PDF and CDF of a spinner

Mos5180d

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Question:
Consider an anti-clockwise spinner which can land at any angle between 0 and 360. Each spin is recorded electronically to 3dp for example 38.764.

what is the pdf and cdf for the values of the result from one spin of the spinner?

this is my working but I’m sure it’s terribly wrong

D3A44EE0-F9B5-4827-BC87-A2C28F221ECD.jpeg
 
hrm...

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]=1360000, k{1,2,,359999}\displaystyle p[k] = \dfrac{1}{360000},~k \in \{1, 2, \dots ,359999\}

can you figure out the CDF given the above pmf?
 
hrm...

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]=1360000, k{1,2,,359999}\displaystyle p[k] = \dfrac{1}{360000},~k \in \{1, 2, \dots ,359999\}

can you figure out the CDF given the above pmf?
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 CDF
 
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]
 
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?
 
Well 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.
 
Well 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.
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 much
 
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