Math Problem Statement

4) A variable X is distributed at random between the value 0 and 4, and its probability density function is given by f(x) = kx^3(4 - x^2). Find the value of k. 5) In four tosses of a coin, let X be the number of heads. Tabulate the 16 possible outcomes with the corresponding values of X. By simple counting, derive the probability distribution of X and hence find the expected value and variance of X. 6) If X is the number scored in a throw of a fair die, show that Chebyshev's inequality gives P(|X - μ| > 2.5) < 0.47, where μ is the mean of X. Also, find the actual probability.

Solution

Let's break down and solve the problems one by one from the image you provided.


Question 4:

Problem: A variable XX is distributed at random between 0 and 4, with its probability density function given by:

f(x)=kx3(4x2)f(x) = kx^3(4 - x^2)

Find the value of kk.

Solution: To find kk, we use the property that the total probability of a continuous random variable over its range must be 1, i.e.,

04f(x)dx=1\int_0^4 f(x) dx = 1

Substitute f(x)=kx3(4x2)f(x) = kx^3(4 - x^2) and solve for kk.


Question 5:

Problem: In four tosses of a coin, let XX be the number of heads. Tabulate the 16 possible outcomes and the corresponding values of XX. By simple counting, derive the probability distribution of XX, and hence find the expected value and variance of XX.

Solution: The total number of outcomes is 24=162^4 = 16. You can list all possible outcomes (e.g., HHHH, HHHT, etc.) and count the number of heads in each.

Once we have the distribution, we can calculate the expected value E(X)E(X) and variance Var(X)\text{Var}(X) of XX using the formulas:

E(X)=xP(X=x)E(X) = \sum x \cdot P(X = x)

Var(X)=E(X2)(E(X))2\text{Var}(X) = E(X^2) - (E(X))^2


Question 6:

Problem: If XX is the number scored in a throw of a fair die, show that Chebyshev's inequality gives:

P(Xμ>2.5)<0.47P(|X - \mu| > 2.5) < 0.47

where μ\mu is the mean of XX. Also, find the actual probability.

Solution: For a fair die, the possible values of XX are 1,2,3,4,5,61, 2, 3, 4, 5, 6, each with a probability of 16\frac{1}{6}. The mean μ\mu and variance σ2\sigma^2 are:

μ=E(X)=1+2+3+4+5+66=3.5\mu = E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5

Var(X)=E(X2)(E(X))2\text{Var}(X) = E(X^2) - (E(X))^2

Next, use Chebyshev's inequality:

P(Xμ>kσ)1k2P(|X - \mu| > k\sigma) \leq \frac{1}{k^2}

Set k=2.5k = 2.5, then compare this result with the actual probability, which can be found by directly computing the probability.


Would you like me to go into more detail for any specific question or help with any further clarification?


Here are 5 related questions to explore further:

  1. How do you solve a general probability density function problem?
  2. What is the concept of expected value and variance in probability distributions?
  3. How can Chebyshev's inequality be applied in real-life problems?
  4. How would the expected value change for different types of coin toss experiments?
  5. What is the significance of the integral in probability density functions?

Tip: When solving probability problems, always start by ensuring that the total probability sums up to 1, especially when dealing with probability density functions (PDFs).

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Math Problem Analysis

Mathematical Concepts

Probability Density Function
Expected Value
Variance
Chebyshev's Inequality

Formulas

∫ f(x) dx = 1
E(X) = Σ x * P(X = x)
Var(X) = E(X^2) - (E(X))^2
P(|X - μ| > kσ) ≤ 1/k^2

Theorems

Chebyshev's Inequality
Probability Distribution
Expected Value Theorem

Suitable Grade Level

University Level (First-Year Probability and Statistics)