Example 6.31. Three continuous random variables have different distributions: a) uniform; b) exponential; c) normal. For all three distributions, $M\left(X_{i}\right)=\sigma\left(X_{i}\right)=4$.

Find for all distribution laws the probability that in the result of the trial, the RV $X_{i}$ will take a value contained in the interval $(2 ; 12)$.
## Solution.

a) Uniform distribution.

The differential function of the uniform distribution of the random variable $X_{1}$ is:

$$
\begin{aligned}
& 0 \quad \text { for } \quad xb \text {. }
\end{aligned}
$$

The parameters $a$ and $b$ are found from the condition that for a uniform distribution

$$
\begin{aligned}
& M\left(X_{1}\right)=\frac{b+a}{2}, \quad \frac{b+a}{2}=4 \\
& D\left(X_{1}\right)=\frac{(b-a)^{2}}{12} \quad \text { or } \quad \frac{(b-a)^{2}}{12}=4^{2} .
\end{aligned}
$$

## Hence

$$
\begin{aligned}
& b+a=8, \quad b+a=8, \\
& (b-a)^{2}=192 \quad \text { or } \quad b-a=8 \sqrt{3} .
\end{aligned}
$$

Solving this system, we find that

$$
a=4(1-\sqrt{3}) \cong 2.93 ; \quad b=4(1+\sqrt{3}) \cong 10.93
$$

When taking the square root in the second equation of the system, the sign “+” is taken into account, considering that $a < 4(1+\sqrt{3})$.

The probability of the random variable $X_{1}$ falling into the interval $(\alpha ; \beta)$ is determined by the formula

$$
P\left(\alpha<X_{1}<\beta\right)=\int_{\alpha}^{\beta} f(x) d x
$$

For our case, the desired probability is:

$$
\begin{aligned}
P(5< & \left.X_{1}<12\right)=\int_{5}^{12} f(x) \cdot d x=\int_{5}^{4(1+\sqrt{3})} \frac{1}{8 \sqrt{3}} d x+\int_{4(1+\sqrt{3})}^{12} 0 \cdot d x= \\
& =\left.\frac{1}{8 \sqrt{3}} x\right|_{5} ^{4(1+\sqrt{3})}=\frac{1}{8 \sqrt{3}}(4+4 \sqrt{3}-5)=0.428
\end{aligned}
$$

Thus,

$$
P\left(5<X_{1}<12\right)=0.428 .
$$

b) Exponential distribution

The integral function of the exponential distribution of the random variable $X_{2}$ is:

$$
F(x)=1-e^{-\lambda x} .
$$

It is known that $M\left(X_{2}\right)=\frac{1}{\lambda}$ or $\frac{1}{\lambda}=4$ according to the condition. Hence, $\lambda=0.25$. Then

$$
F(x)=1-e^{-0.25 x} .
$$

The desired probability:

$$
P\left(5<X_{2}<12\right)=e^{-0.25 \cdot 5}-e^{-0.25 \cdot 12}=e^{-1.25}-e^{-3}=0.2367 .
$$

c) Normal distribution.

The probability that the random variable $X_{3}$, which follows a normal distribution, falls into the interval $(\alpha ; \beta)$, is:

$$
P\left(\alpha<X_{3}<\beta\right)=\Phi \frac{\beta-a}{\sigma}-\Phi \frac{\alpha-a}{\sigma}
$$

Here, $\Phi(x)$ is the Laplace function, $a$ is the expected value, and $\sigma$ is the standard deviation. Since, according to the condition of the problem, $a=\sigma=4$, the desired probability is:

$$
P\left(5<X_{3}<12\right)=\Phi \frac{12-4}{4}-\Phi \frac{5-4}{4}=\Phi(2)-\Phi \frac{1}{4}
$$

We find from the tables that

$$
\Phi(2)=0.4772 ; \quad \Phi \quad \frac{1}{4}=\Phi(0.25)=0.0987
$$

Then

$$
P\left(5<X_{3}<12\right)=0.4772-0.0987=0.3785 .
$$

## 6.4. Basic concepts of mathematical statistics