Math Problem Statement

1) A variable X is distributed at random between the value 0 and 4, and its probability density function is given by f(x) = kx^2(4 - x^2). Find the value of k. 2) 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. 3) 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 go through the questions one by one:

Question 4:

A variable XX is distributed randomly between the values 0 and 4, with the probability density function (PDF) given by:

f(x)=kx2(4x2)f(x) = kx^2(4 - x^2)

To find the value of kk, we must use the fact that the total probability for a PDF is 1, meaning:

04f(x)dx=1\int_{0}^{4} f(x) \, dx = 1

Substituting the given PDF:

04kx2(4x2)dx=1\int_{0}^{4} kx^2(4 - x^2) \, dx = 1

This will give us the value of kk.

Question 5:

In four tosses of a coin, let XX be the number of heads. You are asked to:

  • Tabulate the 16 possible outcomes and the corresponding values of XX.
  • Find the probability distribution of XX, the expected value, and the variance.

Here, XX follows a binomial distribution with parameters:

  • n=4n = 4 (number of tosses),
  • p=0.5p = 0.5 (probability of getting heads in each toss).

We can calculate the probabilities for each value of X=0,1,2,3,4X = 0, 1, 2, 3, 4 using the binomial formula: P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1 - p)^{n-k}

After determining the distribution, we can compute the expected value and variance:

  • Expected value E[X]=npE[X] = np,
  • Variance Var(X)=np(1p)\text{Var}(X) = np(1 - p).

Question 6:

XX is the number scored in a throw of a fair die. The task involves applying Chebyshev's inequality and finding the actual probability.

Chebyshev's inequality gives the probability that XX deviates from its mean μ\mu by more than 2.5: P(Xμ>2.5)<0.47P(|X - \mu| > 2.5) < 0.47

For a fair die, XX takes values from 1 to 6, so the mean μ\mu is: μ=1+2+3+4+5+66=3.5\mu = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5

We'll also find the actual probability using the distribution of XX.


Would you like me to solve one of these in more detail? Here are 5 related questions to deepen your understanding:

  1. How do you calculate the variance of a continuous random variable using a PDF?
  2. What is the interpretation of Chebyshev's inequality in practical terms?
  3. How does binomial distribution relate to the probability of heads in multiple coin tosses?
  4. What are the steps to compute the actual probability in question 6?
  5. How would the expected value change if the coin in question 5 was biased?

Tip: Always verify the normalization condition when dealing with probability density functions (i.e., the total probability should integrate to 1).

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Math Problem Analysis

Mathematical Concepts

Probability Density Function
Binomial Distribution
Chebyshev's Inequality
Expected Value
Variance

Formulas

f(x) = kx^2(4 - x^2)
P(X = k) = (n choose k) * p^k * (1 - p)^(n-k)
Chebyshev's Inequality: P(|X - μ| > kσ) ≤ 1/k^2

Theorems

Chebyshev's Inequality
Binomial Distribution Theorem

Suitable Grade Level

University Level