Statistics Marathon & Questions (2 Viewers)

davidgoes4wce · Apr 7, 2017

Re: University Statistics Discussion Marathon

Id rule out A straight up as well, which would love me down to B or C.

davidgoes4wce · Apr 8, 2017

Re: University Statistics Discussion Marathon

Statistical validity refers to whether a statistical study is able to draw conclusions that are in agreement with statistical and scientific laws. This means if a conclusion is drawn from a given data set after experimentation, it is said to be scientifically valid if the conclusion drawn from the experiment is scientific and relies on mathematical and statistical laws.

id also refer out C, and would go for B in this question.

Flop21 · Apr 10, 2017

Re: Statistics

is there an easy way to find the probability mass function??

right now im just looking at example P(X=x) like P(X=1... etc.) and trying to find a pattern

but this is hard for tricky ones

He-Mann · Apr 10, 2017

Re: Statistics

Flop21 said:
is there an easy way to find the probability mass function??

right now im just looking at example P(X=x) like P(X=1... etc.) and trying to find a pattern

but this is hard for tricky ones

Got an example question?

Flop21 · Apr 10, 2017

Re: Statistics

He-Mann said:
Got an example question?

A box contains four red and two black balls. Two balls are drawn. Let X be
the number of red balls obtained. Find fX(x)

Okay so if I haven't screwed up, P(X=0) = 1/15, P(X=1) = 4/5, P(X=2) = 2/5, and any other value of X, P = 0.

So how do I put this into a proper answer for this question?

InteGrand · Apr 10, 2017

Re: Statistics

Flop21 said:
A box contains four red and two black balls. Two balls are drawn. Let X be
the number of red balls obtained. Find fX(x)

Okay so if I haven't screwed up, P(X=0) = 1/15, P(X=1) = 4/5, P(X=2) = 2/5, and any other value of X, P = 0.

So how do I put this into a proper answer for this question?

$\noindent Those answers aren't right (assuming sampling without replacement). Note your pmf values don't sum to $1$, which they should. This'll just be a hypergeometric distribution, but you don't really need to know that here. We have, letting $f(k) \equiv f_{X}(k) = \mathbb{P}\left(X = k\right),$

$\mathbb{P}\left(X = 0\right) = \frac{1}{\binom{6}{2}} = \frac{1}{15} \quad (\text{so you got this one right})$

$\mathbb{P}\left(X = 1\right) = \frac{2\times 4}{\binom{6}{2}} = \frac{8}{15}$

$\noindent and$

$\mathbb{P}\left(X = 2\right) =\frac{\binom{4}{2}}{\binom{6}{2}} = \frac{6}{15} = \frac{2}{5} \quad (\text{you got this one right too}).$

$\noindent (Perhaps you just typoed one of your answers since you got the other two right.) So the pmf is $f(0) = \frac{1}{15}$, $f(1) = \frac{8}{15}$ and $f(2) = \frac{2}{5}$, and of course $f(k) = 0$ for $k \notin \left\{0,1,2\right\}$.$

Flop21 · Apr 10, 2017

Re: Statistics

InteGrand said:
$\noindent Those answers aren't right (assuming sampling without replacement). Note your pmf values don't sum to $1$, which they should. This'll just be a hypergeometric distribution, but you don't really need to know that here. We have, letting $f(k) \equiv f_{X}(k) = \mathbb{P}\left(X = k\right),$

$\mathbb{P}\left(X = 0\right) = \frac{1}{\binom{6}{2}} = \frac{1}{15} \quad (\text{so you got this one right})$

$\mathbb{P}\left(X = 1\right) = \frac{2\times 4}{\binom{6}{2}} = \frac{8}{15}$

$\noindent and$

$\mathbb{P}\left(X = 2\right) =\frac{\binom{4}{2}}{\binom{6}{2}} = \frac{6}{15} = \frac{2}{5} \quad (\text{you got this one right too}).$

$\noindent (Perhaps you just typoed one of your answers since you got the other two right.) So the pmf is $f(0) = \frac{1}{15}$, $f(1) = \frac{8}{15}$ and $f(2) = \frac{2}{5}$, and of course $f(k) = 0$ for $k \notin \left\{0,1,2\right\}$.$

yeah whoops typo'd something in my calculations so that stuffed up,

so your final sentence, is that enough for an answer or do I need to know how to put it in a form with x ?

