Jika median dari data terurut 5, 6, 6, k, 10, 10, 11, 16 adalah 9, maka simpangan baku data tersebut adalah ....

Pertanyaan

Jika median dari data terurut 5, 6, 6, k, 10, 10, 11, 16 adalah 9, maka simpangan baku data tersebut adalah .... $\;\;$

$begin mathsize 14px style 3 over 2 square root of 5 end style$
$begin mathsize 14px style 2 over 3 square root of 5 end style$
$begin mathsize 14px style square root of 13 end style$
$begin mathsize 14px style square root of 14 end style$
$begin mathsize 14px style square root of 15 end style$

Pembahasan:

Diketahui ada 8 data dan mediannya adalah 9, maka

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell M e end cell equals cell fraction numerator x subscript 4 plus x subscript 5 over denominator 2 end fraction end cell row 9 equals cell fraction numerator k plus 10 over denominator 2 end fraction end cell row 18 equals cell k plus 10 end cell row cell 18 minus 10 end cell equals k row 8 equals k end table end style$

Selanjutnya, kita cari rata-ratanya terlebih dahulu.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell x with minus on top end cell equals cell fraction numerator 5 plus 2 left parenthesis 6 right parenthesis plus 8 plus 2 left parenthesis 10 right parenthesis plus 11 plus 16 over denominator 8 end fraction end cell row blank equals cell 72 over 8 end cell row blank equals 9 end table end style$

Sehingga kita peroleh simpangan bakunya adalah

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sigma squared end cell equals cell fraction numerator sum from i equals 1 to n of open parentheses x subscript i minus x with minus on top close parentheses squared over denominator n end fraction end cell row cell sigma squared end cell equals cell fraction numerator open parentheses 5 minus 9 close parentheses squared plus 2 open parentheses 6 minus 9 close parentheses squared plus open parentheses 8 minus 9 close parentheses squared over denominator 8 end fraction end cell row blank blank cell blank plus fraction numerator 2 open parentheses 10 minus 9 close parentheses squared plus open parentheses 11 minus 9 close parentheses squared plus open parentheses 16 minus 9 close parentheses squared over denominator 8 end fraction end cell row cell sigma squared end cell equals cell fraction numerator 16 plus 18 plus 1 plus 2 plus 4 plus 49 over denominator 8 end fraction end cell row cell sigma squared end cell equals cell 90 over 8 end cell row cell sigma squared end cell equals cell 45 over 4 end cell row sigma equals cell 3 over 2 square root of 5 end cell end table end style$

Jadi, jawaban yang tepat adalah A.

Jawaban terverifikasi

Dijawab oleh:

Terakhir diupdate 04 Oktober 2021