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A quantity x has cumulative distribution function p(x) = x − x2/4 for 0 ≤ x ≤ 2 and p(x) = 0 for x < 0 and p(x) = 1 for x > 2. find the mean and median of x.

Respuesta :

LammettHash
[tex]F_X(x)=\begin{cases}0&\text{for }x<0\\\\x-\dfrac{x^2}4&\text{for }0\le x\le2\\\\1&\text{for }x>2\end{cases}[/tex]

Recall that the PDF is given by the derivative of the CDF:

[tex]f_X(x)=\dfrac{\mathrm dF_X(x)}{\mathrm dx}=\begin{cases}1-\dfrac x2\\\\0&\text{otherwise}\end{cases}[/tex]

The mean is given by

[tex]\mathbb E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^2\left(x-\dfrac{x^2}2\right)\,\mathrm dx=\frac23[/tex]

The median is the number [tex]M[/tex] such that [tex]F_X(x)=\mathbb P(X\le M)=\dfrac12[/tex]. We have

[tex]F_X(M)=M-\dfrac{M^2}4=\dfrac12\implies M=2\pm\sqrt2[/tex]

but both roots can't be medians. As a matter of fact, the median must satisfy [tex]0\le M\le2[/tex], so we take the solution with the negative root. So [tex]M=2-\sqrt2[/tex] is the median.
MrRoyal

The mean of x is 2/3, and the median of x is [tex]2 - \sqrt 2[/tex]

How to calculate the mean and the median

The cumulative distribution function is given as:

[tex]p(x) = \left[\begin{array}{ccc}0&for&x < 0\\x - \frac{x^2}{4}&for&0 \le x \le 2\\1&for&x>2\end{array}\right[/tex]

Differentiate the cumulative distribution function.

So, we have:

[tex]f(x) = \left[\begin{array}{ccc}1 - \frac{x}{2}&for&0 \le x \le 2\\0&&Otherwise\end{array}\right[/tex]

The mean is then calculated as:

[tex]E(x) = \int\limits^a_b {x * f(x)} \, dx[/tex]

So, we have:

[tex]E(x) = \int\limits^2_0 {x *[1 - \frac{x}{2}]} \, dx[/tex]

Expand

[tex]E(x) = \int\limits^2_0 {x - \frac{x^2}{2}} \, dx[/tex]

Integrate

[tex]E(x) = [\frac{x^2}{2} - \frac{x^3}{6}]|\limits^2_0[/tex]

Expand

[tex]E(x) = \frac{2^2}{2} - \frac{2^3}{6}[/tex]

[tex]E(x) = 2 - \frac{8}{6}[/tex]

[tex]E(x) = \frac{12-8}{6}[/tex]

[tex]E(x) = \frac{4}{6}[/tex]

Reduce fraction

[tex]E(x) = \frac{2}{3}[/tex]

So, the mean of x is 2/3

To calculate the median, we have:

[tex]p(x) = x - \frac{x^2}{4}[/tex]

Rewrite as:

[tex]p(M) = M - \frac{M^2}{4}[/tex]

The median element at the mid-position (i.e. 1/2).

So, we have:

[tex]M - \frac{M^2}{4} =\frac 12[/tex]

Multiply through by 4

[tex]4M - M^2 = 2[/tex]

Rewrite as:

[tex]M^2 - 4M + 2 = 0[/tex]

Solve for M

[tex]M = 2 \pm \sqrt 2[/tex]

Recall that:

The domain is given as: [tex]0 \le x \le 2[/tex]

[tex]2 + \sqrt 2[/tex] is outside the domain.

Hence, the median of x is [tex]2 - \sqrt 2[/tex]

Read more about mean and median at:

https://brainly.com/question/14532771

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