107. The random variable $X$ is given by the distribution function

$$
F(x)=\left\{\begin{array}{lll}
0 & \text { if } & x \leq 2 \\
(x-2)^{2} & \text { if } & 2 < x \leq 3 \\
1 & \text { if } & x > 3
\end{array}\right.
$$

Find:

a) the probability density function $f(x)$;

b) the graphs of the functions $F(x)$ and $f(x)$;

c) the probability that in the experiment $X$ will take a value not less than 2.1 and less than 2.5, using both the known function $F(x)$ and the found function $f(x)$.

Provide a geometric interpretation of the found probability $P(2.1 \leq x < 2.5)$.
Solution. a) The probability density function $f(x)$ is the derivative of the distribution function $F(x)$, so for $x \leq 2$ and $x > 3$, the function $f(x)=0$. Therefore, the probability density function is characterized by the expression:

$$
f(x)=\left\{\begin{array}{lll}
0 & \text { if } & x \leq 2 \\
2(x-2) & \text { if } & 2 < x \leq 3 \\
0 & \text { if } & x > 3
\end{array}\right.
$$

b) The graphs of the functions $F(x)$ and $f(x)$ are shown in Fig. 15 and Fig. 16.

![](https://cdn.mathpix.com/cropped/2024_05_22_c311c9dd990dc55c716eg-133.jpg?height=466&width=648&top_left_y=92&top_left_x=374)

Fig. 15

![](https://cdn.mathpix.com/cropped/2024_05_22_c311c9dd990dc55c716eg-133.jpg?height=504&width=734&top_left_y=621&top_left_x=347)

Fig. 16

c) The probability of the random variable falling into the interval $[2.1, 2.5)$ is determined as the increment of the distribution function over this interval using formula (28):

$$
P(2.1 \leq X < 2.5) = F(2.5) - F(2.1) = (0.5)^2 - (0.1)^2 = 0.24
$$

We can also find this probability using the known probability density function, applying formula (25):

$$
P(2.1 \leq X < 2.5) = \int_{2.1}^{2.5} f(x) \, dx = 2 \int_{2.1}^{2.5} (x-2) \, dx = \left. (x-2)^2 \right|_{2.1}^{2.5} = 0.24
$$

The obtained probability $P(2.1 \leq X < 2.5)$ can be geometrically interpreted as the area of the shaded curvilinear trapezoid shown in Fig. 16.