Advertisements
Advertisements
प्रश्न
Given the p. d. f. (probability density function) of a continuous random variable x as :
`f(x)=x^2/3, -1`
= 0 , otherwise
Determine the c. d. f. (cumulative distribution function) of x and hence find P(x < 1), P(x ≤ -2), P(x > 0), P(1 < x < 2)
उत्तर
c.d.f. of the continous random variable is given by
`F(x)=int_-1^xy^2/3dx`
`=[y^3/9]_-1^x`
`=(x^3+1)/9, x in R`
Consider P(X<1)=F(1)=(13+1)/9=2/9
`P(x<=-2)=0`
`P(X>0)=1-P(X<=0)`
`=1-F(0)`
`=1-(0/9+1/9)`
`=8/9`
`P(1<x<2)=F(2) - F(1)`
`=1-(1/9+1/9)`
=`7/9`
APPEARS IN
संबंधित प्रश्न
The following is the p.d.f. (ProbabiIity Density Function) of a continuous random variable X :
`f(x)=x/32,0<x<8`
= 0 otherwise
(a) Find the expression for c.d.f. (Cumulative Distribution Function) of X.
(b) Also find its value at x = 0.5 and 9.
If the p.d.f. of a continuous random variable X is given as
`f(x)=x^2/3` for -1< x<2
=0 otherwise
then c.d.f. fo X is
(A) `x^3/9+1/9`
(B) `x^3/9-1/9`
(C) `x^2/4+1/4`
(D) `1/(9x^3)+1/9`
The p.d.f of a random variable x is given by
f(x) = `1/"4a"`, 0 < x < 4a, (a > 0)
= 0, otherwise
and `"P"(x < "3a"/2) = "kp"(x > "5a"/2)` then k = ?
The pdf of a random variable X is
f(x) = 3(1 - 2x2), 0 < x < 1
= 0 otherwise
The P`(1/4 < "X" < 1/3)` = ?
If the probability density function continuous random variable X is
f(x) = k(1 - x2); 0 < x < 1
= 0 ; otherwise
Then P`(0 < X < 1/2)` = ?
If random variable waiting time in minutes for bus and probability density function of x is given by
f(x) = `{{:(1/5",", 0 ≤ x ≤ 5),(0",", "otherwise,"):}`
then probability of waiting time not more than 4 minutes is equal to ______.
If the probability density function of a random variable X is given as
| x1 | –2 | –1 | 0 | 1 | 2 |
| P(X = x1) | 0.2 | 0.3 | 0.15 | 0.25 | 0.1 |
then F(0) is equal to
