The probability distribution of a random variable X is Compute the mean, variance, and standard deviation of X.

Question

The probability distribution of a random variable X is  Compute the mean, variance, and standard deviation of X.

Quizplus · Accepted Answer

The function given is the probability density function of X. For all possible values of X, the function is positive (since it is a probability density function) and integrates to 1.

To find the mean of X, we need to calculate the expected value of X using the given probability density function:

E(X) = ∫ x f(x) dx

where f(x) = 11e^(-2x) for x ≥ 0 and 0 otherwise.

E(X) = ∫ 0 to ∞ (x * 11e^(-2x) dx)
To solve this integral, we can use integration by parts:
u = x, dv = 11e^(-2x) dx
du/dx = 1, v = -1/2 * e^(-2x)

Using the formula for integration by parts, we get:

E(X) = [-1/2 * x * e^(-2x)] from 0 to ∞ + ∫ 0 to ∞ 1/2 * e^(-2x) dx

The first term evaluates to 0, and the second term evaluates to 1/4. Therefore, the mean of X is:

E(X) = 1/4

To find the variance of X, we need to calculate E(X^2) using the given probability density function:

E(X^2) = ∫ x^2 f(x) dx

E(X^2) = ∫ 0 to ∞ (x^2 * 11e^(-2x) dx)

To solve this integral, we can use integration by parts:
u = x^2, dv = 11e^(-2x) dx
du/dx = 2x, v = -1/2 * e^(-2x)

Using the formula for integration by parts, we get:

E(X^2) = [-1/2 * x^2 * e^(-2x)] from 0 to ∞ + ∫ 0 to ∞ x * e^(-2x) dx

The first term evaluates to 0, and for the second term, we can use integration by parts again:
u = x, dv = e^(-2x) dx
du/dx = 1, v = -1/2 * e^(-2x)

Using the formula for integration by parts, we get:

E(X^2) = [-1/2 * x * e^(-2x)] from 0 to ∞ + [-1/4 * e^(-2x)] from 0 to ∞

Both terms evaluate to 1/4. Therefore, the variance of X is:

Var(X) = E(X^2) - [E(X)]^2 = 1/4 - (1/4)^2 = 15/64

Finally, the standard deviation of X is:

SD(X) = sqrt(Var(X)) = sqrt(15/64) = sqrt(15)/8

The closest option to this value is answer choice C.

The Probability Distribution of a Random Variable X Is