Probability and Statistics (Quartiles)Q: Find the upper and lower quartiles of the random variable. f ( x ) = 1 x ln ⁡ ( 1.5 ) for 4 ≤ x ≤ 6I set F ( x ) = 0.25 and F ( x ) = 0.75 to denote Q 1 and Q 3This is what I got for the answers:For F ( X ) = 0.25 , x = 9.86For F ( X ) = 0.75 , x = 3.28However, these answers are wrong as compared with my textbook. It should be:For F ( X ) = 0.25 , x = 4.43For F ( X ) = 0.75 , x = 5.42Please help me. I don't know what I'm doing wrong.

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Probability and Statistics (Quartiles)Q: Find the upper and lower quartiles of the random variable.  f  (  x  )  =      1          x      ln      ⁡      (      1.5      )      for   4  ≤  x  ≤  6I set   F  (  x  )  =  0.25 and   F  (  x  )  =  0.75 to denote   Q  1 and   Q  3This is what I got for the answers:For  F  (  X  )  =  0.25  ,  x  =  9.86For  F  (  X  )  =  0.75  ,  x  =  3.28However, these answers are wrong as compared with my textbook. It should be:For  F  (  X  )  =  0.25  ,  x  =  4.43For  F  (  X  )  =  0.75  ,  x  =  5.42Please help me. I don&#039;t know what I&#039;m doing wrong.

Diya Bass · Accepted Answer

For   x between   4 and   6 we have  F  (  x  )  =  Pr  (  X  ≤  x  )  =      ∫    4    x        1          ln      ⁡      (      1.5      )      t          d  t  =      1          ln      ⁡      (      1.5      )        (  ln  ⁡  x  −  ln  ⁡  4  )  .For the first quartile       q    1  , we want      1          ln      ⁡      (      1.5      )        (  ln  ⁡  (      q    1    )  −  ln  ⁡  4  )  =  0.25.To solve, note that the above equation is equivalent to  ln  ⁡  (      q    1        /    4  )  =  (  0.25  )  ln  ⁡  (  1.5  )  .Equivalently,            q      1        4    =  exp  ⁡  (  (  0.25  )  ln  ⁡  (  1.5  )  )  =  (  1.5      )          0.25        .We get       q    1    ≈  4.427. I am sure you can now take care of       q    3

Probability and Statistics (Quartiles) Q: Find the upper and lower quartiles of the random variabl

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