Question

The probability that a certain plant will die within x hours in a certain environment is
estimated to be [ 1–(1+x²)^–1]. Determine the probabilities that the plant will die
within 2 hours and that it will survive more than 3 hours. Find the corresponding density
function.

Answer 1

Let 𝑋 be a continuous random variable. The cumulative distribution function (CDF) for a random variable 𝑋 is defined by

$$F(x)=P(X \leq x)=\displaystyle\int_{-\infin}^\infin f(x)dx$$

We have that

$$F(x) = \begin{cases} 0 &\text{,\ } x<0 \\ {1-\dfrac{1}{1+x^2}} &\text{, \ } x\geq0 \end{cases}$$

$$x_2>x_1>0 => F(x_2)=1-\dfrac{1}{1+{x_2}^2}>1-\dfrac{1}{1+{x_1}^2}=F(x_1)$$

The CDF is non-decreasing function.

$$F(0)=1-\dfrac{1}{1+(0)^2}=0$$

$$\lim\limits_{x\rarr0^+}F(x)=\lim\limits_{x\rarr0^+}(1-\dfrac{1}{1+{x}^2})=1-\dfrac{1}{1+(0)^2}=0$$

$$\lim\limits_{x\rarr\infin}F(x)=\lim\limits_{x\rarr\infin}(1-\dfrac{1}{1+{x}^2})=1-0=1$$

The probability that the plant will die within 2 hours is equal to

$$P(X\leq2)=F(2)-F(0)=1-\dfrac{1}{1+(2)^2}-(1-\dfrac{1}{1+(0)^2})=$$ $$=\dfrac{4}{5}=0.8$$

The probability that the plant will survive more than 3 hours is equal to

$$P(X>3)=1-P(X\leq3)=1-(1-\dfrac{1}{1+(3)^2})=$$ $$=\dfrac{1}{10}=0.1$$

The function f(x) is the so- called density function (PDF) if

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

The cumulative distribution function (CDF)

$$F(x)=P(X \leq x)=\displaystyle\int_{-\infin}^\infin f(x)dx$$

Then

$$f(x)=F'(x)$$

We have that

$$F(x) = \begin{cases} 0 &\text{,\ } x<0 \\ {1-\dfrac{1}{1+x^2}} &\text{, \ } x\geq0 \end{cases}$$

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

Thus, the corresponding density function is

$$f(x) = \begin{cases} 0 &\text{,\ } x<0 \\ {\dfrac{2x}{(1+{x}^2)^2}} &\text{, \ } x\geq0 \end{cases}$$

$P(X\leq2)=\dfrac{4}{5}=0.8$

$P(X>3)=\dfrac{1}{10}=0.1$

$f(x) = \begin{cases} 0 &\text{,\ } x<0 \\ {\dfrac{2x}{(1+{x}^2)^2}} &\text{, \ } x\geq0 \end{cases}$