Uniform Distribution: PDF, Mean, and Standard Deviation

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The discussion centers on understanding the uniform distribution of a continuous random variable X within the interval [2, 6]. The probability density function (PDF) is established as f(x) = 1/4, with the mean calculated as 4 and the standard deviation as 4/3. The conversation shifts to the distribution function, where it is clarified that the cumulative distribution function (CDF) is the integral of the PDF, representing the probability that X is less than or equal to a certain value. Participants confirm that the CDF, F(x), is indeed the same as the cumulative density function, measuring probabilities up to a specified point. Overall, the discussion emphasizes the relationship between the PDF and CDF in uniform distributions.
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Unifrom distribution ?

a continuous random variable X is uniformly distributed in the interval 2<=X<=6 and can take no other value
1.what is the probability density function ?
2.find mean and standard deviation of X ?
3.obtain distribution function of X?

yr obviously i can answer for first two question ...

1.for a uniform function
PDF is f(x)=1/(b-a)====>f(x)=1/4

2.mean=a+b/2=2+6/2=4;
SD= (b-a)^2/12=4^2/12=16/12=4/3

but i have no idea how to do the third part ...please can anybody help me out...what is distribution function of X ...Not only the answer ...theory behind it ...then i may be able to get the answer for my self
 
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Look up the uniform distribution function
 


i don't understand u mate ...can u explain it a bit !it will be really helpful thanks!
 


hey no help for 3 hours ...i have provided all ma works and waiting for someone to help!please tell me what is the distribution function ...how it is ddifer from PDF ?any idea ?//and how to approach it ?
 


Which of the words in "Look up uniform distribution function" did you not understand?
You seem to not know what a "distribution function" is. Surely that definition is given in your textbook or, if not, on the internet.


If you had done that you would have found that the "probability density function" is the derivative of distribution function and that the distribution function, the probability that the result is between a and b, is the integral, from a to b, of the density function.
 


yr thanks !i did think like that ...but i was getting 1 as the answer ...is it correct?if it is correct then all the uniform distributions have the same distribution function that is 1...if that so thanks for helping me ...
 


You misinterpreted what I said. In fact, now that I look at it, I didn't say it very well! The "distribution function", F(x), gives the probability a value of between a and b by F(b)- F(a). F(x) itself is the anti-derivative of f(x) the constant of integration determined by F(2)= 0. In other words, it is F(x)= \int_2^x f(x) dx.
 


Ok then answer should be F(x)=x/4-1/2; am i correct now ...ii think so ...so i have another question ...is it the same function which is caled as cumulative density function (CDF)??
 


rclakmal said:
Ok then answer should be F(x)=x/4-1/2; am i correct now ...ii think so ...so i have another question ...is it the same function which is caled as cumulative density function (CDF)??

The F(X) is called the CDF or cumulative distribution function. What this means is that the CDF measures P(X <= A) instead of the usual PDF measure which is P(X = A) or as in the continuous case P(A < X < B).

So the CDF measures probability up till a certain point of X. So its like saying that if we wanted to find P(X < A) we sum all probabilities up to but less than A instead of say at just A.
 
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