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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?

Answer 1

Grace Agate

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#

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Answer 2

Anthony Amato

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).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.