# Unsolved statistics questions from other sites...

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#### chisigma

##### Well-known member
This thread is opened by me with the purpose to answer to statistic questions proposed in other sites that didn't receive answer for not less that three days. Let's start from that question posted on mathhelpforum.com by Frida on 04 22 2012...

A variable of two populations has a mean of 7.9 and a standard deviation of 5.4 for one of the populations and a mean of 7.1 and a standard deviation of 4.6 for the other populations Can you conclude that the variable x1-x2 is normally distributed and why?...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
This thread is opened by me with the purpose to answer to statistic questions proposed in other sites that didn't receive answer for not less that three days. Let's start from that question posted on mathhelpforum.com by Frida on 04 22 2012...

A variable of two populations has a mean of 7.9 and a standard deviation of 5.4 for one of the populations and a mean of 7.1 and a standard deviation of 4.6 for the other populations Can you conclude that the variable x1-x2 is normally distributed and why?...

Kind regards

$\chi$ $\sigma$
No, not as it stands. Even now I find myself interpolating information that is not included in the question, like x1 and x2 are drawn from the first and second populations respectivly.

You can see that this is the case if in population 1 we value 7.9-5.4 with probability 0.5 and value 7.9+5.4 with probability 0.5, and population 2 we have 7.1-4.6 with probabilty 0.5 and 7.1+4.6 with probability 0.5. Then x1-x2 is discrete and so not normal (and the same sort of trick but a bit more subtle can be played if we insist on continuous distributions).

More likely is that the OP has missed the stipulation that the two populations be normal, when the answer is yes.

CB

#### chisigma

##### Well-known member
If one indicates mean and standard deviation of one random variable and nothing other, probably he implies that the random variable is normal distributed. If that is true, the $X_{1}$ and $X_{2}$ have p.d.f. ...

$\displaystyle f_{1}(x)= \frac{1}{\sqrt{2\ \pi}\ \sigma_{1}}\ e^{- \frac{(x - \mu_{1})^{2}}{2\ \sigma_{1}^{2}}}$

$\displaystyle f_{2}(x)= \frac{1}{\sqrt{2\ \pi}\ \sigma_{2}}\ e^{- \frac{(x - \mu_{2})^{2}}{2\ \sigma_{2}^{2}}}$ (1)

If we define a new random variable $X=X_{1}+X_{2}$, the X has p.d.f. $f(x)=f_{1}(x)\ *\ f_{2}(x)$ , when '*' means 'convolution'. Applying a basic property of Fourier Transform, if we set...

$\displaystyle F_{1}(\omega)= \mathcal {F} \{f_{1}(x)\} = e^{-i\ 2\ \pi\ \mu_{1}\ \omega}\ e^{-2\ \sigma_{1}^{2}\ \omega^{2}}$

$\displaystyle F_{2}(\omega)= \mathcal {F} \{f_{2}(x)\} = e^{-i\ 2\ \pi\ \mu_{2}\ \omega}\ e^{-2\ \sigma_{2}^{2}\ \omega^{2}}$ (2)

... is...

$\displaystyle \mathcal {F} \{f(x)\} = F_{1}(\omega)\ F_{2}(\omega) = e^{-i\ 2\ \pi\ (\mu_{1}+ \mu_{2} )\ \omega}\ e^{-2\ (\sigma_{1}^{2}+\sigma_{2}^{2})\ \omega^{2}}$ (3)

From (3) it is clear that X is a normal random variable with mean $\mu=\mu_{1}+\mu_{2}$ and standard deviation $\sigma= \sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}$. In Frida's example is $\mu_{1}=7.9$, $\mu_{2}=-7.1$, $\sigma_{1}=5.4$, $\sigma_{2}=4.6$. so that is $\mu=.8$ and $\sigma=7.09$. The extension to the case of number of r.v. greater than two is simple and is left to the reader...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
If one indicates mean and standard deviation of one random variable and nothing other, probably he implies that the random variable is normal distributed.
That implication is not acceptable, it gets people into bad habits like assuming you can word a question carelessly and still expect a correct answer.

