UTS CLO 2 Attempt Review [PDF]

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11/19/21, 1:40 PM



UTS CLO 2: Attempt review



TEORI PELUANG IF-44-08 [IPL] Dashboard / My courses / CII2G3-IF-44-08 / UTS / UTS CLO 2



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Completed on



Time taken



Monday, 8 November 2021, 2:53 PM



Finished



Monday, 8 November 2021, 3:47 PM



53 mins 47 secs



Marks



5.75/16.00



Grade



35.94 out of 100.00



Question 1



Complete Mark -0.25 out of 1.00



[ID] Diketahui X adalah variabel random dengan fungsi peluang : [EN} Given the probability function of discrete random variable X : x



0



1



2



3



4



p(x) 0.41 0.37 0.16 0.05 0.01



Select one: F(2) = P(X < 2) = 0.78 F(2) = 0.06 P(X > 2) = 0.06 P(X = 2) = F(2) - P(X = 1) P(X > 2) = 1 - P(X < 2)



The correct answer is: P(X > 2) = 0.06



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UTS CLO 2: Attempt review



Question 2



Complete Mark 0.50 out of 1.00



[ID] X merupakan variabel acak kontinyu. Diketahui E(X) = 2 dan E(X



2



[EN] Assumed that X is continuous random variable E(X) = 2 and E(X



. Pilihlah dari pernyataan berikut yang benar.



) = 8



2



, which the following are true.



) = 8



Select one or more: A. 2



E[(2 + 4X) ] = 164



B. V ar(−4X + 12) = 16 ⋅ V ar(X) = 48



C. V ar(X) = 3



 D. V ar(X) = 4



E. V ar(X) = 6



The correct answers are: V ar(X) = 4



, 2



E[(2 + 4X) ] = 164



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UTS CLO 2: Attempt review



Question 3



Complete Mark 0.17 out of 1.00



[ID] Diketahui fungsi padat peluang dari peubah acak kontinu X. Manakah pernyataan yang benar. [EN] Given the probability density function of continuous random variable X . Which the following are true.



Select one or more: a.



b.



 c.



 d.



e.



The correct answers are: ,



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UTS CLO 2: Attempt review



Question 4



Complete Mark 0.00 out of 1.00



[ID] Diketahui X dan Y adalah variabel acak dengan fungsi kepadatan peluang bersama sebagai berikut: [EN] Given the joint probability function of X and Y , f (x, y) = {



2,



0 < x < y < 1



0,



x lainnya



. Hint: calculate ρxy



Select one or more: A. fungsi marjinal X dari f(x,y), 1



g(x) = ∫



2dy = 2(1 − x) untuk 0 < x < 1,  dan 0 untuk xlainnya x



sehingga V ar(X) = E[X



2



2



] − (E[X])



1 = 18



.  B. E[XY ] =



1 4



, sehingga C OV (X, Y ) =



dan ρ



1 36



xy



=



1 2



C. fungsi marjinal Y dari f(x,y), y



h(y) = ∫



2dx = 2y untuk 0 < y < 1,  dan 0 untuk y lainnya 0



sehingga V ar(Y ) = E[Y



2



2



] − (E[Y ])



1 = 18



.



D. fungsi marjinal Y dari f(x,y), 1



h(y) = ∫



2dx = 2 untuk 0 < y < 1,  dan 0 untuk y lainnya 0



sehingga V ar(Y ) = E[Y



2



2



] − (E[Y ])



1 = 6



.



 E.



fungsi marjinal X dari f(x,y), 1



g(x) = ∫



2dy = 2 untuk 0 < x < 1,  dan 0 untuk xlainnya 0



sehingga



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4/16



11/19/21, 1:40 PM



UTS CLO 2: Attempt review V ar(X) = E[X



2



2



] − (E[X])



1 = 6



.



