QUESTION 1
The discrete random variable X has the following probability distribution function.
x 0 1 2 3 4
P(X=x) 0.3 0.2 0.1 0.3 0.1
Find
(a) The expected value of X, E(X)
(b) The variance of X, Var(X)
(c) The standard deviation of X, σ
(d) E(2X)
(e) Var(2X)
QUESTION 2
The probability distribution function of a discrete random variable X is
f(x)={(x^2/55 ,x=1,2,3,4,5)/(0 ,otherwise)}
(a) E(X)
(b) Var(X)
(c) E(3X+2)
(d) Var(3X+2)
(e) 5E(2X-1)
(f) E(2X+1)^2
QUESTION 3
The probability distribution function of a discrete random variable X is given below
x 1 2 3 4 5
P(X=x) 0.3 c 0.1 3c 2c
(a) Find the value of c.
(b) Var(X)
To find the answers to the questions, we need to use the formulas and properties of expected value, variance, and standard deviation.
QUESTION 1:
(a) To find the expected value of X, E(X), we use the formula: E(X) = Σ(x * P(X=x)), where Σ represents the sum, x represents the possible values of X, and P(X=x) represents the probability of X taking the value x.
For the given probability distribution, we have:
E(X) = (0 * 0.3) + (1 * 0.2) + (2 * 0.1) + (3 * 0.3) + (4 * 0.1) = 0 + 0.2 + 0.2 + 0.9 + 0.4 = 1.7
Therefore, the expected value of X, E(X), is 1.7.
(b) To find the variance of X, Var(X), we use the formula: Var(X) = Σ((x - E(X))^2 * P(X=x)).
Using the values from the given probability distribution and the previously calculated E(X) value, we have:
Var(X) = ((0 - 1.7)^2 * 0.3) + ((1 - 1.7)^2 * 0.2) + ((2 - 1.7)^2 * 0.1) + ((3 - 1.7)^2 * 0.3) + ((4 - 1.7)^2 * 0.1)
= (2.89 * 0.3) + (0.49 * 0.2) + (0.09 * 0.1) + (0.49 * 0.3) + (2.89 * 0.1)
= 0.867 + 0.098 + 0.009 + 0.147 + 0.289
= 1.4
Therefore, the variance of X, Var(X), is 1.4.
(c) To find the standard deviation of X, σ, we take the square root of the variance: σ = √Var(X).
Using the previously calculated variance, we have:
σ = √1.4 ≈ 1.18
Therefore, the standard deviation of X, σ, is approximately 1.18.
(d) To find E(2X), we multiply the expected value of X by 2:
E(2X) = 2 * E(X) = 2 * 1.7 = 3.4
Therefore, E(2X) is 3.4.
(e) To find Var(2X), we multiply the variance of X by 4 (since Var(aX) = a^2 * Var(X), where a is a constant):
Var(2X) = 4 * Var(X) = 4 * 1.4 = 5.6
Therefore, Var(2X) is 5.6.
QUESTION 2:
(a) To find E(X), we use the same formula as in Question 1(a):
E(X) = (1 * (1^2/55)) + (2 * (2^2/55)) + (3 * (3^2/55)) + (4 * (4^2/55)) + (5 * (5^2/55))
= (1/55) + (8/55) + (27/55) + (64/55) + (125/55)
= 225/55
≈ 4.09
Therefore, E(X) is approximately 4.09.
(b) To find Var(X), we use the formula from Question 1(b):
Var(X) = ((1 - 4.09)^2 * (1^2/55)) + ((2 - 4.09)^2 * (2^2/55)) + ((3 - 4.09)^2 * (3^2/55)) + ((4 - 4.09)^2 * (4^2/55)) + ((5 - 4.09)^2 * (5^2/55))
= (-9.81^2 * 1/55) + (-6.09^2 * 4/55) + (-1.09^2 * 9/55) + (0.09^2 * 16/55) + (0.91^2 * 25/55)
= 96.06/55
≈ 1.75
Therefore, Var(X) is approximately 1.75.
(c) To find E(3X + 2), we substitute the formula for E(X) into the expression:
E(3X + 2) = 3 * E(X) + 2 = 3 * 4.09 + 2 = 12.27 + 2 = 14.27
Therefore, E(3X + 2) is 14.27.
(d) To find Var(3X + 2), we substitute the formula for Var(X) into the expression:
Var(3X + 2) = 3^2 * Var(X) = 9 * 1.75 = 15.75
Therefore, Var(3X + 2) is 15.75.
(e) To find 5E(2X - 1), we substitute the formula for E(2X) into the expression:
5E(2X - 1) = 5 * (2 * E(X) - 1) = 5 * (2 * 4.09 - 1) = 5 * (8.18 - 1) = 5 * 7.18 = 35.9
Therefore, 5E(2X - 1) is 35.9.
(f) To find E(2X + 1)^2, we expand the expression:
E(2X + 1)^2 = E(4X^2 + 4X + 1)
= (4 * E(X^2)) + (4 * E(X)) + 1
Calculating E(X^2):
E(X^2) = (1^2 * (1^2/55)) + (2^2 * (2^2/55)) + (3^2 * (3^2/55)) + (4^2 * (4^2/55)) + (5^2 * (5^2/55))
= (1/55) + (8/55) + (27/55) + (64/55) + (125/55)
= 225/55
≈ 4.09
Substituting the values into the expression:
E(2X + 1)^2 = (4 * 4.09) + (4 * 4.09) + 1
= 16.36 + 16.36 + 1
= 33.72
Therefore, E(2X + 1)^2 is 33.72.
QUESTION 3:
(a) To find the value of c, we use the fact that the sum of all probabilities must equal 1:
0.3 + c + 0.1 + 3c + 2c = 1
3.3c = 0.6
c = 0.6/3.3 ≈ 0.1818
Therefore, the value of c is approximately 0.1818.
(b) To find Var(X), we use the formula from Question 1(b):
Var(X) = ((1 - E(X))^2 * P(X=1)) + ((2 - E(X))^2 * P(X=2)) + ((3 - E(X))^2 * P(X=3)) + ((4 - E(X))^2 * P(X=4)) + ((5 - E(X))^2 * P(X=5))
Since E(X) is not specified in the question, we cannot calculate Var(X) without further information.