We must tell posters to post the question as asked, or answer the question as asked and point out this is probably not what they really wanted to ask. Especially in this case like the one here, where the question asks if a particular conclusion is valid, the answer to the question as asked is no.

CB

#### chisigma

##### Well-known member
Posted the 04 15 2012 in the Italian site www.matematicamente.it f by the member lutteo2000 [original in Italian language...] and not yet solved...

On a circumference three points A,B and C are randomly chosen. What's the probability that the center O of the circumference is internal to the triangle ABC?...

Certainly lutteo2000 has very well formulated his question!... how to answer?...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Posted the 04 15 2012 in the Italian site www.matematicamente.it f by the member lutteo2000 [original in Italian language...] and not yet solved...

On a circumference three points A,B and C are randomly chosen. What's the probability that the center O of the circumference is internal to the triangle ABC?...

Certainly lutteo2000 has very well formulated his question!... how to answer?...

Kind regards

$\chi$ $\sigma$
Well I have a method, but it is difficult to explain without being able to sketch, but:

The position of the first point A can be taken to be (in polars) $$\theta=0$$, and if we know the smaller of the angles between the first and second point is $$\phi$$ the probability that the third forms a triangle that enclosed the centre of the circle is $$\phi / (2 \pi)$$ (this last bit is where a sketch is needed). Then as $$\phi$$ is $$\sim U(0, \pi)$$ the required probability is:

$P=\int_0^{\pi} \frac{\phi}{2 \pi} \frac{1}{ \pi}\; d\phi=1/4$

This result is supported by a simple Monte-Carlo calculation without simplifying assumptions.

CB

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#### chisigma

##### Well-known member
Posted the 04 15 2012 in the Italian site www.matematicamente.it f by the member lutteo2000 [original in Italian language...] and not yet solved...

On a circumference three points A,B and C are randomly chosen. What's the probability that the center O of the circumference is internal to the triangle ABC?...

Certainly lutteo2000 has very well formulated his question!... how to answer?...
The solution is surprisingly simple if one uses some tricks...

a) first we suppose that the circle is the unit disk and each point is defined by an angle...

b) second it is not a limitation to suppose that the first point is the point [1,0]...

c) third we normalize the angles to $\pi$ so that the second and third points are random variables uniformely distributed between 0 and 2...

Under these hypotheses, indicating with X and Y the angles of the second and third points, with simple geometric considerations one finds that the requested probability is...

$\displaystyle P=\frac{1}{4}\ \int_{0}^{1} dx\ \int_{x}^{1+x} dy= \frac{1}{4}$

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Posted on 03 20 2012 on www.scienzematematiche.it by the member fry [original in Italian language...] and not yet properly solved...

Let be $X_{n},\ n=1,2,...$ a sequence of [binary] independent random variables with $\displaystyle P\{X_{n}=0\}=P\{X_{n}=1\}=\frac{1}{2}$ and let's define the R.V. $\displaystyle Y=\sum_{n=1}^{\infty} \frac{X_{n}}{2^{n}}$. How to demonstrate that Y is uniformely distributed between 0 and 1?...

For completeness sake I would suggest to analyse the more general case $\displaystyle P\{X_{n}=0\}=p,\ 0<p<1$...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Posted on 03 20 2012 on www.scienzematematiche.it by the member fry [original in Italian language...] and not yet properly solved...

Let be $X_{n},\ n=1,2,...$ a sequence of [binary] independent random variables with $\displaystyle P\{X_{n}=0\}=P\{X_{n}=1\}=\frac{1}{2}$ and let's define the R.V. $\displaystyle Y=\sum_{n=1}^{\infty} \frac{X_{n}}{2^{n}}$. How to demonstrate that Y is uniformely distributed between 0 and 1?...
The demonstration I will give is based on the Fourier Transform and on the so called 'Vieta's product'. In the sixteenth century the French mathematician Francois Vieta was able to solve [first in Wenstern...] a non trivial infinite product. He started iteratively applying n times the trigonometric relationship...