The correct answers are: fungsi marjinal X dari f(x,y), 1



g(x) = ∫



2dy = 2(1 − x) untuk 0 < x < 1,  dan 0 untuk xlainnya x



sehingga V ar(X) = E[X



2



] − (E[X])



2



=



1 18



. , fungsi marjinal Y dari f(x,y), y



h(y) = ∫



2dx = 2y untuk 0 < y < 1,  dan 0 untuk y lainnya 0



sehingga V ar(Y ) = E[Y



2



2



] − (E[Y ])



1 = 18



. , E[XY ] =



1 4



, sehingga C OV (X, Y ) =



1 36



dan ρ



xy



=



1 2



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UTS CLO 2: Attempt review



Question 5



Complete Mark 0.00 out of 1.00



[ID] Probability density function (pdf) dari suatu variable random X diketahui sebagai berikut : −



f (x) = {



ke



1 2



x



 untuk x ≥ 0



0



 untuk x < 0



. Tentukan mana dari pernyataan-pernyataan berikut ini yang benar. [EN] Given the following probability density function of X −



f (x) = {



ke



1 2



x



 untuk x ≥ 0



0



 untuk x < 0



. Which the following are true.



Select one or more:  A. k =



3 2



.



B. E[X] = 2



. Hint: gunakan integral parsial untuk menghitung integral (use partial integral). 



C. k =



1 2



.  D. −1/2



E[X] = 2 − 3 e



. Hint: gunakan integral parsial untuk menghitung integral (use partial integral).  E. E[2X + 3] = 6



The correct answers are: k =



1 2



. , E[X] = 2



. Hint: gunakan integral parsial untuk menghitung integral (use partial integral). 



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UTS CLO 2: Attempt review



Question 6



Complete Mark 0.17 out of 1.00



[ID] Diketahui fungsi padat peluang dari peubah acak kontinu X. Manakah pernyataan yang benar. [EN] Given the probability density function of continuous random variable X . Which the following are true.



Select one or more: a. k = 3/26  b.



 c.



k = 26/3 d.



e.



The correct answers are: k = 26/3,



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UTS CLO 2: Attempt review



Question 7



Complete Mark 0.00 out of 1.00



[ID] Sebuah toko buku mengeluarkan 8 buku baru pada bulan ini, yang terdiri dari 3 novel, 2 komik, dan 3 buku resep masakan. Seseorang membeli 3 buah buku di toko tersebut hari ini. Jika X dan Y masing-masing menyatakan jumlah Novel dan Komik yang dia beli, maka pernyataan yang betul adalah : [EN] A particular store releases eight new books on this month which are three novels, two comics and three cookbooks. This day, someone buy three books at the store. Suppose X and Y are the number of novels and comics that he intends to buy. Which the following are true?



Select one or more: 



Cov (X,Y) =



E (X) = 



6 5



3 4



Var (X) =



9 20



Var (Y) =



2 5



Cov (X,Y) = −



3 10



The correct answers are:



Var (X) =



9 2 , Var (Y) = , 20 5



Cov (X,Y) = −



3 10



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UTS CLO 2: Attempt review



Question 8



Complete Mark 0.67 out of 1.00



[ID] Misalkan  X dan  Y dua peubah acak yang mempunyai pdf gabungan. Pilihlah yang benar dari pernyataan-pernyataan berikut.  [EN] Suppose X and Y are two random variables which have this following joint probability density function. Which one are true. f (x, y) = {



k,



0 < x < y < 1



0,



x, y lainnya



Select one or more:  fungsi marjinal X dari fpp f(x,y) (the marginal probability function of X): ∞



g(x) = ∫



1



f (x, y)dy = ∫ −∞



2dy = 2 ⋅ (1 − x) x



. fungsi marjinal X dari fpp f(x,y) (the marginal probability function of X): ∞



g(x) = ∫



1



f (x, y)dy = ∫ −∞



4dy = 4 0



. fungsi marjinal X dari fpp f(x,y), yaitu g(x) juga merupakan fungsi padat peluang untuk peubah acak X (the marginal probability function of X is g(x) which is the probability density function of X ) 



k = 2



k = 4



The correct answers are: k = 2 , fungsi marjinal X dari fpp f(x,y) (the marginal probability function of X): ∞



g(x) = ∫



1



f (x, y)dy = ∫ −∞



2dy = 2 ⋅ (1 − x) x



., fungsi marjinal X dari fpp f(x,y), yaitu g(x) juga merupakan fungsi padat peluang untuk peubah acak X (the marginal probability function of X is g(x) which is the probability density function of X )



