Let the joint probability density function for X and Y be f(x, y) = 0
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![Let the jaint probability density function for X and Y be
f(z,v) = {6 V
0<z<y<1
otherwise
Find the value of c that would make this a valid joint probability distribution.
C =
number (rtol=0.01, atol=1e-08)
Notes: You can express your answer as a fraction or decimal. If you answer is in decimal, include at least
4 non-zero digits after the decimal point.
Find the marginal probability density function of X, fx(z).
O (a) fx(x) = 9r² - 9z7/2, 0< z <1
ㅇ (b) fx(#) %3D 2 -꼭포기2, 0<z<1
O (c) fx(1) = 9z² – 9z72, 0 <z < y
O (d) fx(=) = 9z², 0<I<y
O (e) fx(2) = 9x², 0<I<1
ㅇ () fx(2) 3D꽃고2-9z3/2, 0<z<y
Find the marginal probability density function of Y, fr(y).
o (a) fr(y) %3D 을y기2, z<y<1
O (b) fy(y) = 9 – 9y/2, I<y<1
ㅇ (0) fr(v)%3DyV2, 0<y<1
ㅇ (d) fr(v) %3D 을y기2, 0<y<1
O (e) fr(3) = 91 – 9y7/2,
0 <y <1
o () fr(u)=D 을l/2, r<y<1
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1d010a9b-2cee-4dae-9b72-ad7f9f7a8cad%2F3ffcc01e-90d1-4207-ad10-f5e954839ea1%2Ffq5h4ug_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Let the jaint probability density function for X and Y be
f(z,v) = {6 V
0<z<y<1
otherwise
Find the value of c that would make this a valid joint probability distribution.
C =
number (rtol=0.01, atol=1e-08)
Notes: You can express your answer as a fraction or decimal. If you answer is in decimal, include at least
4 non-zero digits after the decimal point.
Find the marginal probability density function of X, fx(z).
O (a) fx(x) = 9r² - 9z7/2, 0< z <1
ㅇ (b) fx(#) %3D 2 -꼭포기2, 0<z<1
O (c) fx(1) = 9z² – 9z72, 0 <z < y
O (d) fx(=) = 9z², 0<I<y
O (e) fx(2) = 9x², 0<I<1
ㅇ () fx(2) 3D꽃고2-9z3/2, 0<z<y
Find the marginal probability density function of Y, fr(y).
o (a) fr(y) %3D 을y기2, z<y<1
O (b) fy(y) = 9 – 9y/2, I<y<1
ㅇ (0) fr(v)%3DyV2, 0<y<1
ㅇ (d) fr(v) %3D 을y기2, 0<y<1
O (e) fr(3) = 91 – 9y7/2,
0 <y <1
o () fr(u)=D 을l/2, r<y<1
%3D
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