What are the mean and standard deviation of the probability density function given by #(p(x))/k=x^8-x^16# for # x in [0,1]#, in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

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Answer 1 · 2016-12-15T10:18:53

The mean is #mu=0.7875#
The standard deviation is #sigma=0.33#

The probability density function is #p(x)=k(x^8-x^16)#

We use #intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

Therefore,

#int_0^1k(x^8-x^16)dx=1#

#k [x^9/9-x^17/17]_0^1=1 #

#k(1/9-1/17)=1#

#k=17*9/8=19.125#

The mean is #E(x)=mu=int_0^1xp(x)dx#

#=19.125*int_0^1(x^9-x^17)#

#=19.125* [x^10/10-x^17/17] _0^1#

#=19.125*(1/10-1/17)#

#=19.125*7/170=0.7875#

The standard deviation is

#sigma=sqrt(int_0^1x^2p(x)dx-mu^2)#

#int_0^1x^2p(x)dx=kint_0^1(x^10-x^18)dx#

#=k[x^11/11-x^19/19]_0^1#

#=19.125*[1/11-1/19]#

#=19.125*8/(19*11)=0.73#

Therefore,

#sigma=sqrt(0.73-0.7875^2)=0.33#