Math Problem Statement

please no mistakes

Let X be a continuous random variable with a density function f(x) = { x^2 (2x + 3/2) for 0 < x < 1, 0 elsewhere }. Then P(X > 1/3) is equal to:

Solution

To solve for P(X>13)P(X > \frac{1}{3}) given the probability density function (PDF) f(x)=x2(2x+32)f(x) = x^2 \left(2x + \frac{3}{2}\right) for 0<x<10 < x < 1, we need to integrate the PDF from 13\frac{1}{3} to 11 as follows:

Step-by-Step Solution

  1. Set Up the Probability: P(X>13)=131f(x)dxP\left(X > \frac{1}{3}\right) = \int_{\frac{1}{3}}^1 f(x) \, dx

  2. Substitute the Density Function: P(X>13)=131x2(2x+32)dxP\left(X > \frac{1}{3}\right) = \int_{\frac{1}{3}}^1 x^2 \left(2x + \frac{3}{2}\right) \, dx

  3. Expand the Integrand: f(x)=x2(2x+32)=2x3+32x2f(x) = x^2 \left(2x + \frac{3}{2}\right) = 2x^3 + \frac{3}{2}x^2 Therefore, P(X>13)=131(2x3+32x2)dxP\left(X > \frac{1}{3}\right) = \int_{\frac{1}{3}}^1 \left(2x^3 + \frac{3}{2}x^2\right) \, dx

  4. Integrate Each Term:

    • For 2x3dx=2x44=x42\int 2x^3 \, dx = \frac{2x^4}{4} = \frac{x^4}{2}
    • For 32x2dx=32x33=x32\int \frac{3}{2}x^2 \, dx = \frac{3}{2} \cdot \frac{x^3}{3} = \frac{x^3}{2}

    So, P(X>13)=[x42+x32]131P\left(X > \frac{1}{3}\right) = \left[\frac{x^4}{2} + \frac{x^3}{2}\right]_{\frac{1}{3}}^1

  5. Evaluate the Integral at the Bounds:

    • At x=1x = 1: 142+132=12+12=1\frac{1^4}{2} + \frac{1^3}{2} = \frac{1}{2} + \frac{1}{2} = 1
    • At x=13x = \frac{1}{3}: (13)42+(13)32=1812+1272=1162+154=1+3162=4162=281\frac{\left(\frac{1}{3}\right)^4}{2} + \frac{\left(\frac{1}{3}\right)^3}{2} = \frac{\frac{1}{81}}{2} + \frac{\frac{1}{27}}{2} = \frac{1}{162} + \frac{1}{54} = \frac{1 + 3}{162} = \frac{4}{162} = \frac{2}{81}
  6. Calculate the Probability: P(X>13)=1281=81281=7981P\left(X > \frac{1}{3}\right) = 1 - \frac{2}{81} = \frac{81 - 2}{81} = \frac{79}{81}

Final Answer

P(X>13)=7981P\left(X > \frac{1}{3}\right) = \frac{79}{81}

Would you like more details on any step or have additional questions?


Related Questions

  1. How would you find P(X<13)P(X < \frac{1}{3}) using the same PDF?
  2. What is the expectation E(X)E(X) of this continuous random variable?
  3. How would you verify if f(x)f(x) is a valid PDF?
  4. What if XX had a different range, say 0<x<20 < x < 2?
  5. How does this probability change if the function f(x)f(x) is modified slightly?

Tip

When calculating probabilities with continuous distributions, remember that integrating over the PDF within the specified bounds gives the probability directly.

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Math Problem Analysis

Mathematical Concepts

Probability
Continuous Random Variables
Integration

Formulas

P(X > a) = ∫[a, b] f(x) dx for continuous random variable X with PDF f(x)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Undergraduate Level (Introductory Statistics or Probability)