Answer to Question #218939 in Statistics and Probability for max

Question

A random variable T has a probability density function given by

f(t)={1/2,0<t<1 both inclusive

{x-1/2,1<t<c both inclusive

{0,elsewhere

where c is a positive constant

1. sketch f(t) hence using geometric shapes or otherwise show that

c=(1+ 5^1/2)/2

2.Write down the median of T and the mode of T hence describe the skewness of the distribution(Give a reason for your answer)

Answer 1

1.

$$\displaystyle\int_{-\infin}^{\infin}f(x)dx=1$$

$$\displaystyle\int_{0}^{1}\dfrac{1}{2}dx+\displaystyle\int_{1}^{c}(x-\dfrac{1}{2})dx=\dfrac{1}{2}[x]\begin{matrix} 1 \\ 0 \end{matrix}+\dfrac{1}{2}[x^2-x]\begin{matrix} c \\ 1 \end{matrix}$$

$$=\dfrac{1}{2}+\dfrac{1}{2}(c^2-c-(1-1))=1$$

$$c^2-c-1=0$$

$$c=\dfrac{1\pm\sqrt{5}}{2}$$

Since $c\geq 1,$ we take $c=\dfrac{1+\sqrt{5}}{2}.$

$$f(x) = \begin{cases} \dfrac{1}{2}, &0\leq x\leq 1 \\ x-\dfrac{1}{2}, &1\leq x\leq\dfrac{1+\sqrt{5}}{2}\\ 0, & elsewhere \end{cases}$$

2. The mode of a continuous probability distribution is the point at which the probability density function attains its maximum value.

$$mode=\dfrac{1+\sqrt{5}}{2}>1$$

$$F(x) = \begin{cases} 0, & x<0 \\ \dfrac{1}{2}x, & 0\leq x<1 \\ \dfrac{x^2}{2}-\dfrac{1}{2}x+ \dfrac{1}{2}, &1\leq x<\dfrac{1+\sqrt{5}}{2}\\ 1, & x\geq \dfrac{1+\sqrt{5}}{2}\\ \end{cases}$$

The median of a continuous probability distribution

$$F(median)= \dfrac{1}{2}$$

$$median= 1$$

$$mean=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=$$

$$=\displaystyle\int_{0}^{1}\dfrac{1}{2}xdx+\displaystyle\int_{1}^{(1+\sqrt{5})/2}x(x-\dfrac{1}{2})dx$$

$$=\dfrac{1}{2}[\dfrac{1}{2}x^2]\begin{matrix} 1 \\ 0 \end{matrix}+[\dfrac{1}{3}x^3-\dfrac{1}{4}x^2]\begin{matrix} (1+\sqrt{5}/2 \\ 1 \end{matrix}$$

$$=\dfrac{1}{4}+\dfrac{1}{2}(\dfrac{1+\sqrt{5}}{2})^3-\dfrac{1}{2}(\dfrac{1+\sqrt{5}}{2})^2-\dfrac{1}{3}+\dfrac{1}{4}$$

$$=\dfrac{11+5\sqrt{5}}{24}\approx0.92<1$$

We have a negative (left) skewness

$$mean<median<mode$$