A random variable has the probability density function f(x) = 2x/5² on 0 ≤ x ≤ 5 and zero elsewhere. Verify that it integrates to one, find P(1.5 ≤ X ≤ 3.5), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^5 2x/5^{2} dx = [x^{2}/5^{2}]_{0}^5 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (3.5^{2} - 1.5^{2})/5^{2} = 0.4000$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(5)/3 = 3.333$$

6. Median — F(m) = 0.5 → m²/5² = 0.5 → m = 5/√2 = 3.536

Why the other options are there

• μ = 2.50 (assumed a uniform density)
• P = 0.400 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/5² on 0 ≤ x ≤ 5 and zero elsewhere. Verify that it integrates to one, find P(2.5 ≤ X ≤ 4.0), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^5 2x/5^{2} dx = [x^{2}/5^{2}]_{0}^5 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (4.0^{2} - 2.5^{2})/5^{2} = 0.3900$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(5)/3 = 3.333$$

6. Median — F(m) = 0.5 → m²/5² = 0.5 → m = 5/√2 = 3.536

Why the other options are there

• μ = 2.50 (assumed a uniform density)
• P = 0.300 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/4² on 0 ≤ x ≤ 4 and zero elsewhere. Verify that it integrates to one, find P(2.0 ≤ X ≤ 3.5), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^4 2x/4^{2} dx = [x^{2}/4^{2}]_{0}^4 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (3.5^{2} - 2.0^{2})/4^{2} = 0.5156$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(4)/3 = 2.667$$

6. Median — F(m) = 0.5 → m²/4² = 0.5 → m = 4/√2 = 2.828

Why the other options are there

• μ = 2.00 (assumed a uniform density)
• P = 0.375 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/6² on 0 ≤ x ≤ 6 and zero elsewhere. Verify that it integrates to one, find P(1.5 ≤ X ≤ 4.0), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^6 2x/6^{2} dx = [x^{2}/6^{2}]_{0}^6 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (4.0^{2} - 1.5^{2})/6^{2} = 0.3819$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(6)/3 = 4.000$$

6. Median — F(m) = 0.5 → m²/6² = 0.5 → m = 6/√2 = 4.243

Why the other options are there

• μ = 3.00 (assumed a uniform density)
• P = 0.417 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/4² on 0 ≤ x ≤ 4 and zero elsewhere. Verify that it integrates to one, find P(0.5 ≤ X ≤ 2.5), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^4 2x/4^{2} dx = [x^{2}/4^{2}]_{0}^4 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (2.5^{2} - 0.5^{2})/4^{2} = 0.3750$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(4)/3 = 2.667$$

6. Median — F(m) = 0.5 → m²/4² = 0.5 → m = 4/√2 = 2.828

Why the other options are there

• μ = 2.00 (assumed a uniform density)
• P = 0.500 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/3² on 0 ≤ x ≤ 3 and zero elsewhere. Verify that it integrates to one, find P(1.5 ≤ X ≤ 2.0), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^3 2x/3^{2} dx = [x^{2}/3^{2}]_{0}^3 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (2.0^{2} - 1.5^{2})/3^{2} = 0.1944$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(3)/3 = 2.000$$

6. Median — F(m) = 0.5 → m²/3² = 0.5 → m = 3/√2 = 2.121

Why the other options are there

• μ = 1.50 (assumed a uniform density)
• P = 0.167 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/6² on 0 ≤ x ≤ 6 and zero elsewhere. Verify that it integrates to one, find P(2.5 ≤ X ≤ 4.5), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^6 2x/6^{2} dx = [x^{2}/6^{2}]_{0}^6 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (4.5^{2} - 2.5^{2})/6^{2} = 0.3889$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(6)/3 = 4.000$$

6. Median — F(m) = 0.5 → m²/6² = 0.5 → m = 6/√2 = 4.243

Why the other options are there

• μ = 3.00 (assumed a uniform density)
• P = 0.333 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/8² on 0 ≤ x ≤ 8 and zero elsewhere. Verify that it integrates to one, find P(3.5 ≤ X ≤ 7.5), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^8 2x/8^{2} dx = [x^{2}/8^{2}]_{0}^8 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (7.5^{2} - 3.5^{2})/8^{2} = 0.6875$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(8)/3 = 5.333$$

6. Median — F(m) = 0.5 → m²/8² = 0.5 → m = 8/√2 = 5.657

Why the other options are there

• μ = 4.00 (assumed a uniform density)
• P = 0.500 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/6² on 0 ≤ x ≤ 6 and zero elsewhere. Verify that it integrates to one, find P(2.5 ≤ X ≤ 3.5), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^6 2x/6^{2} dx = [x^{2}/6^{2}]_{0}^6 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (3.5^{2} - 2.5^{2})/6^{2} = 0.1667$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(6)/3 = 4.000$$

6. Median — F(m) = 0.5 → m²/6² = 0.5 → m = 6/√2 = 4.243

Why the other options are there

• μ = 3.00 (assumed a uniform density)
• P = 0.167 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function

A random variable has the probability density function f(x) = 2x/3² on 0 ≤ x ≤ 3 and zero elsewhere. Verify that it integrates to one, find P(1.0 ≤ X ≤ 2.0), the mean, and the median (0.50 fractile).

Start with the thinking

• A valid probability density function must integrate to exactly 1 over its support.
• Interval probabilities come from the cumulative distribution F(x) = x²/b².

Step-by-step solution

1. Normalisation

   $$\int_{0}^3 2x/3^{2} dx = [x^{2}/3^{2}]_{0}^3 = 1 ✓$$

2. Formula

   $$P(x_{1} \le X \le x_{2}) = F(x_{2}) - F(x_{1}) = (x_{2}^{2} - x_{1}^{2})/b^{2}$$

3. Substituting

   $$P = (2.0^{2} - 1.0^{2})/3^{2} = 0.3333$$

4. Formula

   $$\mu = \int x f(x) dx = 2b/3$$

5. Substituting

   $$\mu = 2(3)/3 = 2.000$$

6. Median — F(m) = 0.5 → m²/3² = 0.5 → m = 3/√2 = 2.121

Why the other options are there

• μ = 1.50 (assumed a uniform density)
• P = 0.333 (used a uniform density)

Reference: FE Reference Handbook — Probability and Statistics → Probability Density Function