Question

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?

Answer 1

`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`