Answer :
Let's solve the problem step by step.
### Step 1: Determine the formula for [tex]\( P(x \leq x_0) \)[/tex]
Given the exponential probability density function:
[tex]\[ f(x) = \frac{1}{5} e^{-\frac{x}{5}} \quad \text{for } x \geq 0 \][/tex]
To find the cumulative distribution function (CDF), [tex]\( F(x) \)[/tex], which gives us [tex]\( P(x \leq x_0) \)[/tex], we integrate the probability density function from 0 to [tex]\( x_0 \)[/tex]:
[tex]\[ F(x_0) = \int_0^{x_0} \frac{1}{5} e^{-\frac{x}{5}} \, dx \][/tex]
Solving this integral, we get:
[tex]\[ F(x_0) = \left[ -e^{-\frac{x}{5}} \right]_0^{x_0} = \left(1 - e^{-\frac{x_0}{5}}\right) \][/tex]
Therefore, the correct formula for [tex]\( P(x \leq x_0) \)[/tex] is:
[tex]\[ P(x \leq x_0) = 1 - e^{-\frac{x_0}{5}} \][/tex]
Thus, the correct answer is option 2.
### Step 2: Find [tex]\( P(x \leq 1) \)[/tex]
Using the formula from Step 1, for [tex]\( x_0 = 1 \)[/tex]:
[tex]\[ P(x \leq 1) = 1 - e^{-\frac{1}{5}} \][/tex]
Given the final result, this value is:
[tex]\[ P(x \leq 1) = 0.1813 \, \text{ (to 4 decimal places)} \][/tex]
### Step 3: Find [tex]\( P(x \geq 3) \)[/tex]
To find [tex]\( P(x \geq 3) \)[/tex], we can use the complementary rule [tex]\( P(x \geq x_0) = 1 - P(x \leq x_0) \)[/tex].
Using the formula from Step 1, for [tex]\( x_0 = 3 \)[/tex]:
[tex]\[ P(x \leq 3) = 1 - e^{-\frac{3}{5}} \][/tex]
Therefore:
[tex]\[ P(x \geq 3) = 1 - P(x \leq 3) = e^{-\frac{3}{5}} \][/tex]
Given the final result, this value is:
[tex]\[ P(x \geq 3) = 0.5488 \, \text{ (to 4 decimal places)} \][/tex]
### Step 4: Find [tex]\( P(x \leq 5) \)[/tex]
Using the formula from Step 1, for [tex]\( x_0 = 5 \)[/tex]:
[tex]\[ P(x \leq 5) = 1 - e^{-\frac{5}{5}} = 1 - e^{-1} \][/tex]
Given the final result, this value is:
[tex]\[ P(x \leq 5) = 0.6321 \, \text{ (to 4 decimal places)} \][/tex]
### Step 5: Find [tex]\( P(1 \leq x \leq 5) \)[/tex]
This can be found using the difference of two CDFs:
[tex]\[ P(1 \leq x \leq 5) = P(x \leq 5) - P(x \leq 1) \][/tex]
From the steps above:
[tex]\[ P(x \leq 5) = 0.6321 \][/tex]
[tex]\[ P(x \leq 1) = 0.1813 \][/tex]
Therefore:
[tex]\[ P(1 \leq x \leq 5) = 0.6321 - 0.1813 \][/tex]
Given the final result, this value is:
[tex]\[ P(1 \leq x \leq 5) = 0.4509 \, \text{ (to 4 decimal places)} \][/tex]
### Summary
- The formula for [tex]\( P(x \leq x_0) \)[/tex] is [tex]\( 1 - e^{-\frac{x_0}{5}} \)[/tex]
- [tex]\( P(x \leq 1) \)[/tex] is 0.1813
- [tex]\( P(x \geq 3) \)[/tex] is 0.5488
- [tex]\( P(x \leq 5) \)[/tex] is 0.6321
- [tex]\( P(1 \leq x \leq 5) \)[/tex] is 0.4509
### Step 1: Determine the formula for [tex]\( P(x \leq x_0) \)[/tex]
Given the exponential probability density function:
[tex]\[ f(x) = \frac{1}{5} e^{-\frac{x}{5}} \quad \text{for } x \geq 0 \][/tex]
To find the cumulative distribution function (CDF), [tex]\( F(x) \)[/tex], which gives us [tex]\( P(x \leq x_0) \)[/tex], we integrate the probability density function from 0 to [tex]\( x_0 \)[/tex]:
[tex]\[ F(x_0) = \int_0^{x_0} \frac{1}{5} e^{-\frac{x}{5}} \, dx \][/tex]
Solving this integral, we get:
[tex]\[ F(x_0) = \left[ -e^{-\frac{x}{5}} \right]_0^{x_0} = \left(1 - e^{-\frac{x_0}{5}}\right) \][/tex]
Therefore, the correct formula for [tex]\( P(x \leq x_0) \)[/tex] is:
[tex]\[ P(x \leq x_0) = 1 - e^{-\frac{x_0}{5}} \][/tex]
Thus, the correct answer is option 2.
