جواب تمرین صفحه 93 درس 3 آمار و احتمال یازدهم ریاضی (آمار توصیفی)

میانگین \(\bar x = \frac{{\sum {{x_i}} }}{n} = \frac{{504}}{{11}} = 45.81 \approx 46\)

واریانس \({\sigma ^2} = \frac{{\sum {{{({x_i} - \bar x)}^2}} }}{n} = \frac{{3534}}{{11}} = 321/27\)

انحراف معیار \(\sigma = \sqrt {321/27} = 18/09\)

ضریب تغییرات \(CV = \frac{\sigma }{{\bar x}} = \frac{{18/09}}{{46}} = 0/29\)

 میانگین \(\bar x = \frac{{\sum {{x_i}} }}{n} = \frac{{108/2}}{9} \simeq 12\)

واریانس \({\sigma ^2} = \frac{{\sum {{{({x_i} - \bar x)}^2}} }}{n} = \frac{{442/664}}{9} = 49/18\)

انحراف معیار \(\sigma = \sqrt {49/18} = 7/01\)

ضریب تغییرات \(CV = \frac{\sigma }{{\bar x}} = \frac{{7/01}}{{12}} = 0/58\)

ضریب تغییرات \(CV = \frac{\sigma }{{\bar x}} \Rightarrow 2 = \frac{\sigma }{4} \Rightarrow \sigma = 8 \Rightarrow {\sigma ^2} = 64\)

\(\begin{array}{l}{y_1} = c{x_1},\quad {y_2} = c{x_2},\quad {y_3} = c{x_3},\; \cdots ,\;{y_n} = c{x_n}\\\\\bar y = \frac{{\sum {{y_i}} }}{n} = \frac{{\sum c {x_i}}}{n} = c\frac{{\sum {{x_i}} }}{n} = c\bar x\\\\\sigma _y^2 = \frac{{\sum {{{({y_i} - \bar y)}^2}} }}{n} = \frac{{\sum {{{(c{x_i} - c\bar x)}^2}} }}{n} = \frac{{\sum {{c^2}} {{({x_i} - \bar x)}^2}}}{n} = {c^2}\frac{{\sum {{{({x_i} - \bar x)}^2}} }}{n} = {c^2}\sigma _x^2\\\\ \Rightarrow {\sigma _y} = |c|{\sigma _x}\\\\C{V_y} = \frac{{{\sigma _y}}}{{\bar y}} = \frac{{|c|{\sigma _x}}}{{c\bar x}} = \pm C{V_x}\end{array}\)

\(\begin{array}{l}1\,,\,1\,,\,1\,,\,1\,,\,1\,,\,2\,,\,2\,,\,2\,,\,2\,,\,2\,,\,2\,,\,3\,,\,3\,,\,4\,,\,4\,,\,4\,,\,6\,5\,,\,5\,,\,6\,,\,7\,,\,7\,,\,8\,\\\\\min = 1,\qquad {Q_1} = 2,\qquad {Q_2} = \frac{{2 + 3}}{2} = 2/5,\qquad {Q_3} = 5,\qquad \max = 8\end{array}\)

الف

با توجه به نمودار، میانگین شاخص توده بدنی در اقایان بیشتر است.

ب

به نظر می رسد پراکندگی یکسان است.

\(\begin{array}{l}10/4\,,\,10/4\,,\,\,10/5\,,\,11/3\,,\,11/5\,,\,11/9\,,\,12/2\,,\,12/3\,,\,13/5\,,\,30/1\\\\\min = 10/4,\quad {Q_1} = 10/5,\quad {Q_2} = \frac{{11/5 + 11/9}}{2} = 11/7,\quad {Q_3} = 13/3,\quad \max = 30/1\end{array}\)