Calculating histogram function, $h_X(75.5)$ and $int_{54.2}^{67.8} h_X(x)$












0












$begingroup$


A population $prod$ and a measure $X$ on $prod$ produced the data set below



$71, 80, 58, 60, 63, 64, 51, 55, 48, 67$



Let $h_X$ denote the density histogram function of $X$ based on the four intervals



$(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$



$(a)$ Calculate $h_X(75.5)$



$(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$





attempt



$(a)$ I made a table



begin{array}{|c|c|c|c|}
hline
interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
frequency & 2/10& 4/10& 2/10 & 2/10\ hline
length & 7.2 & 9.2 & 4.4 & 13.2\ hline
h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
end{array}



Therefore $h_X(75.5)$ lies in $(67.8, 81]$



$h_X(75.5) = frac{0.2}{13.2}$



(b)



$int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$



is this correct?










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    A population $prod$ and a measure $X$ on $prod$ produced the data set below



    $71, 80, 58, 60, 63, 64, 51, 55, 48, 67$



    Let $h_X$ denote the density histogram function of $X$ based on the four intervals



    $(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$



    $(a)$ Calculate $h_X(75.5)$



    $(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$





    attempt



    $(a)$ I made a table



    begin{array}{|c|c|c|c|}
    hline
    interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
    frequency & 2/10& 4/10& 2/10 & 2/10\ hline
    length & 7.2 & 9.2 & 4.4 & 13.2\ hline
    h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
    end{array}



    Therefore $h_X(75.5)$ lies in $(67.8, 81]$



    $h_X(75.5) = frac{0.2}{13.2}$



    (b)



    $int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$



    is this correct?










    share|cite|improve this question









    $endgroup$















      0












      0








      0


      1



      $begingroup$


      A population $prod$ and a measure $X$ on $prod$ produced the data set below



      $71, 80, 58, 60, 63, 64, 51, 55, 48, 67$



      Let $h_X$ denote the density histogram function of $X$ based on the four intervals



      $(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$



      $(a)$ Calculate $h_X(75.5)$



      $(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$





      attempt



      $(a)$ I made a table



      begin{array}{|c|c|c|c|}
      hline
      interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
      frequency & 2/10& 4/10& 2/10 & 2/10\ hline
      length & 7.2 & 9.2 & 4.4 & 13.2\ hline
      h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
      end{array}



      Therefore $h_X(75.5)$ lies in $(67.8, 81]$



      $h_X(75.5) = frac{0.2}{13.2}$



      (b)



      $int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$



      is this correct?










      share|cite|improve this question









      $endgroup$




      A population $prod$ and a measure $X$ on $prod$ produced the data set below



      $71, 80, 58, 60, 63, 64, 51, 55, 48, 67$



      Let $h_X$ denote the density histogram function of $X$ based on the four intervals



      $(47, 54.2], (54.2, 63.4], (63.4, 67.8]$ and $(67.8, 81]$



      $(a)$ Calculate $h_X(75.5)$



      $(b)$ Calculate $int_{54.2}^{67.8} h_X(x) dx$





      attempt



      $(a)$ I made a table



      begin{array}{|c|c|c|c|}
      hline
      interval& (47, 54.2] & (54.2, 63.4] & (63.4, 67.8] & (67.8, 81] \ hline
      frequency & 2/10& 4/10& 2/10 & 2/10\ hline
      length & 7.2 & 9.2 & 4.4 & 13.2\ hline
      h_X(x) & frac{0.2}{7.2} & frac{0.4}{9.2} & frac{0.2}{4.4} & frac{0.2}{13.2} \ hline
      end{array}



      Therefore $h_X(75.5)$ lies in $(67.8, 81]$



      $h_X(75.5) = frac{0.2}{13.2}$



      (b)



      $int_{54.2}^{67.8} h_X(x) = F_X(67.8) - F_X(54.2) = frac{8}{10} - frac{2}{10} = frac{6}{10}$



      is this correct?







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      asked Jan 30 at 20:04









      Bas basBas bas

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