What are the mean and standard deviation of the probability density function given by #p(x)=k/x # for # x in [3,8]#, with k being a constant such that the cumulative density is equal to 1?

1 Answer
Aug 25, 2016

#mu=5.1#
#sigma=1.43#

Explanation:

the mean is just the expected value defined as
#int p(x) xdx#
and expected standard deviation
#sqrt(int p(x) (mu- x)^2dx)#
finally we know that
#int_3^8 p(x) dx= 1#

first solve for #k#
now we solve for k and find out what the standard deviation and mean should be.

#int_3^8 k/x dx= 1#
#int_3^8 1/x dx= 1/k#
#[ln(x)]_3^8=1/k#

#k=1/(ln(8)-ln(3)) = 1.02#

now solving for mean
#int_3^8 kx/x dx = k(8-3) = 5k =5.1#

now for standard deviation

#sqrt(int_3^8 k((x-5.1)^2)/x dx )=sqrt( kint_3^8 x-10.2+26.01/x dx)#
#= sqrt( k(int_3^8 xdx-int_3^8 10.2dx+int_3^8 26.01/x dx))#
#= sqrt(k([x^2/2]_3^8 -[10.2x]_3^8+[26.01ln(x)]_3^8))= sqrt(k(27.5-51+25.51) )= sqrt(2.05) = 1.43#