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UTS CLO 2: Attempt review



Question 9



Complete Mark 0.00 out of 1.00



[ID] Kovariansi dari dua variabel acak X dan Y dapat dirumuskan dengan operator ekspektasi sebagai berikut: [EN] Covariance of two random variables X and Y can be expressed using expectation as follow C OV (X, Y ) = E[(X − E[X]) ⋅ (Y − E[Y ])] = E[XY ] − E[X] ⋅ E[Y ]



.



Select one: True False



The correct answer is 'True'.



Question 10



Complete Mark 1.00 out of 1.00



[ID] Sebuah toko menjual obral 15 radio, diantara radio tersebut terdapat 5 radio yang rusak. Jika seorang calon pembeli melakukan test terhadap 3 radio yang dipilih secara acak, maka …… Catatan : peubah acak X = jumlah radio yang rusak [EN] A store gives 15 radio sales including five damaged radios. Suppose X is the number of damaged radios. If someone picks three radios randomly then ... 



Select one:



E (X ) = ∑ x p (x) 0 ≤ E (X ) ≤ 1 E (X ) =



115 91



E (3X − 1) = 3 E (X ) =



∫ x f (x) dx



The correct answer is: E (X ) = ∑ x p (x)



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UTS CLO 2: Attempt review



Question 11



Complete Mark 0.17 out of 1.00



[ID] Fungsi peluang dari suatu peubah acak diskret X dinyatakan sebagai berikut : [EN} Given the probability function of discrete random variable X :



⎧⎪ ⎪⎪ ⎪⎪ ⎪⎪ P (X = x) = ⎪⎨⎪⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎩



2p p 4p



; x =1



0



; x lainnya



; x =2 ; x =3



Select one or more:



P (X > 1) = F (2) −F (1)  P (1 ≤ X < 3) = F (3) −F (1)



P (X ≤ 1) = 1 −P (X = 1) F (2) = F (3) − p (3) 



P (X < 2) = 1 −



5 7



The correct answers are:



F (2) = F (3) − p (3) , P (X < 2) = 1 −



5 7



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UTS CLO 2: Attempt review



Question 12



Complete Mark 1.00 out of 1.00



[ID] Jika variabel acak X  dan Y  saling bebas dengan h (y) >0  distribusi marjinal terhadap Y  dari  f (x,y) , maka peluang bersyarat dari X  diberikan  Y = y :



f (x | y) =



f (x,y) h (y)



tidak lain merupakan distribusi marjinal g (x)  dari fungsi distribusi f (x,y) .



[EN] If the random variable X and Y are independent and h(y) > 0 is marginal probability function of Y , then the conditional probability of X given Y



= y



or f (x, y) f (x|y) = h(y)



is the marginal probability function of X or g(x) .



Select one: True False



The correct answer is 'True'.



Question 13



Complete Mark 1.00 out of 1.00



[ID] Diketahui X adalah variabel random dengan fungsi peluang : [EN] Given the probability function of discrete random variable X : x



0



1



2



3



4



p(x) 0.41 0.37 0.16 0.05 0.01



Select one:



Var (2X − 1) = 2 Var (X ) = 1 . 6912 Var (2X − 1) = 4Var (X ) = 3 . 3824 Var (2X − 1) = E (X 2) − [E (X )]2 Var (2X − 1) = 2Var (X ) − 1 = 0,6912 Var (2X − 1) = 4 var (X ) − 1 = 2 . 3824



The correct answer is: Var (2X − 1) = 4Var (X ) = 3 . 3824



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UTS CLO 2: Attempt review



Question 14



Complete Mark 0.67 out of 1.00



[ID] Diberikan fungsi massa peluang gabungan (joint pmf) dari bivariat X,Y. Manakah pernyataan yang betul  [EN] Given the joint probability mass function of X and Y . Which ones are true.



