###### Advertisements

###### Advertisements

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.

###### Advertisements

#### Solution 1

(a) c.d.f. of a continuous random variable X is given by

`F(x)=int_(-oo)^xf(y)dy`

In the given density function f(x), range of X starts at '0'.

`therefore F(x)=int_0^xf(y)dy=int_0^xy/32dy=[y^2/64]^2=x^2/64`

`"Thus ",F(x)=x^2/64, AA x in R`

(b) `F(0.5) =(0.5)^2/64=1/256`

For any value of x ≥ 8, F(x)=1

∴ F(9)=1

#### Solution 2

Given p.d.f

`f(x)=x/32,0<x<8`

`c.d.f. = F(X)=int_0^xf(x)dx`

`=int_0^x x/32dx`

`=[x^2/64]_0^x`

`=X^2/64`

`"at "x = 0.5`

`F(X)=F(0.5)=(0.5)^2/64=0.25/64=0.0039`

For any value of x greater than 8

`F(X)=1`

`therefore F(9)=1`

#### APPEARS IN

#### RELATED QUESTIONS

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)

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 - 2x^{2}), 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 - x_{2}); 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

x_{1} |
–2 | –1 | 0 | 1 | 2 |

P(X = x_{1}) |
0.2 | 0.3 | 0.15 | 0.25 | 0.1 |

then F(0) is equal to