Uniform PDF for continuous variable, why does the probability values increase to 1, when its normalized? Consider a "spinner": an object like an unmagnetized compass needle that can pivots freely around an axis, and is stable pointing in any direction. You give it a spin and see where it comes to rest, measuring the resulting angle (divided by 2π) as a number from 0 to 1.
I am bit confused, when i look into the PDF for this distribution, when each outcome its divides by 2π, the probability of each outcome turns out be 1. I mean when we draw a PDF we get a horizontal straight line at 1. However the chance of getting a value within the range 0 to 360 (2π) = 1/ 360, so when I plot the PDF for 0 to 360, it is a straight line at 0.0028, where as when i divide the same by 360 its a straight line at 1. I know the area under the curve for PDF 1 but, just becoz I divided by 360, why is the chance of occurrence or probability for each out come increases to 1 rather than 0.0028?
 A: If you're using radians, then you have $U_1 \sim Unif(0, 2\pi).$
Then the density function must have height $1/2\pi$ over $(0, 2\pi).$
If you're using degrees, then you have $U_2 \sim Unif(0, 360),$
which has density $f_{U_2}(x) = 1/360,$ for $0 < x < 360\,$ (and $0$
elsewhere).
If you define $U_3 = U_1/2\pi$ then you have $U_3 \sim Unif(0,1),$
which has density $f_{U_3}(x) = 1,$ for $0 < x < 1$ (and $0$
elsewhere. Because the density function is positive over the
(long) interval $(0, 360)$, the height of the density function
must be (small) $1/360$ in order to enclose the area 1.
And if you define $U_4 = U_2/360$ then you also
have $U_b \sim Unif(0,1).$
Notice that each of these density functions encloses an area of 1, as
must be the case for any density function. 
In general for $a < b$, $Unif(a, b)$ has
$$ \int_{- \infty}^\infty f(x)\,dx
= \int_{-\infty}^0 0\,dx + \int_a^b \frac{1}{b-a}\,dx
+ \int_b^\infty 0\,dx = \int_a^b \frac{1}{b-a}\,dx = 1.$$
Also, in each of your cases, the probability the spinner stops in the first
quadrant is $1/4,$ whether you denote the 'first quadrant' as
$(0, \pi/2)$, as $(0, 90)$, or $(0, 1/4).$ The definition of the
density function must match your definition of 'first quadrant'.
A: 1) Remember, that for a continuous random variable, $Pr(X=x) = 0 $. To derive the PDF write the CDF as : 
$$ Pr[X \leq x] = \int_{0}^{x} \frac{1}{2\pi}dx' = \frac{x}{2\pi} $$
Hence PDF is:
$$ f(x) = \frac{1}{2\pi}$$
2) It is inappropriate to think that there are exactly 360 points between 0 and 360 or 1 point between 0 and 1.
3) Even if you choose to take an experiment between $[0,1]$ the r.v. follows a uniform
A: You can depict this uniform distribution in many different ways:
(a) Angle measured in radian and divided by $2\pi$ results in a uniform distribution over the interval $[0,1)$. In this case the pdf is
$$f_1(x)=\begin{cases}
1,& \text{ if } 0\le x <1\\
0,& \text{ otherwise. }
\end{cases}$$
(b) Angle measured in radian (not divided by $2\pi$) results in a uniform distribution over $[0,2\pi)$. In this case the pdf is
$$f_{2\pi}(x)=\begin{cases}
\frac1{2\pi},& \text{ if } 0\le x <2\pi\\
0,& \text{ otherwise. }
\end{cases}$$
(c) Angle measured in degrees results in a uniform distribution over $[0^{\circ},360^{\circ})$. This method results in a uniform distribution over $[0^{\circ},360^{\circ})$ and the pdf is
$$f_{\circ}(x)=\begin{cases}
\frac1{360},& \text{ if } 0\le x <360\\
0,& \text{ otherwise. }
\end{cases}$$
The question is if these pdf's give different results if one wants to calculate the probability that the point of the compass points in an interval $[\alpha,\beta)$?
No, the results will be the same if one transforms $\alpha$ and $\beta$ and uses the corresponding pdf. For instance, let $\alpha=15^{\circ}$ and $\beta=30^{\circ}$.
Then
$$P(15^{\circ}\le \text{ angle } \le 30^{\circ})=\int_{15^{\circ}}^{30^{\circ}}f_{\circ}(x)\ dx=\frac{15}{360}.$$
Having converted the angles to radians we have $\alpha=\frac{15^{\circ}}{360^{\circ}}2\pi$ and $\beta=\frac{30^{\circ}}{360^{\circ}}2\pi$. Then
$$P\left(\frac{30^{\circ}}{360^{\circ}}2\pi\le \text{ angle } \le 30^{\circ}\right)=\int_{\frac{15^{\circ}}{360^{\circ}}2\pi}^{\frac{30^{\circ}}{360^{\circ}}2\pi}f_{2\pi}(x)\ dx=\frac{15}{360}.$$
The result would be the same if we used the angles $\alpha=\frac{15^{\circ}} {360^{\circ}}$ and $\beta=\frac{30^{\circ}} {360^{\circ}}$ and $f_1$.
