Math Problem Statement

Consider the continuous probability density function defined by y = 0.5x + 0.2 on -0.4 ≤ x ≤ 1.6. Compute the following probabilities: P(X < 1.2) = 0.64 and P(X > 1.2).

Solution

The problem in the image asks to compute probabilities based on a continuous probability density function (PDF) given by y=0.5x+0.2y = 0.5x + 0.2, for 0.4x1.6-0.4 \leq x \leq 1.6. The graph shows a line representing this function.

Let’s break down the tasks:

  1. The first part asks for P(X<1.2)P(X < 1.2).

    • To find this probability, you need to integrate the PDF from the lower bound of xx (i.e., 0.4-0.4) to 1.2. This will give the area under the PDF curve, which represents the probability.
  2. The second part asks for P(X>1.2)P(X > 1.2).

    • This is equal to 1P(X1.2)1 - P(X \leq 1.2), which means you subtract the probability found in the first part from 1.

I’ll guide you through solving these:

Step 1: Find the normalization constant

Since this is a continuous PDF, the total area under the curve from x=0.4x = -0.4 to x=1.6x = 1.6 must equal 1. We can integrate the given PDF to check and normalize if necessary.

0.41.6(0.5x+0.2)dx\int_{-0.4}^{1.6} (0.5x + 0.2) \, dx

Step 2: Calculate P(X<1.2)P(X < 1.2)

To find P(X<1.2)P(X < 1.2), we will integrate the PDF from x=0.4x = -0.4 to x=1.2x = 1.2:

P(X<1.2)=0.41.2(0.5x+0.2)dxP(X < 1.2) = \int_{-0.4}^{1.2} (0.5x + 0.2) \, dx

Step 3: Calculate P(X>1.2)P(X > 1.2)

This can be computed as:

P(X>1.2)=1P(X1.2)P(X > 1.2) = 1 - P(X \leq 1.2)

Let me solve these for you using integration.The total area under the probability density function (PDF) from x=0.4x = -0.4 to x=1.6x = 1.6 is 1, confirming that the PDF is correctly normalized.

Here are the results for the probabilities:

  1. P(X<1.2)=0.64P(X < 1.2) = 0.64
  2. P(X>1.2)=0.36P(X > 1.2) = 0.36

These values represent the areas under the PDF curve for the respective ranges.

Do you want more details or have any questions?

Here are some related questions for further practice:

  1. How would you compute P(0<X<1.2)P(0 < X < 1.2)?
  2. What is the cumulative distribution function (CDF) for this PDF?
  3. How would you find the median value of XX given this PDF?
  4. Can you determine the expected value (mean) of XX based on the PDF?
  5. How would the probabilities change if the PDF were modified to y=0.5x+0.3y = 0.5x + 0.3?

Tip: The area under the PDF curve between any two points represents the probability that the random variable falls within that range.

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Math Problem Analysis

Mathematical Concepts

Probability Density Function
Integration
Area under the curve

Formulas

P(X < a) = ∫ (PDF) dx from x = lower bound to x = a
P(X > a) = 1 - P(X ≤ a)

Theorems

Properties of continuous probability density functions
Fundamental Theorem of Calculus

Suitable Grade Level

College Level or Advanced High School