$\displaystyle \sin 2\alpha=2\ \sin \alpha\ \cos \alpha$ (1)

… and obtaining…

$\displaystyle \sin\ 2^{n}\ \alpha = 2^{n}\ \sin \alpha\ \cos \alpha\ \cos^{2} \alpha\ …\cos 2^{n-1} \alpha$ (2)

Setting in (2) $\displaystyle \theta= 2^{n} \alpha$ he obtained first…

$\displaystyle \sin \theta= 2^{n}\ \sin \frac{\theta}{2^{n}}\ \cos \frac{\theta}{2^{n}}\ \cos \frac{\theta}{2^{n-1}}\ …\ \cos \frac{\theta}{2}$ (3)

… and from (3)…

$\displaystyle\frac{\sin \theta}{\theta}=\frac{\sin \frac{ \theta}{2^{n}}}{\frac{\theta}{2^{n}}}\ \prod_{k=1}^{n} \cos \frac{\theta}{2^{k}}$ (4)

… and finally leaving n to tend to infinity he obtained…

$\displaystyle \frac{\sin \theta}{\theta}= \prod_{k=1}^{\infty} \cos \frac{\theta}{2^{k}}$ (5)

… a very remarkable result for the sixteenth century!...

Now we consider that each $X_{n}$ has [discrete] p.d.f. $\displaystyle f_{n}(x)= \frac{\delta(x)+ \delta(x-\frac{1}{2^{n}})}{2}$, the Fourier Transform of which is…

$\displaystyle F_{n}(\omega)= \mathcal{F} \{f_{n}(x)\}= \frac{1}{2}\ (1+e^{- i\ \frac{\omega}{2^{n}}})= e^{-i\ \frac{\omega}{2^{n+1}}}\ \cos \frac{\omega}{2^{n+1}}$ (6)

From (6) it is possible to derive the F.T. of the p.d.f. of Y applying the convolution theorem and the (5)…

$\displaystyle F(\omega)= \prod_{n=1}^{\infty}F_{n}(\omega)= \prod_{n=1}^{\infty} (e^{-i\ \frac{\omega}{2^{n+1}}}\ \cos\frac{\omega}{2^{n+1}})= e^{-i\ \omega\ \sum_{n>0} \frac{1}{2^{n+1}}}\ \prod_{n=1}^{\infty} \frac{\cos \omega}{2^{2n+1}}= e^{-i\ \frac{\omega}{2}}\ \frac{\sin \frac{\omega}{2}}{\frac{\omega}{2}}$ (7)

... and performing the inverse F.T. of (7) we obtain finally...

$\displaystyle f(x)=\mathcal{F}^{-1}\{F(\omega)\}= \mathcal{U}(x)-\mathcal{U}(x-1)$ (8)

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Posted on 04 19 2012 on www.matematicamente.it by the member caporock [original in Italian language...] and not yet properly solved...

In a post office there is a queue of five waiting persons in front of the window. The service time $T$ is a R.V. uniformly distributed from 0 to 1 minute. Let be $T_{f}$ the time required to serve the five persons. Compute expected value and variance of $T_{f}$...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Posted on 04 19 2012 on www.matematicamente.it by the member caporock [original in Italian language...] and not yet properly solved...

In a post office there is a queue of five waiting persons in front of the window. The service time $T$ is a R.V. uniformly distributed from 0 to 1 minute. Let be $T_{f}$ the time required to serve the five persons. Compute expected value and variance of $T_{f}$...

