Advertisements
Advertisements
Question
For the random variable X with the given probability mass function as below, find the mean and variance.
`f(x) = {{:(2(x - 1), 1 < x ≤ 2),(0, "otherwise"):}`
Solution
Mean: `mu = "E"("X")`
= `int_(1/2)^2 x f(x) "d"x`
= `int_1^(1/2) x xx 2(x - 1) "d"x`
= `2 int_1^2 (x^2 - x) "d"x`
= `2[x^3/3 - x^2/2]_1^2`
= `2[8/3 - 4/2 - 1/3 + 1/2]`
= `2(7/3 - 3/2)`
= `2 xx 5/6`
= `5/3`
Variance: `"E"("X"^2)`
= `int_1^2 x^2 f(x) "d"x`
= `2int_1^2 (x^3 - x^2) "d"x`
= `2[x^4/4 - x^3/3]_1^2`
= `2[16/4 - 8/3 - 1/4 + 1/3]`
= `2[15/4 - 7/3]`
= `2 xx 17/12`
= `17/6`
Var(X) = `"E"("X"^2) - ["E"("X")]^2`
= `17/6 - (5/3)^2`
= `17/6 - 25/9`
= `(51 - 50)/18`
= `1/18`
APPEARS IN
RELATED QUESTIONS
The probability density function of the random variable X is given by
`f(x) = {{:(16x"e"^(-4x), x > 0),(0, x ≤ 0):}`
find the mean and variance of X
A lottery with 600 tickets gives one prize of ₹ 200, four prizes of ₹ 100, and six prizes of ₹ 50. If the ticket costs is ₹ 2, find the expected winning amount of a ticket
Choose the correct alternative:
Consider a game where the player tosses a six-sided fair die. If the face that comes up is 6, the player wins ₹ 36, otherwise he loses ₹ k2, where k is the face that comes up k = {1, 2, 3, 4, 5}. The expected amount to win at this game in ₹ is
The following table is describing about the probability mass function of the random variable X
| x | 3 | 4 | 5 |
| P(x) | 0.2 | 0.3 | 0.5 |
Find the standard deviation 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
How do you defi ne variance in terms of Mathematical expectation?
State the definition of Mathematical expectation using continuous random variable
A person tosses a coin and is to receive ₹ 4 for a head and is to pay ₹ 2 for a tail. Find the expectation and variance of his gains
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:
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:
Which of the following is not possible in probability distribution?
Choose the correct alternative:
A discrete probability distribution may be represented by
Choose the correct alternative:
A probability density function may be represented by
Choose the correct alternative:
E[X – E(X)] is equal to
Choose the correct alternative:
If p(x) = `1/10`, x = 10, then E(X) is
Choose the correct alternative:
An expected value of a random variable is equal to it’s
Choose the correct alternative:
A discrete probability function p(x) is always non-negative and always lies between
Choose the correct alternative:
The distribution function F(x) is equal to
