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?

Question

Grace Agate · Accepted Answer

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#

Consequently,

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

So,

The average is

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

The difference 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#

Consequently,

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

The deviation from the mean is

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

Anthony Amato · Answer

undefined To find the mean and standard deviation of the probability density function $\frac{p(x)}{k} = \ln(x^3 + 1)$ for $x$ in $[0, 1]$, we first need to find the value of $k$ such that the cumulative density across the range of $x$ is equal to 1.

The cumulative density function (CDF) is given by the integral of the probability density function (PDF) from the lower bound to the upper bound of the interval:

$$ \int_{0}^{1} \frac{p(x)}{k} \, dx = 1 $$

$$ \int_{0}^{1} \ln(x^3 + 1) \, dx = k $$

Now, we need to find the value of $k$ by evaluating this integral.

After finding the value of $k$, we can calculate the mean ($\mu$) and standard deviation ($\sigma$) using the formulas:

$$ \mu = \int_{0}^{1} x \cdot \frac{p(x)}{k} \, dx $$
$$ \sigma = \sqrt{ \int_{0}^{1} (x - \mu)^2 \cdot \frac{p(x)}{k} \, dx } $$

By integrating these expressions, we can find the mean and standard deviation in terms of $k$.