Kind regards

$\chi$ $\sigma$
$$T_f$$ is the sum of $$5$$ $$U(0,1)$$ RV, so its mean is $$5 \times 0.5$$, and its variance is $$5 \times \frac{1}{12}$$

That is means and variances sum for independednt RVs (note I have assumed service times are independent)

And just to show off we can do a Monte-Carlo experiment to show this:

Code:
>N=1000000;
>tt=random(N,5);  ..generate N 5 person service times ~U(0,1)
>
>tf=(sum(tt))';   .. compute N total service times for 5 person queues
>
>{m,s}=meandev(tf);[m,s^2], .. compute mean and variance
2.50008      0.416111
>
>[5*0.5,5/12],   .. reference values
2.5      0.416667
>
CB

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#### bincybn

##### Member
Does the service time to serve each persons iid? Is that what uniform distribution?

If yes, exp(Tf)=5*exp(T)=2.5min
and var(Tf)=5*var(T)=5/12 min.

But i doubt it is not the Qn.

Sorry, i didn't see the prev. post

#### chisigma

##### Well-known member
Posted on 11 22 2011 on www.scienzematematiche.it [original in Italian...]...

What is the probability P that the distance between two random points inside an n-sphere of radious 1 is less than 1?...

I suggest to start with n=2, i.e. the 'sphere' is a circle...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Posted on 11 22 2011 on www.scienzematematiche.it [original in Italian...]...

What is the probability P that the distance between two random points inside an n-sphere of radious 1 is less than 1?...

I suggest to start with n=2, i.e. the 'sphere' is a circle...

Clearly it is not limitative to suppose that the first point $x_{0}$ and in this case, as illustrated in the figure...

... the probability that, setting $x_{1}$ the second random point, is $|x_{1}-x_{0}|<1$ is the ratio between the two circular segments separated by the line $x=\frac{x_{0}}{2}$ and the area of the unit circle, i.e. ...

$\displaystyle P\{|x_{1}-x_{0}|<1\} = \frac{2}{\pi}\ (\cos ^{-1} \frac{x_{0}}{2} - \frac{x_{0}}{2}\ \sqrt{1-\frac{x_{0}^{2}}{4}})$ (1)

But $x_{0}$ is uniformly distributed from 0 to 1, so that the requested probability is...

$\displaystyle P= \frac{4}{\pi} \int_{0}^{\frac{1}{2}} (\cos^{-1} x -x\ \sqrt{1-x^{2}})\ dx = \frac{4}{\pi}\ |x\ \cos^{-1} x -\sqrt{1-x^{2}} + \frac{(1-x^{2})^{\frac{3}{2}}}{3}|_{0}^{\frac{1}{2}}= .688849968669...$ (2)

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Clearly it is not limitative to suppose that the first point $x_{0}$ and in this case, as illustrated in the figure...
View attachment 170

... the probability that, setting $x_{1}$ the second random point, is $|x_{1}-x_{0}|<1$ is the ratio between the two circular segments separated by the line $x=\frac{x_{0}}{2}$ and the area of the unit circle, i.e. ...

$\displaystyle P\{|x_{1}-x_{0}|<1\} = \frac{2}{\pi}\ (\cos ^{-1} \frac{x_{0}}{2} - \frac{x_{0}}{2}\ \sqrt{1-\frac{x_{0}^{2}}{4}})$ (1)

But $x_{0}$ is uniformly distributed from 0 to 1, so that the requested probability is...

$\displaystyle P= \frac{4}{\pi} \int_{0}^{\frac{1}{2}} (\cos^{-1} x -x\ \sqrt{1-x^{2}})\ dx = \frac{4}{\pi}\ |x\ \cos^{-1} x -\sqrt{1-x^{2}} + \frac{(1-x^{2})^{\frac{3}{2}}}{3}|_{0}^{\frac{1}{2}}= .688849968669...$ (2)

Kind regards

$\chi$ $\sigma$
$$x_0$$ is not uniformly distributed from 0 to 1 since it is the radial component of a random point in the unit circle in polars.

$$\displaystyle p(x_0)= \frac{x_0}{\pi}$$ for $$x_0 \in [0,1]$$ and zero otherwise.