InteGrand · Apr 10, 2017

Re: Statistics

Flop21 said:
yeah whoops typo'd something in my calculations so that stuffed up,

so your final sentence, is that enough for an answer or do I need to know how to put it in a form with x ?

$\noindent What do you mean a form with $x$? You don't need to express it as a formula like $f(x) = \frac{\binom{4}{x}\binom{2}{2-x}}{\binom{6}{x}}$ or something (this is the formula for it though if you're curious) if that's what you mean. Writing it as I wrote it is fine, since there's only three pmf values, so it's easy to list them out. If we had a more general situation or many pmf values though, it might be easier to give a formula for the pmf then, since it might not be possible or practical to list out all pmf values separately (there may even be infinitely many values, so you'd really want a formula then generally speaking).$

Flop21 · Apr 10, 2017

Re: Statistics

extremely helpful thanks!

Flop21 · Apr 10, 2017

Re: Statistics

I have notes saying if X is discrete, cdf is SUM t<=x fX(t)

I might not be looking hard enough but I cannot find examples of this being used anywhere,

say we are given:

fX(x) = P(X = x) = p(1 − p)^x, x = 0, 1, 2, . . . ; 0 < p < 1

how do I calculate the FX(x) = P(X <= x)?

Or to start me off, how do I use what I'm given in the formula? Thanks.

InteGrand · Apr 10, 2017

Re: Statistics

Flop21 said:
I have notes saying if X is discrete, cdf is SUM t<=x fX(t)

I might not be looking hard enough but I cannot find examples of this being used anywhere,

say we are given:

fX(x) = P(X = x) = p(1 − p)^x, x = 0, 1, 2, . . . ; 0 < p < 1

how do I calculate the FX(x) = P(X <= x)?

Or to start me off, how do I use what I'm given in the formula? Thanks.

$\noindent [Aside: the distribution with that pmf is the \emph{geometric distribution on } $\mathbb{N}:= \left\{ 0,1,2,\ldots \right\}$ with parameter $p$.]$

$\noindent It is really only of interest to find the CDF for non-negative integer values. For each $k\in \mathbb{N}$, the CDF is$

$\begin{align*}F_{X}(k) \equiv \mathbb{P}\left(X \leq k\right) &= \sum_{j = 0}^{k}f_{X}(j) = \sum_{j=0}^{k} \mathbb{P}\left(X = j\right)\\ &= \sum_{j=0}^{k} p(1-p)^{j}, \end{align*}$

$\noindent and you can calculate this sum using the geometric series formula (note there are $k+1$ terms).$

Flop21 · Apr 11, 2017

Re: Statistics

InteGrand said:
$\noindent [Aside: the distribution with that pmf is the \emph{geometric distribution on } $\mathbb{N}:= \left\{ 0,1,2,\ldots \right\}$ with parameter $p$.]$

$\noindent It is really only of interest to find the CDF for non-negative integer values. For each $k\in \mathbb{N}$, the CDF is$

$\begin{align*}F_{X}(k) \equiv \mathbb{P}\left(X \leq k\right) &= \sum_{j = 0}^{k}f_{X}(j) = \sum_{j=0}^{k} \mathbb{P}\left(X = j\right)\\ &= \sum_{j=0}^{k} p(1-p)^{j}, \end{align*}$

$\noindent and you can calculate this sum using the geometric series formula (note there are $k+1$ terms).$

-edited because I just realised there's solutions

Thanks btw super useful.

