Let x be a random variable with probability density function fx(x) = ( c(1 − x 2 ) if − 1 < x < 1 0 otherwise. a) what is the value of c? b) what is the cumulative distribution function of x? c) compute e(x) and var(x).

a)   c=3/4

b)  for -1<x<1

 F(x) = (3/4) ( x + (2-x^3)/3 )

c)

  E(x)=0

  Var(x) = 1/5

Step-by-step explanation:

Hi!

a)

In order to f(x) be a probability density function it must be normalized, which means that its integral must be equal to 1:

Since f(x) is zero for x>1 and x<-1

Therefore:

c=3/4

b)

The cumulative distribution fucntion F(x) can be obtained integrating f(x) from -∞ to x:

Since f(x) = 0 for x<-1      

        F(x) = 0 for x<-1

for -1<x<1:

for x>1

  F(x)=1

c)

The mean E(x) can be found integrating  xf(x)

We can easily infer that the mean must be zero, since f(x) is an even function and thus xf(x) is an odd function integrated in a simetric interval:

E(x) = 0

The variance of x, Var(x), can be evaluated integrating  (x-E(x))^2f(x), since E(x)=0:

Since c=3/4

Var(x) = 1/5

hey seth! : )

i’m not 100% sure, but i think it’s b

hope this ya : p

miss u!

- nicole