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Diketahui fungsi probabilitas:     Tentukan: b. Median

Pertanyaan

Diketahui fungsi probabilitas: 

begin mathsize 14px style f open parentheses x close parentheses equals open curly brackets table attributes columnalign left end attributes row cell 1 half minus 1 over 8 x comma space pada space open square brackets 0 comma 4 close square brackets end cell row cell 0 comma space di space tempat space lain end cell end table close end style  

Tentukan:

b. Median 

Pembahasan Soal:

Berdasarakan sifat fungsi probabilitas kontinu, jika median dari fungsi prababilitas tersebut adalah begin mathsize 14px style straight m end style maka begin mathsize 14px style integral subscript a superscript m f open parentheses x close parentheses d x equals 1 half end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral subscript a superscript m f open parentheses x close parentheses d x end cell equals cell 1 half end cell row cell integral subscript 0 superscript m open parentheses 1 half minus 1 over 8 x close parentheses d x end cell equals cell 1 half end cell row cell open square brackets 1 half x minus 1 over 16 x squared close square brackets subscript o superscript m end cell equals cell 1 half end cell row cell open parentheses 1 half m minus 1 over 16 m squared close parentheses minus open parentheses 1 half 0 minus 1 over 16 0 squared close parentheses end cell equals cell 1 half end cell row cell 1 half m minus 1 over 16 m squared end cell equals cell 1 half end cell row cell 8 m minus m squared end cell equals 8 row cell m squared minus 8 m plus 8 end cell equals 0 row m equals cell fraction numerator negative open parentheses negative 8 close parentheses plus-or-minus square root of open parentheses negative 8 close parentheses squared minus 4 open parentheses 1 close parentheses open parentheses 8 close parentheses end root over denominator 2 open parentheses 1 close parentheses end fraction end cell row m equals cell fraction numerator 8 plus-or-minus square root of 32 over denominator 2 end fraction end cell row m equals cell 4 plus-or-minus 2 square root of 2 end cell row cell m subscript 1 end cell equals cell 4 plus 2 square root of 2 equals 6 comma 8 end cell row cell m subscript 2 end cell equals cell 4 minus 2 square root of 2 equals 1 comma 2 end cell end table end style  

Karena begin mathsize 14px style 6 comma 8 end style tidak masuk dalam interval undefined maka mediannya adalah begin mathsize 14px style m equals 1 comma 2 end style

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 15 Maret 2021

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