CB

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#### chisigma

##### Well-known member
$$x_0$$ is not uniformly distributed from 0 to 1 since it is the radial component of a random point in the unit circle in polars.

$$\displaystyle p(x_0)= \frac{x_0}{\pi}$$ for $$x_0 \in [0,1]$$ and zero otherwise.

CB
Of course we have to agree on the definition of 'random point inside the unit circle'. The definition I adopted is a complex number of the form...

$\displaystyle x_{0}= \rho\ e^{i\ \theta}$ (1)

... where $\rho$ is a R.V. uniformly distributed from 0 to 1 and $\theta$ is a R.V. uniformly distributed from $-\pi$ to $+\pi$. As in all problem with circular symmetry You can set $\theta=0$ so that $x_{0}$ is a real R.V. uniformly distributed from 0 to 1. If the definition is different, of course all must be revised... but what is that 'different definition'?...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Of course we have to agree on the definition of 'random point inside the unit circle'. The definition I adopted is a complex number of the form...

$\displaystyle x_{0}= \rho\ e^{i\ \theta}$ (1)

... where $\rho$ is a R.V. uniformly distributed from 0 to 1 and $\theta$ is a R.V. uniformly distributed from $-\pi$ to $+\pi$. As in all problem with circular symmetry You can set $\theta=0$ so that $x_{0}$ is a real R.V. uniformly distributed from 0 to 1. If the definition is different, of course all must be revised... but what is that 'different definition'?...

Kind regards

$\chi$ $\sigma$
The definition of a uniform distribution on a region $$R$$ of $$\mathbb{R}^n$$ containing a $$n$$ dimensional ball, is something like that any sub-region has probability of occurring equal to the ratio of its (hyper)volume to that of $$R$$.

It is not my definition, it is the definition (with concessions to approximating the measure theory form of the definition)

CB

#### CaptainBlack

##### Well-known member
Posted on 11 22 2011 on www.scienzematematiche.it [original in Italian...]...

What is the probability P that the distance between two random points inside an n-sphere of radious 1 is less than 1?...

I suggest to start with n=2, i.e. the 'sphere' is a circle...

Kind regards

$\chi$ $\sigma$
It is fairly easy to provide some estimates of the required probability using Monte-Carlo methods. My calculations give:

Code:
            N         p_est      SE
-----------------------------
1        0.7441    0.0043
2        0.5866    0.0049
3        0.4706    0.0050
4        0.3712    0.0048
5        0.3153    0.0046
6        0.2608    0.0044
7        0.2154    0.0041
8        0.1714    0.0038
9        0.1445    0.0035
10        0.1180    0.0032
where N is the dimension of the problem, p_est is the MC estimate of the probability and SE is the approximate standard error of the estimate.

CB

#### chisigma

##### Well-known member
The original question was about 'the distance between two random point in an n-sphere of radius 1'... I proposed a precise definition of 'random point' in the case n=2 and someone seems don't agree with me... never mind!... but I don't understand what is the correct definition of 'random point' in a 2-sphere of radius 1... can someone solve my doubt, please!...

Kind regards

$\chi$ $\sigma$

#### chisigma

##### Well-known member
Clearly it is not limitative to suppose that the first point $x_{0}$ and in this case, as illustrated in the figure...
View attachment 170

... the probability that, setting $x_{1}$ the second random point, is $|x_{1}-x_{0}|<1$ is the ratio between the two circular segments separated by the line $x=\frac{x_{0}}{2}$ and the area of the unit circle, i.e. ...

$\displaystyle P\{|x_{1}-x_{0}|<1\} = \frac{2}{\pi}\ (\cos ^{-1} \frac{x_{0}}{2} - \frac{x_{0}}{2}\ \sqrt{1-\frac{x_{0}^{2}}{4}})$ (1)

But $x_{0}$ is uniformly distributed from 0 to 1, so that the requested probability is...

