Math Problem Statement

Bir sürekli rastgele değişken X'in olasılık dağılım fonksiyonu 0 < X < 8 için f(x) = 3(8x - x^2)/256 şeklinde verilmiştir. Kümülatif dağılım fonksiyonunu ve aşağıdaki olasılıkları hesaplayınız: (a) P(X < 2), (b) P(X < 9), (c) P(2 < X < 4), (d) P(X > 6), (e) P(X < x) = 0.95 ise x=?

Solution

To solve this problem, let's go through each part systematically.

Given Data:

The probability density function (PDF) of a continuous random variable XX is provided as: f(x)=3(8xx2)256,0<x<8f(x) = \frac{3(8x - x^2)}{256}, \quad 0 < x < 8

Objective:

  1. Find the cumulative distribution function (CDF) F(x)F(x).
  2. Calculate the probabilities:
    • (a) P(X<2)P(X < 2)
    • (b) P(X<9)P(X < 9)
    • (c) P(2<X<4)P(2 < X < 4)
    • (d) P(X>6)P(X > 6)
    • (e) Find xx such that P(X<x)=0.95P(X < x) = 0.95.

Step-by-Step Solution

Step 1: Find the Cumulative Distribution Function (CDF)

To find F(x)F(x), the cumulative distribution function, we integrate f(x)f(x) over the range 00 to xx: F(x)=0xf(t)dt=0x3(8tt2)256dtF(x) = \int_{0}^{x} f(t) \, dt = \int_{0}^{x} \frac{3(8t - t^2)}{256} \, dt

  1. Integrate f(t)f(t): Calculate 3(8tt2)256dt\int \frac{3(8t - t^2)}{256} \, dt.

  2. Evaluate the definite integral from 00 to xx: Substitute the limits of integration.

Step 2: Calculate the Probabilities

Once we have the CDF F(x)F(x), we can use it to calculate each probability as follows:

  1. (a) P(X<2)P(X < 2): Calculate F(2)F(2).
  2. (b) P(X<9)P(X < 9): Since the maximum of XX is 8, P(X<9)=P(X<8)=F(8)=1P(X < 9) = P(X < 8) = F(8) = 1.
  3. (c) P(2<X<4)P(2 < X < 4): Calculate F(4)F(2)F(4) - F(2).
  4. (d) P(X>6)P(X > 6): Calculate 1F(6)1 - F(6).
  5. (e) Find xx such that F(x)=0.95F(x) = 0.95: Solve for xx in the equation F(x)=0.95F(x) = 0.95.

I’ll perform the calculations for these integrals and probabilities now.Here are the results for each part:

  1. CDF F(x)F(x): F(x)=x2(12x)256F(x) = \frac{x^2(12 - x)}{256}

  2. Calculated Probabilities:

    • (a) P(X<2)=F(2)=0.15625P(X < 2) = F(2) = 0.15625
    • (b) P(X<9)=F(8)=1P(X < 9) = F(8) = 1 (since the maximum range of XX is 8, so P(X<9)=1P(X < 9) = 1)
    • (c) P(2<X<4)=F(4)F(2)=0.34375P(2 < X < 4) = F(4) - F(2) = 0.34375
    • (d) P(X>6)=1F(6)=0.15625P(X > 6) = 1 - F(6) = 0.15625
  3. (e) To find xx such that P(X<x)=0.95P(X < x) = 0.95:

    • The solution to F(x)=0.95F(x) = 0.95 yields x6.92x \approx 6.92.

Let me know if you would like a detailed explanation of any step, or if you have further questions!

Here are some related questions for deeper understanding:

  1. How is the cumulative distribution function derived from a probability density function?
  2. What does it mean when P(X<9)=1P(X < 9) = 1 in this context?
  3. How can you interpret the probability P(X>6)=0.15625P(X > 6) = 0.15625 in terms of the distribution?
  4. Why is the solution for xx when F(x)=0.95F(x) = 0.95 significant in probability?
  5. What are some applications of calculating specific probabilities for continuous distributions?

Tip: When integrating probability density functions, always check that the function integrates to 1 over the defined interval, confirming it as a valid PDF.

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Math Problem Analysis

Mathematical Concepts

Probability
Continuous Probability Distributions
Cumulative Distribution Function (CDF)
Integration

Formulas

f(x) = 3(8x - x^2) / 256
F(x) = ∫f(t) dt from 0 to x
P(a < X < b) = F(b) - F(a)
P(X > x) = 1 - F(x)

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

University Level