A chemist test involves measuring two independent quantities A and B, the expressing the results of the test as C such that C= 3A-0.5B. Give that A is normally distributed with mean 10 and standard deviation of 1, and B is also normally distributed with mean 34 and standard deviation of 2. Find E(C)and var(C)
To find the expected value (E(C)) and variance (Var(C)) of C, we need to use the properties of expectation and variance.
First, let's calculate the expected value (E(A)) and variance (Var(A)) of A:
E(A) = mean of A = 10
Var(A) = standard deviation of A squared = 1^2 = 1
Similarly, the expected value (E(B)) and variance (Var(B)) of B are:
E(B) = mean of B = 34
Var(B) = standard deviation of B squared = 2^2 = 4
Now, let's find the expected value of C (E(C)):
E(C) = E(3A - 0.5B)
= 3 * E(A) - 0.5 * E(B)
Substituting the values we calculated earlier:
E(C) = 3 * 10 - 0.5 * 34
= 30 - 17
= 13
So, E(C) = 13.
Next, let's find the variance of C (Var(C)):
Var(C) = Var(3A - 0.5B)
= 3^2 * Var(A) + (-0.5)^2 * Var(B)
= 9 * Var(A) + 0.25 * Var(B)
Substituting the values we calculated earlier:
Var(C) = 9 * 1 + 0.25 * 4
= 9 + 1
= 10
So, Var(C) = 10.
Therefore, the expected value of C (E(C)) is 13, and the variance of C (Var(C)) is 10.