Question

Find (a) the mean and (b) the median of the random variable with the given pdf. F(x) = 3x^2, 0 ≤ x ≤ 1

Answer 1

@Loser66 can u he;p?

Answer 2

@onegirl sorry, I really dont' understand what is mean or median.

Answer 3

ok

Answer 4

@terenzreignz

Answer 5

the area under the curve is int(3x^2) = x^3 {between 0 & 1} = 1
ergo the mean is 1 because 1x1 = 1. the integration converts the parabola to a box!

the median is the halfway value - plug in x= 0.5 you get y = 0.75.

is this useful?

Answer 6

In the discrete case, recall that the mean of a random variable is merely its expected value.$$\bar{x}=E[X]=\sum_{k}^nx_kp(x_k)$$where \(x_1,\dots,x_n\) are the possible values of our random variable (i.e. the elements of our state space) and \(p(x)\) is our probability mass function.

In our continuous case, this sum is generalized to an integral:$$\bar{x}=E[X]=\int_\Omega xf(x)\,\mathrm{d}x$$where \(\Omega\) is our state space and \(f(x)\) is our probability density function.

In our case, \(\Omega=[0,1]\) and \(f(x)=3x^2\), hence $$\bar{x}=\int_0^13x^3\,\mathrm{d}x=3\int_0^1x^3\,\mathrm{d}x=3\left[\frac14x^4\right]_0^1=\frac34$$

Answer 7

For the median, you want there to be 50% chance the random variable is less than the median and 50% chance it is greater than the median. This can be visualized like so:



The area under our probability density function corresponds to probability; in fact, $$P(X<\tilde{x})=\int_0^{\tilde{x}}f(x)\,\mathrm{d}x$$hence$$P(X<\tilde{x})=\frac12\quad\implies\quad\int_0^{\tilde{x}}f(x)\,\mathrm{d}x=\frac12$$ Plug in our probability density function \(f\) and solve:$$\int_0^{\tilde{x}}3x^2\,\mathrm{d}x=\frac12\\3\int_0^{\tilde{x}}x^2\,\mathrm{d}x=\frac12\\3\left[\frac13x^3\right]_0^{\tilde{x}}=\frac12\\\tilde{x}^3=\frac12\\x=\sqrt[3]{\frac12}=\frac1{\sqrt[3]2}\approx0.794$$