac{5}{18}\right)\tag{formula for variance of binomial}\\ &= \frac{13}{6840}\\ \implies \sigma_{\hat{p}}&= \sqrt{\frac{13}{6840}}\end{align*}$
_____________________________
$\text{Assuming everything else is right, at this point all we need to do}\\ \text{is basically compute a }z\text{-score}\\ \text{for our observed value of }\hat{p}$
$i.e.\text{ plug into the formula }\frac{x - \mu}{\sigma}\\ \text{where }\mu\text{ is the mean}$
$\text{No. of standard deviations away} = \frac{67/100 - 13/18}{\sqrt{13/6840}} = -1.165926505\dots$

RuiAce · « **Reply #118 on:** September 09, 2017, 01:50:52 pm »

Quote from: anotherworld2b on September 09, 2017, 11:56:16 am

I was wondering would it be correct to say that
Population proportion: is the experimental probability
sample proportion: is practical probability?

I was also wondering for part d when I typed it into my calculator it gave -1.1978 but when I put 0.0448 instead of the square root of 13/6840 it gave me -1.1656?

I thought "experimental" and "practical" meant the same thing (unless I was taught that part wrong).

I call the first one the "theoretical" probability.
_______________________________________

There's a typo. 6840 should be 6480.

anotherworld2b · « **Reply #119 on:** September 09, 2017, 04:52:49 pm »

Thank you for your help.
Can I also have some help with this question please?

Quote from: RuiAce on September 09, 2017, 01:50:52 pm

I thought "experimental" and "practical" meant the same thing (unless I was taught that part wrong).

I call the first one the "theoretical" probability.
_______________________________________

There's a typo. 6840 should be 6480.

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Author Topic: Maths 3A/3B (Read 83812 times) Tweet Share

jamonwindeyer

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