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Question:
\n" ); document.write( "If X has the distribution function given by;
\n" ); document.write( " 0 for x<1
\n" ); document.write( " 1/3 for 1≤x<4
\n" ); document.write( " F(X) = 1/2 for4≤x<6
\n" ); document.write( " 5/6 for 6≤x<10
\n" ); document.write( " 1 for x≥10
\n" ); document.write( "Find;
\n" ); document.write( "The probability density function of X
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\n" ); document.write( "Solution:
\n" ); document.write( "To solve the problem, one needs to focus on the relationship between p(x) (pdf) and F(x) (cdf).
\n" ); document.write( "read up the textbook if necessary to make sure you are aware of the terms and how they are related.\r
\n" ); document.write( "\n" ); document.write( "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.
\n" ); document.write( "So it is interpreted that all values given are F(X).
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\n" ); document.write( "Given:
\n" ); document.write( "X F(x)
\n" ); document.write( "<1 0
\n" ); document.write( "1-4 1/3
\n" ); document.write( "4-6 1/2
\n" ); document.write( "6-10 5/6
\n" ); document.write( ">=10 1
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\n" ); document.write( "This means that the pdf distribution is non-zero between 1 and 10 (where F(x) ranges between 0 and 1).
\n" ); document.write( "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.
\n" ); document.write( "At X=1, p(1)=1/3
\n" ); document.write( "At X=4, p(4)=(1/2-1/3)=1/6
\n" ); document.write( "at X=6, p(6)=(5/6-1/2)=1/3
\n" ); document.write( "at x=10,p(10)=(1-5/6)=1/6
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\n" ); document.write( "Sum all probabilities (1/3+1/6+1/3+1/6)=1, => p(x) is indeed a pmf.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "