1) Measurements for the length and width of rectangular plastic covers for CDs a
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Question
1) Measurements for the length and width of rectangular plastic covers for CDs are rounded to the nearest mm (so they are discrete). Let X denote the width, and Y denote the length. The possible values of X are 15 and 16 mm. The possible values of Y are 129, 130, and 131 mm. We show the probability for each pair in the following table: Width Width Width 129 15 0.12 0.08 Length 130 0.42 0.28 131 0.06 0.04 16 Given the above table for the joint probability mass function for length (X) and Width randomly selected CD cover. Calculate the following: () for aExplanation / Answer
Let X be the width and Y denote the length.
X takes values 15,16.
Y takes values 129, 130 and 131.
E(X) = xp(x)
= 15*0.6 + 16*0.4 = 15.4
E(X2) = x2*p(x)
= 152*0.6 + 162*0.4 = 237.4
V(X) = E(X2) - (E(X))2 = 237.4 - 15.42 = 237.4 - 237.16 = 0.24
E(Y) = 129*0.2 + 130*0.7 + 131*0.1 = 129.9
E(Y2) = 1292*0.2 + 1302*0.7 + 1312*0.1 = 16874.3
V(Y) = 16874.3 - 129.92 = 16874.3 - 16874.01 = 0.29
E(XY) = xyp(x,y)
= 15*129*0.12 + 15*130*0.42 + 15*131*0.06 + 16*129*0.08 + 16*130*0.28 + 16*131*0.04
= 232.2 + 819 + 117.9 + 165.12 + 582.4 + 83.84
= 2000.46
Cov(X,Y) = E(XY) - E(X)E(Y)
= 2000.46 - 15.4*129.9
= 2000.46 - 2000.46
= 0
Correlation coefficient = Cov(X,Y) / sqrt(var(X)*var(Y))
= 0 / sqrt(0.24*0.29) = 0
Correlation coefficient = 0
There is no correlation between X and Y.
Cov(X,Y) = 0 implies that X and Y are independent.
Y X 129 130 131 total 15 0.12 0.42 0.06 0.6 16 0.08 0.28 0.04 0.4 total 0.2 0.7 0.1 1Related Questions
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