Question 1189369: A discrete random variable x has the following probability distribution:
x 0 1 2 3
P(X=x) q 4p^2 p 0.7-4p^2
a. Find an expression for q in term of p
b. Find the value of p which gives the largest value of E(X)
c. Hence find the largest value of E(X)
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! **a. Find an expression for q in terms of p**
The sum of all probabilities in a probability distribution must equal 1. Therefore,
q + 4p² + p + 0.7 - 4p² = 1
q + p + 0.7 = 1
q = 1 - 0.7 - p
**q = 0.3 - p**
**b. Find the value of p which gives the largest value of E(X)**
The expected value of X, E(X), is given by:
E(X) = 0*q + 1*(4p²) + 2*p + 3*(0.7 - 4p²)
E(X) = 4p² + 2p + 2.1 - 12p²
E(X) = -8p² + 2p + 2.1
To find the value of p that maximizes E(X), we can take the derivative of E(X) with respect to p and set it to zero.
d(E(X))/dp = -16p + 2
Setting the derivative to zero:
-16p + 2 = 0
16p = 2
p = 2/16
**p = 0.125**
To confirm that this is a maximum, we can take the second derivative:
d²(E(X))/dp² = -16
Since the second derivative is negative, this confirms that p = 0.125 gives a maximum value for E(X).
**c. Hence find the largest value of E(X)**
Substitute p = 0.125 into the expression for E(X):
E(X) = -8(0.125)² + 2(0.125) + 2.1
E(X) = -8(0.015625) + 0.25 + 2.1
E(X) = -0.125 + 0.25 + 2.1
**E(X) = 2.225**
Therefore, the largest value of E(X) is 2.225.
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