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Sagot :
To solve this problem, we need to determine the probability [tex]\( P(-3 < X < 5) \)[/tex] for the continuous random variable [tex]\( X \)[/tex] with the given probability density function (pdf):
[tex]\[ f(x) = \begin{cases} 0 & x < 2 \\ \frac{k}{x^2} & x \geq 2 \end{cases} \][/tex]
### Step-by-Step Solution
1. Determine the constant [tex]\( k \)[/tex]:
The constant [tex]\( k \)[/tex] is determined by the requirement that the total area under the probability density function (pdf) must equal 1. This ensures that the pdf is valid.
We need to integrate [tex]\( f(x) \)[/tex] over all possible values of [tex]\( x \)[/tex] and set the result equal to 1:
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \][/tex]
Since [tex]\( f(x) = 0 \)[/tex] for [tex]\( x < 2 \)[/tex], we only need to integrate from 2 to [tex]\(\infty\)[/tex]:
[tex]\[ \int_{2}^{\infty} \frac{k}{x^2} \, dx = 1 \][/tex]
Evaluate the integral:
[tex]\[ \int_{2}^{\infty} \frac{k}{x^2} \, dx = k \left[ -\frac{1}{x} \right]_{2}^{\infty} = k \left( 0 - \left( -\frac{1}{2} \right) \right) = k \cdot \frac{1}{2} = \frac{k}{2} \][/tex]
Set the integral equal to 1 and solve for [tex]\( k \)[/tex]:
[tex]\[ \frac{k}{2} = 1 \implies k = 2 \][/tex]
2. Calculate the probability [tex]\( P(-3 < X < 5) \)[/tex]:
The value of [tex]\( X \)[/tex] must be at least 2 for [tex]\( f(x) \)[/tex] to be non-zero. Therefore, we need to calculate [tex]\( P(2 < X < 5) \)[/tex] since [tex]\( P(-3 < X < 2) \)[/tex] is 0:
[tex]\[ P(2 < X < 5) = \int_{2}^{5} \frac{2}{x^2} \, dx \][/tex]
Evaluate the integral:
[tex]\[ \int_{2}^{5} \frac{2}{x^2} \, dx = 2 \left[ -\frac{1}{x} \right]_{2}^{5} = 2 \left( -\frac{1}{5} - \left( -\frac{1}{2} \right) \right) = 2 \left( \frac{1}{2} - \frac{1}{5} \right) \][/tex]
Simplify the expression:
[tex]\[ 2 \left( \frac{1}{2} - \frac{1}{5} \right) = 2 \left( \frac{5}{10} - \frac{2}{10} \right) = 2 \left( \frac{3}{10} \right) = \frac{6}{10} = 0.6 \][/tex]
Therefore, the probability [tex]\( P(-3 < X < 5) \)[/tex] is [tex]\( 0.6 \)[/tex].
[tex]\[ f(x) = \begin{cases} 0 & x < 2 \\ \frac{k}{x^2} & x \geq 2 \end{cases} \][/tex]
### Step-by-Step Solution
1. Determine the constant [tex]\( k \)[/tex]:
The constant [tex]\( k \)[/tex] is determined by the requirement that the total area under the probability density function (pdf) must equal 1. This ensures that the pdf is valid.
We need to integrate [tex]\( f(x) \)[/tex] over all possible values of [tex]\( x \)[/tex] and set the result equal to 1:
[tex]\[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \][/tex]
Since [tex]\( f(x) = 0 \)[/tex] for [tex]\( x < 2 \)[/tex], we only need to integrate from 2 to [tex]\(\infty\)[/tex]:
[tex]\[ \int_{2}^{\infty} \frac{k}{x^2} \, dx = 1 \][/tex]
Evaluate the integral:
[tex]\[ \int_{2}^{\infty} \frac{k}{x^2} \, dx = k \left[ -\frac{1}{x} \right]_{2}^{\infty} = k \left( 0 - \left( -\frac{1}{2} \right) \right) = k \cdot \frac{1}{2} = \frac{k}{2} \][/tex]
Set the integral equal to 1 and solve for [tex]\( k \)[/tex]:
[tex]\[ \frac{k}{2} = 1 \implies k = 2 \][/tex]
2. Calculate the probability [tex]\( P(-3 < X < 5) \)[/tex]:
The value of [tex]\( X \)[/tex] must be at least 2 for [tex]\( f(x) \)[/tex] to be non-zero. Therefore, we need to calculate [tex]\( P(2 < X < 5) \)[/tex] since [tex]\( P(-3 < X < 2) \)[/tex] is 0:
[tex]\[ P(2 < X < 5) = \int_{2}^{5} \frac{2}{x^2} \, dx \][/tex]
Evaluate the integral:
[tex]\[ \int_{2}^{5} \frac{2}{x^2} \, dx = 2 \left[ -\frac{1}{x} \right]_{2}^{5} = 2 \left( -\frac{1}{5} - \left( -\frac{1}{2} \right) \right) = 2 \left( \frac{1}{2} - \frac{1}{5} \right) \][/tex]
Simplify the expression:
[tex]\[ 2 \left( \frac{1}{2} - \frac{1}{5} \right) = 2 \left( \frac{5}{10} - \frac{2}{10} \right) = 2 \left( \frac{3}{10} \right) = \frac{6}{10} = 0.6 \][/tex]
Therefore, the probability [tex]\( P(-3 < X < 5) \)[/tex] is [tex]\( 0.6 \)[/tex].
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