Advertisements
Advertisements
Question
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
Advertisements
Solution
Let x be a continuous random variable.
In the probability density function
Mean E(x) = `int_oo^oo x"f"(x) "d"x`
Here E(x) = `int_1^oo x(3/x^4)`
= `3int_1^oo 1/x^3 "d"x`
Here E(x) = `int_1^oo x^-3 "d"x`
= `3[x^(-3 + 1)/(-3 + 1)]_1^oo`
= `3int_1^oo x^-3 "d"x`
= `3(x^(-3 + 1)/(-3 + 1))_1^oo`
= `3/(-2) (x^2)_1^oo`
= `(-3)/2 [1/x^2]_1^oo`
= `3/(-2) [1/oo^2 - 1/(1)^2]`
= `- 3/2 [0 - 1]`
= `3/2`
∴ E(x) = `3/2`
`"E"(x^2) = int_(-oo)^oo x^2"f"(x) "d"x`
= `int_1^oo x^2(3/x^4) "d"x`
= `3/x^2 "d"x`
= `3int_1^oo x^2 "d"x`
= `3 [(x^-2 + 1)/(-2 + 1)]`
= `3[x^-1/(-1)]`
= `-3[1/x]_1^oo`
= `-3[1/oo - 1/1]`
= `-3[0 - 1]`
= 3
`"E"(x^2)` = 3
Var(x) = `"E"(x^2) - "E"(x))^2`
= `3 - (3/2)^2`
= `3 - 9/4`
= `(12 - 9)/4`
∴ Var(x) = `3/4`
APPEARS IN
RELATED QUESTIONS
For the random variable X with the given probability mass function as below, find the mean and variance.
`f(x) = {{:(1/2 "e"^(x/2), "for" x > 0),(0, "otherwise"):}`
The time to failure in thousands of hours of an electronic equipment used in a manufactured computer has the density function
`f(x) = {{:(3"e"^(-3x), x > 0),(0, "eleswhere"):}`
Find the expected life of this electronic equipment
Choose the correct alternative:
On a multiple-choice exam with 3 possible destructive for each of the 5 questions, the probability that a student will get 4 or more correct answers just by guessing is
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 random variable defining number of students getting A grade. Find the expected value of X from the given table:
| X = x | 0 | 1 | 2 | 3 |
| P(X = x) | 0.2 | 0.1 | 0.4 | 0.3 |
In a business venture a man can make a profit of ₹ 2,000 with a probability of 0.4 or have a loss of ₹ 1,000 with a probability of 0.6. What is his expected, variance and standard deviation of profit?
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
Choose the correct alternative:
E[X – E(X)]2 is
Choose the correct alternative:
A discrete probability function p(x) is always non-negative and always lies between
Prove that V(aX) = a2V(X)
