Statistics Marathon & Questions (3 Viewers)

davidgoes4wce · Apr 14, 2017

Re: University Statistics Discussion Marathon

The company Atherton uses a particular machine to fill 1 kg bags with sugar. The weight X of the machine processed bags follows a normal distribution with mean $\mu$ and standard deviation $\sigma$ =0.24. The machine mean weight can be set by the operator to any value between 0 kg and 50 kg. Regulations allow to sell sugar bags as "1 kg bags" if at least 97.5% of the production is 1 kg or heavier. Answer the following question, round off the final results to 2 dp, but without rounding numbers during your calculations.

What mean weight $\mu$ should the machine be set to in order to produce 97.5% of the bags weighting at least 1kg?

davidgoes4wce · Apr 14, 2017

Re: University Statistics Discussion Marathon

Anyway I had a go at. If someone could confirm that would be great.

davidgoes4wce · Apr 14, 2017

Re: University Statistics Discussion Marathon

Doing Statistics on a Good Friday is insanely awesome.

leehuan · Apr 16, 2017

Re: Statistics

The content isn't really examinable but it was taught anyway, so I want to grasp some of the intuition behind it for future reference.

What would be some examples of convergence almost surely and convergence in probability? And with or without the mathematical definition, how is almost sure convergence stronger than convergence in probability?

InteGrand · Apr 16, 2017

Re: Statistics

leehuan said:
The content isn't really examinable but it was taught anyway, so I want to grasp some of the intuition behind it for future reference.

What would be some examples of convergence almost surely and convergence in probability? And with or without the mathematical definition, how is almost sure convergence stronger than convergence in probability?

Almost sure convergence is stronger than convergence in probability because the former implies the latter, but the latter generally does not imply the former. You can either try proving this as an exercise or finding a proof online, there should be one on Wikipedia even, I think. You can read here for an intuitive interpretation of convergence in probability vs. almost surely: https://stats.stackexchange.com/questions/2230/convergence-in-probability-vs-almost-sure-convergence .

You can find some intuitive examples of various convergence modes here: https://en.wikipedia.org/wiki/Convergence_of_random_variables .

leehuan · Apr 17, 2017

Re: Statistics

$\text{Given }f_{X,Y}(x,y)=2(1+x+y)^{-3}, \, x,y >0\\ \text{find the density function of }X+Y$

$\text{The solutions presented the method of letting }U=X+Y\text{ and }V=Y\\ \text{which resulted in }f_{U,V}(u,v)=2(1+u)^{-3}\\ \text{and then they just found the marginal density }f_U(u),\\\text{Which I understand, and for reference their final answer was }f_U(u)=\frac{2u}{(1+u)^3}$
_________________________________________
$\text{But I tried using the formula for a sum of independent random variables. For }Z=X+Y:\\ f_Z(z)=\int_{\mathbb{R}} f_X(z-y)f_Y(u)\, dy$

$\text{I get }f_X(x)=(1+x)^{-2}\text{ and }f_Y(y)=(1+y)^{-2}\text{ and then things became a disaster}\\ \text{Am I doing something wrong?}$

(Aside: I also picked 0 and inf to be my boundaries of integration.)

InteGrand · Apr 17, 2017

Re: Statistics

leehuan said:
$\text{Given }f_{X,Y}(x,y)=2(1+x+y)^{-3}, \, x,y >0\\ \text{find the density function of }X+Y$

$\text{The solutions presented the method of letting }U=X+Y\text{ and }V=Y\\ \text{which resulted in }f_{U,V}(u,v)=2(1+u)^{-3}\\ \text{and then they just found the marginal density }f_U(u),\\\text{Which I understand, and for reference their final answer was }f_U(u)=\frac{2u}{(1+u)^3}$
_________________________________________
$\text{But I tried using the formula for a sum of independent random variables. For }Z=X+Y:\\ f_Z(z)=\int_{\mathbb{R}} f_X(z-y)f_Y(u)\, dy$

$\text{I get }f_X(x)=(1+x)^{-2}\text{ and }f_Y(y)=(1+y)^{-2}\text{ and then things became a disaster}\\ \text{Am I doing something wrong?}$

(Aside: I also picked 0 and inf to be my boundaries of integration.)

In this case X and Y are not independent, because the product of the marginal densities is not the same as the joint density.

leehuan · Apr 17, 2017

Re: Statistics

InteGrand said:
In this case X and Y are not independent, because the product of the marginal densities is not the same as the joint density.

Ouch. Thanks, overlooked that completely.

So I should always use the substitution method when they aren't independent?

InteGrand · Apr 17, 2017

Re: Statistics

leehuan said:
Ouch. Thanks, overlooked that completely.

So I should always use the substitution method when they aren't independent?

$\noindent Depends on the question at hand. There might be faster methods sometimes.$

$\noindent For example, another possible approach is to compute the CDF of $Z:= X+Y$, $F_{Z}(z) := \mathbb{P}\left(X+Y \leq z\right)$, and then differentiate this wrt $z$ to get the PDF. (Or use rules for differentiating integrals.) This may or may not be faster than the other approach, depending on the problem at hand.$

davidgoes4wce · Apr 17, 2017

Re: University Statistics Discussion Marathon

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final results to 2d.p but dont round numbers during your calculation.

