Math Problem Statement
solve this step by step with explantions
Solution
Let's solve the problem step by step:
Problem Description:
The probability density function is given as:
\begin{cases} K(x - x^2), & 0 \leq x \leq 6 \\ 0, & \text{otherwise} \end{cases}$$ We need to answer the following: --- ### **(a) Find the value of $$K$$:** #### Explanation: The total probability for a probability density function (PDF) must equal 1: $$\int_{-\infty}^\infty f(x) \, dx = 1$$ For this function, $$f(x) \neq 0$$ only in the range $$0 \leq x \leq 6$$. Thus, the integral simplifies to: $$\int_0^6 K(x - x^2) \, dx = 1$$ **Step 1: Simplify the integral** $$\int_0^6 K(x - x^2) \, dx = K \int_0^6 (x - x^2) \, dx$$ Split the integral: $$K \left( \int_0^6 x \, dx - \int_0^6 x^2 \, dx \right)$$ **Step 2: Solve each integral** 1. $$\int_0^6 x \, dx = \left[\frac{x^2}{2}\right]_0^6 = \frac{6^2}{2} - \frac{0^2}{2} = 18$$ 2. $$\int_0^6 x^2 \, dx = \left[\frac{x^3}{3}\right]_0^6 = \frac{6^3}{3} - \frac{0^3}{3} = 72$$ Substitute into the equation: $$K \left( 18 - 72 \right) = 1$$ **Step 3: Solve for $$K$$** $$K(-54) = 1 \quad \implies \quad K = -\frac{1}{54}$$ --- ### **(b) Find the median of $$x$$:** #### Explanation: The median of a probability distribution is the value $$m$$ such that: $$P(X \leq m) = 0.5$$ This is equivalent to solving: $$\int_0^m K(x - x^2) \, dx = 0.5$$ Substitute $$K = -\frac{1}{54}$$: $$\int_0^m -\frac{1}{54}(x - x^2) \, dx = 0.5$$ Solve this integral: 1. The integral is: $$-\frac{1}{54} \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^m = 0.5$$ 2. Substituting the limits: $$-\frac{1}{54} \left( \frac{m^2}{2} - \frac{m^3}{3} \right) = 0.5$$ Multiply through by $$-54$$: $$\frac{m^2}{2} - \frac{m^3}{3} = -27$$ Simplify: $$3m^2 - 2m^3 = -162$$ Factorize or solve numerically for $$m$$. --- ### **(c) Find the mode of $$x$$:** #### Explanation: The mode is the value of $$x$$ where $$f(x)$$ is maximized. To find it, differentiate $$f(x)$$ and set the derivative to zero: $$f'(x) = K \frac{d}{dx}(x - x^2)$$ $$f'(x) = K(1 - 2x)$$ Set $$f'(x) = 0$$: $$1 - 2x = 0 \quad \implies \quad x = \frac{1}{2}$$ The mode is $$x = \frac{1}{2}$$. --- ### **(d) Find the probability that the student will complete the project between 2 and 4 months:** #### Explanation: The probability is: $$P(2 \leq X \leq 4) = \int_2^4 f(x) \, dx$$ Substitute $$f(x) = -\frac{1}{54}(x - x^2)$$: $$P(2 \leq X \leq 4) = \int_2^4 -\frac{1}{54}(x - x^2) \, dx$$ Solve the integral: $$-\frac{1}{54} \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_2^4$$ 1. Evaluate at $$x = 4$$: $$\frac{4^2}{2} - \frac{4^3}{3} = \frac{16}{2} - \frac{64}{3} = 8 - 21.33 = -13.33$$ 2. Evaluate at $$x = 2$$: $$\frac{2^2}{2} - \frac{2^3}{3} = \frac{4}{2} - \frac{8}{3} = 2 - 2.67 = -0.67$$ 3. Subtract: $$P(2 \leq X \leq 4) = -\frac{1}{54}(-13.33 - (-0.67)) = -\frac{1}{54}(-12.66) = \frac{12.66}{54}$$ Simplify: $$P(2 \leq X \leq 4) \approx 0.2344$$ --- Let me know if you want further details or clarification on any step!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Density Functions
Integration
Median and Mode Calculation
Probability Calculation
Formulas
Integral of PDF equals 1: ∫ f(x) dx = 1
Median: ∫_0^m f(x) dx = 0.5
Mode: Find the maximum of f(x) by solving f'(x) = 0
Probability calculation over an interval: ∫_a^b f(x) dx
Theorems
Fundamental Theorem of Calculus
Properties of Probability Density Functions
Suitable Grade Level
Undergraduate
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