Complete Question
Suppose that the random variable X represents the length of a punched part in
centimeters. Let Y be the length of the part in millimeters. If E(X) = 5 and V(X) = 0.25,
what are the mean and variance of Y?
Answer:
The mean
[tex]E(Y) =50 mm[/tex]
The variance
[tex]V(Y) = 25 mm[/tex]
Step-by-step explanation:
From the question we are told that
Length in cm is X
Length in mm is Y
Generally 10 mm = 1 cm
So
[tex]Y = 10 X[/tex]
Hence the expected mean of Y is
[tex]E(Y) = E (10 X)[/tex]
=> [tex]E(Y) = 10 E(X)[/tex]
From the question [tex]E(X) = 5[/tex]
So
[tex]E(Y) = 10 * 5[/tex]
=> [tex]E(Y) =50 mm[/tex]
Gnerally the variance of X is
[tex]V(X) = E [X^2] -E[X]^2[/tex]
From the question we are told that [tex]V(X) = 0.25[/tex]
=> [tex]0.25 = E [X^2] -E[X]^2[/tex]
Gnerally the variance of Y
[tex]V(Y) = E [Y^2] -E[Y]^2[/tex]
=> [tex]V(Y) = E [10X^2] -E[X]^2[/tex]
=> [tex]V(Y) = 10^2[ E [X^2] -E[X]^2][/tex]
=> [tex]V(Y) = 10^2* V(X)[/tex]
=> [tex]V(Y) = 10^2* 0.25[/tex]
=> [tex]V(Y) = 25 mm[/tex]
