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?

Question

1344 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-13T04:03:30

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)#

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.