I thought that was already the answer I have for the first part? Well the 1/360000 bit. But I still cant get my head around the CDFhrm...
consider the 3 decimal representation of [MATH] 0-359.999[/MATH] degrees multiplied by [MATH]1000[/MATH].
This will be [MATH][0, 1, 2, \dots, 359999][/MATH]
Each spin is equally likely so
p[k]=3600001, k∈{1,2,…,359999}
can you figure out the CDF given the above pmf?
your answer doesn't really get the point across that this is a discrete distribution.
[MATH]CDF(k) = \begin{cases} 0 &k<0\\ \dfrac{k+1}{360000} &k = 0,1,2,\dots, 359999\\ 1 &359999 < k \end{cases}[/MATH][/QUOTE
your answer doesn't really get the point across that this is a discrete distribution.
[MATH]CDF(k) = \begin{cases} 0 &k<0\\ \dfrac{k+1}{360000} &k = 0,1,2,\dots, 359999\\ 1 &359999 < k \end{cases}[/MATH]
You be able to explain how you got this? Did I use the wrong formula?
Ahhh I understand now. The question was throwing me off because it asks for the probability density function and I’ve completed a previous question beforehand which was a continuous uniform distribution and had to approach it differently. Thank you very muchWell for one thing you don't seem to recognize that this is a discrete distribution. You use the word density function.
There are 360000 discrete values the reading of the spinner can assume, 0 to 359999. The actual value of the spinner is this number divided by 1000, but that doesn't matter for the purposes of the distribution.
[MATH]CDF[k] = P[\text{spinner reading}\times 1000 \leq k] = \sum \limits_{j=0}^k ~\dfrac{1}{360000} = \dfrac{k+1}{360000}[/MATH]
The spinner never reads less than 000.000 and never reads higher than 359.999.