Advertisements
Advertisements
प्रश्न
If X is the random variable with probability density function f(x) given by,
`f(x) = {{:(x + 1",", -1 ≤ x < 0),(-x +1",", 0 ≤ x < 1),(0, "otherwise"):}`
then find the distribution function F(x)
उत्तर
Distribution function F(x) = `int_-oo^x f(u) "d"u`
Case 1:
x < – 1
F(x)= `int_-oo^x f(u) "d"u` = 0
Case 2:
– 1 ≤ x <0
F(x) = `int_oo^x f(u) "d"u`
= `int_-oo^-1 f(u) "d"u + int_-1^x f(u) "d"u`
= `0 + int_-1^x (u + 1) "d"u`
= `["u"^2/2 + u]_-1^2`
= `x^2/2 + x - 1/2 + 1`
= `x^2/2 + x + 1/2`
Case 3:
0 ≤ x < 1
F(x) = `int_-oo^x f(u) "d"u`
= `int_oo^-1 f(u) "d"u + int_-1^0 f(u) "d"u + int_0^x f(u) "d"u`
= `0 + int_-1^0 (u + 1) "d"u + int_0^x (- u + 1) "d"u`
= `[u^2/2 + u]_-1^0 + [- u^2/2 + u]_0^x`
= `0 - (1/ - 1) + (- x^2/2 + x - 0)`
= `1/2 - x^2 + x`
Case 4:
x ≥ 1
F(x) = `int_oo^x f(u) "d"u`
= `int_-oo^-1 f(u) "d"u + int_-1^0 f(u) "d"u + int_0^1 f(u) "d"u + int_1^x f(u) "d"u`
= `0 + int_-1^0 (u + 1) du + int_0^1 (- u + 1) "d"u + 0`
= `[u^2/2 + u]_-1^0 + [- u^2/2 + u]_0^1`
= `0 -(1/2 - 1) + (- 1/2 + 1) + 1 - 0`
= `- 1/2 + 1 - 1/2 + 1`
F(x) = `{{:(0",", x ≤ - 1),(x^2/2 + x + 1/2",", - 1 ≤ x < 0),(1/2 - x^2/2 + x",", 0 ≤ x < 1),(1",", x ≥ 1):}`
APPEARS IN
संबंधित प्रश्न
The probability density function of X is given by
`f(x) = {{:(kx"e"^(-2x), "for" x > 0),(0, "for" x ≤ 0):}`
Find the value of k
The probability density function of X is `f(x) = {(x, 0 < x < 1),(2 - x, 1 ≤ x ≤ 2),(0, "otherwise"):}`
Find P(0.2 ≤ X < 0.6)
The probability density function of X is `f(x) = {(x, 0 < x < 1),(2 - x, 1 ≤ x ≤ 2),(0, "otherwise"):}`
Find P(1.2 ≤ X < 1.8)
The probability density function of X is `f(x) = {(x, 0 < x < 1),(2 - x, 1 ≤ x ≤ 2),(0, "otherwise"):}`
Find P(0.5 ≤ X < 1.5)
Suppose the amount of milk sold daily at a milk booth is distributed with a minimum of 200 litres and a maximum of 600 litres with probability density function
`f(x) = {{:(k, 200 ≤ x ≤ 600),(0, "otherwise"):}`
Find the value of k
The probability density function of X is given by
`f(x) = {(k"e"^(- x/3), "for" x > 0),(0,"for" x ≤ 0):}`
Find the value of k
The probability density function of X is given by
`f(x) = {(k"e"^(- x/3), "for" x > 0),(0,"for" x ≤ 0):}`
Find the distribution function
The probability density function of X is given by
`f(x) = {(k"e"^(- x/3), "for" x > 0),(0,"for" x ≤ 0):}`
Find P(5 ≤ X)
The probability density function of X is given by
`f(x) = {(k"e"^(- x/3), "for" x > 0),(0,"for" x ≤ 0):}`
Find P(X ≤ 4)
If X is the random variable with probability density function f(x) given by,
`f(x) = {{:(x + 1",", -1 ≤ x < 0),(-x +1",", 0 ≤ x < 1),(0, "otherwise"):}`
then find P(– 0.5 ≤ x ≤ 0.5)
If X is the random variable with distribution function F(x) given by,
F(x) = `{{:(0",", - oo < x < 0),(1/2(x^2 + x)",", 0 ≤ x ≤ 1),(1",", 1 ≤ x < oo):}`
then find the probability density function f(x)
If X is the random variable with distribution function F(x) given by,
F(x) = `{{:(0",", - oo < x < 0),(1/2(x^2 + x)",", 0 ≤ x ≤ 1),(1",", 1 ≤ x < oo):}`
then find P(0.3 ≤ X ≤ 0.6)
Choose the correct alternative:
Let X be random variable with probability density function
`f(x) = {(2/x^3, x ≥ 1),(0, x < 1):}`
Which of the following statement is correct?
Choose the correct alternative:
If the function f(x) = `1/12` for a < x < b, represents a probability density function of a continuous random variable X, then which of the following cannot be the value of a and b?
Choose the correct alternative:
If `f(x) = {{:(2x, 0 ≤ x ≤ "a"),(0, "otherwise"):}` is a probability density function of a random variable, then the value of a is
