Problem 1 (25 points): (A B) 10pts C) 5pts y, a survey is conducted in College o

Question

Problem 1 (25 points): (A B) 10pts C) 5pts y, a survey is conducted in College of Engineering among 1200 students to collect weight and height statistics. Let X and Y be the RVs for weight and height, respectively. X takes 0, 1, 2 for light, medium and heavy weights whereas Y takes 0 and 1 for short and tall students, respectively. Below are the statistics table for X and Y: X/Y 0 0 100 500 100 200 0 300 (a) Determine the joint pmf of X and Y and the marginal pmf's of X and of Y. (b) Are X and Y independent? Justify (c) Find the Covariance and Correlation of X and Y

Explanation / Answer

Here the joint pdf is calculated by

f(x,y) = Number of their common values/ Total population.

(a)

The joint distribution is given below

f(x) = 1/6 for x = 0

= 5//6 for x = 1

f(y) = 0.5 for y = 0

= 0.25 for y = 1

= 0.25 for y = 2

(b) here to evaluae X and Y are independent , If independent

f(x,y) = f(x) f(y)

so for let say x = 1 and y = 2

f(x,y) = f(0,0) = 1/12

f(x = 0) = 1/4 and f(y = 0) = 1/6

so f(x= 0 ) * f(y =0) = 1/24 not equal to f(x,y)

so we shall say that X and Yare not independent.

(c) here E(X) = 0 * 1/6 + 1 * 5/6 = 5/6

Var(X) = E(X2) - E(X)2

E(X2) = 1/6 * 0 + 5/6 * 12 = 5/6

Var(X) = 5/6 - (5/6)2 = 5/36

E(Y) = 0 * 0.5 + 1 * 0.25 + 2 * 0.25 = 0.75

E(Y2) = 0 * 0.5 + 1 * 0.25 + 4 * 0.25 = 1.25

Var(Y) = 1.25 - 0.752 = 0.6875

Here, Covaraince (X,Y) = E(XY) - E(X) E(Y)

E(XY) = 0 * 0 * 1/12 + 0 * 1 * 1/12 + 0 * 2 * 0 + 1 * 0 * 5/12 + 1 * 1 * 1/6 + 1/4 * 2 * 1 = 2/3

Covaraince (X,Y) = E(XY) - E(X) E(Y) = 2/3- 5/6 * 3/4 = 1/24

Corre(X,Y) = Cov(X,Y)/ sqrt [Var(X) Var(Y) ] = (1/24)/ sqrt( 5/36 * 0.6875) = 0.1348

X/Y 0 1 2 Total 0 0.083333 0.083333 0 0.166667 1 0.416667 0.166667 0.25 0.833333 Total 0.5 0.25 0.25 1

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