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

Answer on Question #84919 – Math – Statistics and Probability

Question

The probability that a certain plant will die within $x$ hours in a certain environment is estimated to be $[1 - (1 + x^2)^{-1}]$. 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.

Solution

Let $X$ be a continuous random variable. The cumulative distribution function (CDF), or briefly the distribution function, for a random variable $X$ is defined by

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

We have that

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

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

The CDF is non-decreasing.

$$F(0) = 1 - \frac{1}{1 + (0)^2} = 0,$$ $$\lim_{x \to 0^+} F(x) = \lim_{x \to 0^+} \left(1 - \frac{1}{1 + x^2}\right) = 1 - \frac{1}{1 + (0)^2} = 0,$$ $$\lim_{x \to \infty} F(x) = \lim_{x \to \infty} \left(1 - \frac{1}{1 + x^2}\right) = 1 - 0 = 1.$$

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

$$P(X \leq 2) = F(2) - F(0) = 1 - \frac{1}{1 + (2)^2} - \left(1 - \frac{1}{1 + (0)^2}\right) = \frac{4}{5} = 0.8$$

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

$$P(X > 3) = 1 - P(X \leq 3) = 1 - \left(1 - \frac{1}{1 + (3)^2}\right) = \frac{1}{10} = 0.1$$

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

$$\int_{-\infty}^{\infty} f(x) \, dx = 1$$

The cumulative distribution function (CDF)

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

Then

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

We have that

$$F(x) = \begin{cases} 0, & x < 0 \\ 1 - \dfrac{1}{1 + x^2}, & x \geq 0 \end{cases}$$ $$\left(1 - \dfrac{1}{1 + x^2}\right)' = -\left(-\dfrac{1}{(1 + x^2)^2}\right)(2x) = \dfrac{2x}{(1 + x^2)^2}$$

Thus, the corresponding density function is

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

Answer:

$$P(X \leq 2) = \frac{4}{5} = 0.8, \quad P(X > 3) = \frac{1}{10} = 0.1, \quad f(x) = \begin{cases} 0, & x < 0, \\ \dfrac{2x}{(1 + x^2)^2}, & x \geq 0. \end{cases}$$

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