SOLUTION: If X has the distribution function given by;
0 for x<1
1/3 for 1≤x<4
F(X) = 1/2 for4≤x<6
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Question 1087223: If X has the distribution function given by;
0 for x<1
1/3 for 1≤x<4
F(X) = 1/2 for4≤x<6
5/6 for 6≤x<10
1 for x≥10
Find;
(i)The probability density function of X
Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!
Question:
If X has the distribution function given by;
0 for x<1
1/3 for 1≤x<4
F(X) = 1/2 for4≤x<6
5/6 for 6≤x<10
1 for x≥10
Find;
The probability density function of X
Solution:
To solve the problem, one needs to focus on the relationship between p(x) (pdf) and F(x) (cdf).
read up the textbook if necessary to make sure you are aware of the terms and how they are related.
From the given information, it seems that p(x)=1/3 between 1 and 4, which is not possible, because area below 1-4 would have been 1 > F(4)=1/2.
So it is interpreted that all values given are F(X).
Given:
X F(x)
<1 0
1-4 1/3
4-6 1/2
6-10 5/6
>=10 1
This means that the pdf distribution is non-zero between 1 and 10 (where F(x) ranges between 0 and 1).
Since F(x) is a horizontal line within the given ranges, cumulative probability (cdf) is therefore a step function, and pmf (probability mass function) have values attributed to the steps.
At X=1, p(1)=1/3
At X=4, p(4)=(1/2-1/3)=1/6
at X=6, p(6)=(5/6-1/2)=1/3
at x=10,p(10)=(1-5/6)=1/6
Sum all probabilities (1/3+1/6+1/3+1/6)=1, => p(x) is indeed a pmf.
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