What are the mean and standard deviation of the probability density function given by #(p(x))/k=ln(x^3+1)# 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 · 2017-11-30T08:11:58

The mean is #=0.78# and the standard deviation is #=0.18#

The function #ln(x^3+1)# is continuous and positive on the interval #[0,1]#, so it's a probability density function.

First , determine #k#

#P(x)=kln(x^3+1)#

#int_0^1p(x)dx=kint_0^1ln(x^3+1)dx=1#

#kxx0.2=1#

#k=1/0.2=5#

Therefore,

#P(x)=5ln(x^3+1)#

So,

The mean is

#E(x)=intxP(x)dx=5int_0^1xln(x^3+1)=0.78#

The variance is

#Var(x)=E(x^2)-(E(x))^2#

#E(x^2)=intx^2P(x)dx=int_0^1 5x^2ln(x^3+1)dx=0.64#

Therefore,

#Var(x)=0.64-0.78^2=0.0316#

The standard deviation is

#sigma(x)=sqrt(Var(x))=sqrt(0.0316)=0.18#