DSE 2010 Q47 ,48

[Khagesh]

Sir, or can anybody expalin to me the solution for d foll question :


Q1: X is  a normally distributed rndm var wid mean 0 n var= sigma. Wats d mean of X^2 ??


Q2: X is a rndm var wid pdf as:


         f(x)={kx,  if x E [0,5)
               
                 k(10-x), if x E [5,10)
               
                0, if x E [10,infinity)

                 }              

[sonudelhi]

hey...1 is quite easy

Mean X2=E[X2]=Var[X]-E[X]
                    =sigma2-0=sigma2

[mukund sharma]

for the Q 48 u need 2 find the value of k,
since it is pdf u need to integrate from -nfinity to + infinity to get the value of k.
integrate the given function over the range which is equal to 1 and u will get the value of k.
if u get the value of k the succeeding qn. becomes easy,
integration of kxdx over the range (0,5)+integrate k(10-x)dx over the range(5,10)=1
and u get the value of k