Question

What are the mean and standard deviation of the probability density function given by `(p(x))/k=x^3-4x` for `x in [0,2]`, 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

The mean is `mu=16/15 approx 1.0667` and the standard deviation is `sigma=2sqrt(11)/15 approx 0.4422`

Explanation:

If `p(x)=k(x^[3}-4x)` for `x in [0,2]`, then `int_{0}^[2}k(x^{3}-4x) dx=k[x^{4}/4-2x^{2}]_{0}^{2}=k(4-8)=-4k`. Therefore, we require `-4k=1` so that `k=-1/4`.

Hence, the density function is `p(x)=1/4(4x-x^{3})=x-x^{3}/4`.

The mean is the expected value of `X`:

`mu=E[X]=int_{0}^{2} x*p(x)\ dx=int_{0}^{2}(x^[2}-x^{4}/4)\ dx`

`=[x^{3}/3-x^{5}/20]_{0}^{2}=8/3-32/20=16/15`.

The expected value of `X^{2}` is

`E[X^{2}]=int_{0}^{2} x^{2}*p(x)\ dx=int_{0}^{2}(x^[3}-x^{5}/4)\ dx`

`=[x^{4}/4-x^{6}/24]_{0}^{2}=4-64/24=4/3`.

Hence, the variance is `sigma^{2}=Var(X)=E[X^{2}]-(E[X])^{2}=4/3-(16/15)^2=44/225`.

The standard deviation is

`sigma=sqrt{sigma^[2}}=sqrt{44/225}=sqrt(44)/15=(2sqrt{11})/15`