$\displaystyle P= \frac{4}{\pi} \int_{0}^{\frac{1}{2}} (\cos^{-1} x -x\ \sqrt{1-x^{2}})\ dx = \frac{4}{\pi}\ |x\ \cos^{-1} x -\sqrt{1-x^{2}} + \frac{(1-x^{2})^{\frac{3}{2}}}{3}|_{0}^{\frac{1}{2}}= .688849968669...$ (2)

Kind regards

$\chi$ $\sigma$
... if 'random point inside a unit circle' means that the probability to find that point inside a region of area A included in the unit circle is $\frac{A}{\pi}$, then $x_{0}$ effectively isn't uniformly distributed from 0 to 1 but its p.d.f. is $2\ x$. In that case the requested probability is...

$\displaystyle P= \frac{4}{\pi}\ \int_{0}^{1} \left(x\ \cos^{-1} \frac{x}{2} -\frac{x}{2}\ \sqrt{1-\frac{x^{2}}{4}}\right)\ dx$
.............$\displaystyle = \frac{4}{\pi}\ \left| \left(\frac{x^{2}}{2}-1\right)\ \cos^{-1} \frac{x}{2} - \frac{x\ \sqrt{4-x^{2}}}{4} + \frac{x\ (4-x^{2})^{\frac{3}{2}}}{16} - \frac{x\ \sqrt{4-x^{2}}}{8} - \frac{1}{2}\ \sin^{-1} \frac{x}{2}\right|_{0}^{1}$

.............$= .586503328433...$

Kind regards

$\chi$ $\sigma$

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#### chisigma

##### Well-known member
Posted on 05 12 2012 on bythe member francifamy [original in Italian language...] and not yet properly solved...

A boy is hitchhiking on a land road where passes a mean of 1 car every 10 minutes,according with Poisson’s statistic. If the probability that a car takes on board the boy is .1, what is the probability that in 30 minutes no car takes on board the boy?...

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Posted on 05 12 2012 on bythe member francifamy [original in Italian language...] and not yet properly solved...

A boy is hitchhiking on a land road where passes a mean of 1 car every 10 minutes,according with Poisson’s statistic. If the probability that a car takes on board the boy is .1, what is the probability that in 30 minutes no car takes on board the boy?...

Kind regards

$\chi$ $\sigma$

The number of cars in a time interval that would give the boy a lift has a Poisson distribution with 1/10 the mean of the number of cars.

CB

#### chisigma

##### Well-known member
The number of cars in a time interval that would give the boy a lift has a Poisson distribution with 1/10 the mean of the number of cars.

CB
Very well!... if we assume the time unit to be 30 minutes, then $\lambda=3$ and the probability that pass k cars in 30 minutes is...

$\displaystyle P_{k}= e^{- \lambda}\ \frac{\lambda^{k}}{k!}= e^{-3}\ \frac{3^{k}}{k!}$ (1)

For each car the probability don't to take on board the boy is $p=.9$. so that the requested probability is...

$\displaystyle P= e^{- \lambda}\ \sum_{k=0}^{\infty} \frac{(\lambda p)^{k}}{k!}= e^{-.3} = .7408182206817...$ (2)

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
Very well!... if we assume the time unit to be 30 minutes, then $\lambda=3$ and the probability that pass k cars in 30 minutes is...

$\displaystyle P_{k}= e^{- \lambda}\ \frac{\lambda^{k}}{k!}= e^{-3}\ \frac{3^{k}}{k!}$ (1)

For each car the probability don't to take on board the boy is $p=.9$. so that the requested probability is...

$\displaystyle P= e^{- \lambda}\ \sum_{k=0}^{\infty} \frac{(\lambda p)^{k}}{k!}= e^{-.3} = .7408182206817...$ (2)

Kind regards

$\chi$ $\sigma$
The mean number of potential lifts in 30 minutes is 0.3

CB

#### chisigma

##### Well-known member
The mean number of potential lifts in 30 minutes is 0.3

CB
Before today it was obvious for me that, if the expected value of arriving cars in 10 minutes is 1, then the expected number of arriving cars in 30 minutes would be 3... may be that recently new regulations have been introduced? ...

Kind regards

$\chi$ $\sigma$

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