⎧⎪ ⎪⎪ x+y ⎪ ; x = 0,1,2, dan y = 0,1,2 p X,Y (x,y) = ⎪⎨⎪⎪ 18 ⎪⎪ ⎪⎩ 0 ; x,y la innya Select one or more: fungsi massa peluang marjinal untuk X (the marginal probability function of X ):



⎧⎪ ⎪⎪ x+1 ⎪ ; x = 0,1,2 p (x) = ⎪⎨⎪⎪ 6 ⎪⎪ ⎪⎩ 0 ; x la innya



 fungsi massa peluang marjinal untuk X (the marginal probability function of X ):



X dan Y saling bebas (X and Y are independent)



fungsi massa peluang marjinal untuk X (the marginal probability function of X ):



 fungsi massa peluang marjinal untuk Y (the marginal probability function of Y ):



⎧⎪ ⎪⎪ 1+y ⎪ ; y = 0,1,2 p (y) = ⎪⎨⎪⎪ 6 ⎪⎪ ⎪⎩ 0 ; y la innya



The correct answers are: fungsi massa peluang marjinal untuk X (the marginal probability function of X ):



⎧⎪ ⎪⎪ x+1 ⎪ ; x = 0,1,2 p (x) = ⎪⎨⎪⎪ 6 ⎪⎪ ⎪⎩ 0 ; x la innya , fungsi massa peluang marjinal untuk X (the marginal probability function of X ):



, fungsi massa peluang marjinal untuk Y (the marginal probability function of Y ):



⎧⎪ ⎪⎪ 1+y ⎪ ; y = 0,1,2 p (y) = ⎪⎨⎪⎪ 6 ⎪⎪ ⎪⎩ 0 ; y la innya



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UTS CLO 2: Attempt review



Question 15



Complete Mark 0.00 out of 1.00



Andaikan (X,Y) adalah dua peubah acak (bivariate), dengan fmp gabungan yang didefinisikan sebagai berikut :  p(1, 1) = p(2, 2) = p(3, 1) = 0.1; p(1, 2) = p(3, 2) = 0.2; p(2, 1) = k



Manakah pernyataan yang betul adalah :



[EN] Suppose (X, Y ) is a bivariate random variable with this joint probability mass function (pmf) as follows. p(1, 1) = p(2, 2) = p(3, 1) = 0.1; p(1, 2) = p(3, 2) = 0.2; p(2, 1) = k



Which one of the following are true.



Select one or more:



Cov (X,Y) = 0  Cov (X,Y) = 3



k = 0.3  E (XY) = 3



E (X) = 1.5



The correct answers are:



k = 0.3 , E (XY) = 3 , Cov (X,Y) = 0



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14/16



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UTS CLO 2: Attempt review



Question 16



Complete Mark 0.67 out of 1.00



[ID] Sebuah toko buku mengeluarkan 8 buku baru pada bulan ini, yang terdiri dari 3 novel, 2 komik, dan 3 buku resep masakan. Seseorang membeli 3 buah buku di toko tersebut hari ini. Jika X dan Y masing-masing menyatakan jumlah Novel dan Komik yang dia beli, maka pernyataan yang betul adalah :



[EN] A particular store releases eight new books on this month which are three novels, two comics and three cookbooks. This day, someone buy three books at the store. Suppose X and Y are the number of novels and comics that he intends to buy. Which the following are true?



Select one or more:  fungsi massa peluang marjinal untuk X (the marginal probability of X ):







p X,Y (0,0) = p X,Y (3,0) =



1 56



fungsi massa peluang marjinal untuk X (the marginal probability of X )



fungsi massa peluang marjinal untuk Y (the marginal probability of Y )



X dan Y saling independen (X and Y are independent)



The correct answers are:



p X,Y (0,0) = p X,Y (3,0) =



1 , 56



fungsi massa peluang marjinal untuk X (the marginal probability of \( X \)):



, fungsi massa peluang marjinal untuk Y (the marginal probability of \( Y \))



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