Computer the probability that an observation 't' drawn from a normal distribution with mean $\mu$ = 24.1 and standard deviation $\sigma = 8.56$ is in the interval (15.05, 32.55) i.e the Prob (15.05 < t < 32.56) .

davidgoes4wce · Apr 18, 2017

Re: University Statistics Discussion Marathon

Just want to confirm if this is the right answer :

InteGrand · Apr 18, 2017

Re: University Statistics Discussion Marathon

davidgoes4wce said:
Just want to confirm if this is the right answer :

$\noindent I haven't checked your numerical calculations but yes that's the right approach, i.e. standardise the random variable using $Z = \frac{X -\mu}{\sigma}$ so that $Z$ becomes standard normal, correspondingly change the inequality bounds, and then use standard normal CDF values to find the required probabilities (e.g. in a table of values).$

Flop21 · Apr 18, 2017

Re: Statistics

How do you find the marginal distribution of a random variable, given a table of the joint probabilities of that random variable and another?

i.e. fY(y).

I see the solution but not sure where they are getting it from.

InteGrand · Apr 18, 2017

Re: Statistics

Flop21 said:
How do you find the marginal distribution of a random variable, given a table of the joint probabilities of that random variable and another?

i.e. fY(y).

I see the solution but not sure where they are getting it from.

You integrate or sum the joint density function over all values of the other variable(s).

leehuan · Apr 18, 2017

Re: Statistics

Flop21 said:
How do you find the marginal distribution of a random variable, given a table of the joint probabilities of that random variable and another?

i.e. fY(y).

I see the solution but not sure where they are getting it from.

$\text{This formula you're just gonna have to know. Because in our courses we only consider (at most) bivariate distributions}\\ \text{I'll just assume that the other r.v. is }X$

$f_Y(y) = \int_{\mathbb{R}} f_{X,Y}(x,y)\,dx\text{ for continuous}$

$\mathbb{P}(Y=y) = \sum_{\forall x}\mathbb{P}(X=x,Y=y)\text{ for discrete}$

$\text{with use of appropriate boundaries/indicator functions/whatever}$

Which is nothing different to what IG said. He basically just described a general case where you have n r.v.s

leehuan · Apr 20, 2017

Re: Statistics

$X_1, X_2 \sim \text{exp}(2)\text{ are iid.}$

And we're using the convention X ~ exp(a) => f_X(x) = 1/a exp(-x/a)

$\text{Determined in part a) }f_{X_1,X_2}(x_1,x_2)=\frac{1}{4}e^{-\frac{x_1+x_2}2}\text{ for }x_1>0, x_2>0$

$\text{Define }Y_1 := \frac12 (X_1-X_2)\text{ and }Y_2:=X_2\\ \text{Found in part b) }f_{Y_1,Y_2}(y_1,y_2) = \frac12 e^{-(y_1+y_2)}$

But how do you find the conditions on y₁ and y₂ where this density is defined for?

davidgoes4wce · Apr 20, 2017

Re: University Statistics Discussion Marathon

50 % of the meals sold at Graylands Hospital contain rice.

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final result to 2dp, but don't round numbers during your calculations.

What is the probability that out of 78 meals sold at Graylands Hospital, 43 or more contain rice?

InteGrand · Apr 20, 2017

Re: Statistics

leehuan said:
$X_1, X_2 \sim \text{exp}(2)\text{ are iid.}$

And we're using the convention X ~ exp(a) => f_X(x) = 1/a exp(-x/a)

$\text{Determined in part a) }f_{X_1,X_2}(x_1,x_2)=\frac{1}{4}e^{-\frac{x_1+x_2}2}\text{ for }x_1>0, x_2>0$

$\text{Define }Y_1 := \frac12 (X_1-X_2)\text{ and }Y_2:=X_2\\ \text{Found in part b) }f_{Y_1,Y_2}(y_1,y_2) = \frac12 e^{-(y_1+y_2)}$

But how do you find the conditions on y₁ and y₂ where this density is defined for?

$\noindent We just need to find what the region for the original sample space, which is $\Omega := \left\{ (x_{1}, x_{2}) \in \mathbb{R}^{2} : x_{1} > 0, x_{2} > 0\right\}$ (i.e. first quadrant), gets mapped to under the transformation in question, which is $\begin{bmatrix}y_{1} \\ y_{2}\end{bmatrix} = \begin{bmatrix}\frac{1}{2}x_{1} -\frac{1}{2}x_{2} \\ x_{2}\end{bmatrix}$. Since this is a linear transformation, this is quite easy to do, if you recall basic properties of how linear maps transform regions.$

InteGrand · Apr 20, 2017

Re: Statistics

$\noindent If you want a ``by inspection'' method, you can observe the new region will be $y_{2} > 0$ and $y_{1} > -\frac{1}{2}y_{2}$. This is essentially because $Y_{2} = X_{2}$, so the range of values for $y_{2}$ is just the same as $x_{2}$, so is $y_{2} > 0$. For $Y_{1}$, we note that $Y_{1} = -\frac{1}{2}X_{2} + \frac{1}{2} X_{1} = -\frac{1}{2}Y_{2} + \frac{1}{2}X_{1}$. Now, $\frac{1}{2}X_{1}$ has range $(0,\infty)$, so the values $Y_{1}$ can take go from than that of $-\frac{1}{2}Y_2$ up to (and not including) $+\infty$. This explains why the other thing was $y_{1} > -\frac{1}{2}y_{2}$.$

He-Mann · Apr 20, 2017

Re: University Statistics Discussion Marathon

davidgoes4wce said:
50 % of the meals sold at Graylands Hospital contain rice.

Answer the following question, keeping in mind that your answer must be a number between 0 and 1, i.e do not use percentages. Round off the final result to 2dp, but don't round numbers during your calculations.

What is the probability that out of 78 meals sold at Graylands Hospital, 43 or more contain rice?

Not sure... how to compute... (p = 0.5)

$X \sim \mathrm{Bin}(78, p) \implies P(X \geq 43) = \sum_{k=43}^{78} \binom{78}{k}p^k (1-p)^{78-k}$

Statistics Marathon & Questions (3 Viewers)

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