Question

What are the mean and standard deviation of the probability density function given by `p(x)=tankx^2` for `x in [pi/8,(5pi)/12]`, in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Answer 1

Mean `E(x)=mu=1/(2k)*[ln sec ((25pi^2k)/144)-ln sec ((pi^2k)/64)]`

Standard deviation `sigma(x)=sqrt(int_(pi/8)^((5pi)/12)(x-mu)^2*tan kx^2 dx)`

Explanation:

Solution:

Mean `= E(x)=mu=int_a^b x *p(x) *dx`

`mu=int_(pi/8)^((5pi)/12) x*tan kx^2* dx`

`mu=1/(2k)*int_(pi/8)^((5pi)/12) 2k*x*tan kx^2* dx`

`mu=1/(2k)*int_(pi/8)^((5pi)/12) tan kx^2*2kx* dx`

`mu=1/(2k)*[ln sec kx^2]_(pi/8)^((5pi)/12)`

`mu=1/(2k)*[ln sec k((25pi^2)/144)-ln sec k((pi^2)/64)]`

Standard deviation

`sigma(x)=sqrt(int_a^b (x-mu)^2 p(x) dx)`

`sigma(x)=sqrt(int_(pi/8)^((5pi)/12)(x-mu)^2*tan kx^2 dx)`

God bless....I hope the explanation is useful.