What are the mean and standard deviation of the probability density function given by #p(x)=x^4-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?

1 Answer
Apr 13, 2016

1) #P(x) = Kint_0^1 x^4-x^3+1 dx#
# = K[x^5/5 - x^4/4 + x]_0^1 = K [1/5-1/4+1]= 19/20#
Thus #K= 20/19#

2) #mu = 20/19 int_0^1 x(x^4-x^3 +1) dx#
# = 29/19int (x^5-x^4+x) dx = 28/57 ~~ 0.491#

3) #var = 20/19 int_0^1 (x-28/57)^2(x^4-x^3 +1) dx#

#var = sqrt[(1,922)/(22,743)] ~~ .291#
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Explanation:

Given : Probability distribution, #p(x)=x^4-x^3+1# for # x in [0,1]#

Required: The mean and standard deviation

Solution Strategy:
1) First ensure you have a probability distribution function in the interval [0,1] by normalizing p(x), #P(x) = int_0^1 p(x) dx = 1 #
2) Use the mean integral equation, #mu= int_0^1 x p(x) dx#
3) Use the variance integral, #var = int_0^1 (x-mu)^2 p(x) dx #

Solution:
1) #P(x) = Kint_0^1 x^4-x^3+1 dx#
# = K[x^5/5 - x^4/4 + x]_0^1 = K [1/5-1/4+1]= 19/20#
Thus #K= 20/19#

2) #mu = 20/19 int_0^1 x(x^4-x^3 +1) dx#
# = 29/19int (x^5-x^4+x) dx = 28/57 ~~ 0.491#

3) #var = 20/19 int_0^1 (x-28/57)^2(x^4-x^3 +1) dx#

#var = sqrt[(1,922)/(22,743)] ~~ .291#