### Step 2: Find [tex]\( P(x \leq 1) \)[/tex]
Using the formula from Step 1, for [tex]\( x_0 = 1 \)[/tex]:
[tex]\[ P(x \leq 1) = 1 - e^{-\frac{1}{5}} \][/tex]
Given the final result, this value is:
[tex]\[ P(x \leq 1) = 0.1813 \, \text{ (to 4 decimal places)} \][/tex]
### Step 3: Find [tex]\( P(x \geq 3) \)[/tex]
To find [tex]\( P(x \geq 3) \)[/tex], we can use the complementary rule [tex]\( P(x \geq x_0) = 1 - P(x \leq x_0) \)[/tex].
Using the formula from Step 1, for [tex]\( x_0 = 3 \)[/tex]:
[tex]\[ P(x \leq 3) = 1 - e^{-\frac{3}{5}} \][/tex]
Therefore:
[tex]\[ P(x \geq 3) = 1 - P(x \leq 3) = e^{-\frac{3}{5}} \][/tex]
Given the final result, this value is:
[tex]\[ P(x \geq 3) = 0.5488 \, \text{ (to 4 decimal places)} \][/tex]
### Step 4: Find [tex]\( P(x \leq 5) \)[/tex]
Using the formula from Step 1, for [tex]\( x_0 = 5 \)[/tex]:
[tex]\[ P(x \leq 5) = 1 - e^{-\frac{5}{5}} = 1 - e^{-1} \][/tex]
Given the final result, this value is:
[tex]\[ P(x \leq 5) = 0.6321 \, \text{ (to 4 decimal places)} \][/tex]
### Step 5: Find [tex]\( P(1 \leq x \leq 5) \)[/tex]
This can be found using the difference of two CDFs:
[tex]\[ P(1 \leq x \leq 5) = P(x \leq 5) - P(x \leq 1) \][/tex]
From the steps above:
[tex]\[ P(x \leq 5) = 0.6321 \][/tex]
[tex]\[ P(x \leq 1) = 0.1813 \][/tex]
Therefore:
[tex]\[ P(1 \leq x \leq 5) = 0.6321 - 0.1813 \][/tex]
Given the final result, this value is:
[tex]\[ P(1 \leq x \leq 5) = 0.4509 \, \text{ (to 4 decimal places)} \][/tex]
### Summary
- The formula for [tex]\( P(x \leq x_0) \)[/tex] is [tex]\( 1 - e^{-\frac{x_0}{5}} \)[/tex]
- [tex]\( P(x \leq 1) \)[/tex] is 0.1813
- [tex]\( P(x \geq 3) \)[/tex] is 0.5488
- [tex]\( P(x \leq 5) \)[/tex] is 0.6321
- [tex]\( P(1 \leq x \leq 5) \)[/tex] is 0.4509
