Final answer:
The mode of the given probability distribution is 2 and 6.
Step-by-step explanation:
The mode of a probability distribution is the value or values that occur with the highest frequency. In this case, we have the probability density function (PDF) of a random variable X which is given by f(x) = (x/108)(6-x)² for 0 ≤ x ≤ 6.
To find the mode, we need to determine which value(s) of x maximize the probability density function. We can do this by finding the derivative of the PDF with respect to x, setting it equal to zero, and solving for x. The x-values that satisfy this condition will be our mode(s).
Let's go through the steps:
- Find the derivative of the PDF: f'(x) = (1/108)(6-x)² - (2/108)(6-x)x
- Set f'(x) = 0 and solve for x: (1/108)(6-x)² - (2/108)(6-x)x = 0
- Simplify and solve the quadratic equation: (6-x)² - 2(6-x)x = 0
- Expand and rearrange the equation: 36 - 12x + x² - 12x + 2x² = 0
- Combine like terms: 3x² - 24x + 36 = 0
- Factor the quadratic equation: 3(x² - 8x + 12) = 0
- Solve for x using the factored form: x = 2, 6
Therefore, the mode of the random variable X is 2 and 6.