Advertisements
Advertisements
प्रश्न
Let X be a continuous random variable with probability density function
`"f"_x(x) = {{:(2x",", 0 ≤ x ≤ 1),(0",", "otherwise"):}`
Find the expected value of X
Advertisements
उत्तर
Let x be a continuous random variable.
In the probability density function,
Expected Value E(x) = `int_(-oo)^oo x"f"(x) "d"x`
Here E(x) = `int_0^1 x(2x) "d"x = int_0^1 2x^2 "d"x`
= `2[x^3/3]_0^1`
= `2/3[x^3]_0^1`
= `2/3 [1 - 0]`
= `2/3`
∴ E(x) = `2/3`
APPEARS IN
संबंधित प्रश्न
For the random variable X with the given probability mass function as below, find the mean and variance.
`f(x) = {{:((4 - x)/6, x = 1"," 2"," 3),(0, "otherwise"):}`
A commuter train arrives punctually at a station every half hour. Each morning, a student leaves his house to the train station. Let X denote the amount of time, in minutes, that the student waits for the train from the time he reaches the train station. It is known that the pdf of X is
`f(x) = {{:(1/30, 0 < x < 30),(0, "elsewhere"):}`
Obtain and interpret the expected value of the random variable X
Choose the correct alternative:
If P(X = 0) = 1 – P(X = 1). If E[X] = 3 Var(X), then P(X = 0) is
Let X be a continuous random variable with probability density function
f(x) = `{{:(3/x^4",", x ≥ 1),(0",", "otherwise"):}`
Find the mean and variance of X
In investment, a man can make a profit of ₹ 5,000 with a probability of 0.62 or a loss of ₹ 8,000 with a probability of 0.38. Find the expected gain
State the definition of Mathematical expectation using continuous random variable
Choose the correct alternative:
Value which is obtained by multiplying possible values of a random variable with a probability of occurrence and is equal to the weighted average is called
Choose the correct alternative:
Given E(X) = 5 and E(Y) = – 2, then E(X – Y) is
Choose the correct alternative:
If X is a discrete random variable and p(x) is the probability of X, then the expected value of this random variable is equal to
Prove that if E(X) = 0, then V(X) = E(X2)