Flop21 · Apr 11, 2017

Re: Statistics

Given the diameter of a trunk fX(x) = 1, 0 < x < 1

I found that the E(X) = 0.5

How do I find the E(X) of the area of the trunk?

I tried doing it this way, area = Pi*r^2, so Pi*(d/2)^2, then I went so d = 1, make LHS = Pi*(d/2)^2 and then you get RHS = Pi/4 which is the area fX(x).

Then I found the E(X) of this, and got Pi/2, which is incorrect.

The correct answer is Pi/12, since they get area = Pi/4 * X^2, and E(A) = Pi/4*E(X^2).

So how did they get that Pi/4 * X^2?

InteGrand · Apr 11, 2017

Re: Statistics

Flop21 said:
Given the diameter of a trunk fX(x) = 1, 0 < x < 1

I found that the E(X) = 0.5

How do I find the E(X) of the area of the trunk?

I tried doing it this way, area = Pi*r^2, so Pi*(d/2)^2, then I went so d = 1, make LHS = Pi*(d/2)^2 and then you get RHS = Pi/4 which is the area fX(x).

Then I found the E(X) of this, and got Pi/2, which is incorrect.

The correct answer is Pi/12, since they get area = Pi/4 * X^2, and E(A) = Pi/4*E(X^2).

So how did they get that Pi/4 * X^2?

$\noindent We want the (expected) \emph{area}, whilst $X$ is the \emph{diameter}, so the area random variable $A$ is $A =\frac{\pi}{4}X^{2}$. (Remember the formula area of a circle given its diameter.) Hence to find the expected area, we need to find the expected value of $X^{2}$, not $X$.$

$\noindent Using expected value formula, this is$

$\mathbb{E}\left[X^{2}\right] = \int_{0}^{1} f_{X}(x)\cdot x^{2}\,\mathrm{d}x =\int_{0}^{1} 1\cdot x^{2}\,\mathrm{d}x = \frac{1}{3}.$

$\noindent Multiplying this by $\frac{\pi}{4}$ gets us the expected area.$

davidgoes4wce · Apr 12, 2017

Re: University Statistics Discussion Marathon

InteGrand · Apr 12, 2017

Re: University Statistics Discussion Marathon

davidgoes4wce said:

Third option

davidgoes4wce · Apr 12, 2017

Re: University Statistics Discussion Marathon

Damn I would have gone for 4th..........

davidgoes4wce · Apr 12, 2017

Re: University Statistics Discussion Marathon

any idea how you would explain it?

$\sum_{i=1} ^n (x_i -1)^2$

$\sum_{i=1} ^n x_i ^2-2x_i+1$

I don't see how you then can get to n-1

InteGrand · Apr 12, 2017

Re: University Statistics Discussion Marathon

davidgoes4wce said:
any idea how you would explain it?

$\sum_{i=1} ^n (x_i -1)^2$

$\sum_{i=1} ^n x_i ^2-2x_i+1$

I don't see how you then can get to n-1

$\noindent Use the sample variance formula: $S^{2} = \frac{1}{n-1}\sum_{i = 1}^{n}\left(x_{i} - \overline{x}\right)^{2}$, where $S^{2}$ is the sample variance and $\overline{x}$ is the sample mean. Note that if the (sample) standard deviation equals the variance, since one is the square of the other, we must have that the sample variance is $S^{2} = 1$ (it can't be $0$ since we are told it is positive). We are told the mean is equal to this too, so the mean is also $\overline{x} = 1$. Hence we have $1 = \frac{1}{n-1}\sum_{i=1}^{n}\left(x_{i} - 1\right)^{2}$, from the sample variance formula. Now rearranging this yields the result.$

InteGrand · Apr 12, 2017

Re: University Statistics Discussion Marathon

davidgoes4wce said:
Damn I would have gone for 4th..........

$\noindent An example of such a dataset: $n = 2$, with $x_{1} = \frac{2-\sqrt{2}}{2}, x_{2} = \frac{2 + \sqrt{2}}{2}$.$

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