Question

Suppose we flip a fair coin 100 times. Define a random variable \(X\) on the underlying sample space \(\Omega\) that counts the number of heads that turn up. (a) What are the mean \(\mu\) and the variance \(a^{2}\) of \(X\) ? (b) Use Theorem 2 to give an upper bound for the probability that \(X\) differs from \(\mu\) by 10 or more.

Short answer

(a) Mean \(\mu = 50\), Variance \(\sigma^2 = 25\). (b) Probability is at most 0.25.

Step-by-step solution

Identify the distribution

The random variable \(X\) counts the number of heads in 100 coin flips. A coin flip is a Bernoulli trial with probability \(p = 0.5\) for heads. Therefore, \(X\) follows a Binomial distribution: \(X \sim \text{Binomial}(n = 100, p = 0.5)\).

Calculate the mean

For a Binomial distribution \(X \sim \text{Binomial}(n, p)\), the mean \(\mu\) is given by \(\mu = np\). Therefore, \(\mu = 100 \times 0.5 = 50\).

Calculate the variance

The variance \(\sigma^2\) of a Binomial distribution \(X \sim \text{Binomial}(n, p)\) is given by \(\sigma^2 = np(1-p)\). Therefore, \(\sigma^2 = 100 \times 0.5 \times 0.5 = 25\).

Apply Theorem 2 (Chebyshev's Inequality)

Chebyshev's Inequality states that for any random variable \(X\) with mean \(\mu\) and variance \(\sigma^2\), the probability that \(X\) deviates from \(\mu\) by \(k\) or more is bounded by \[ P(|X - \mu| \geq k) \leq \frac{\sigma^2}{k^2} \]. For our problem, \(k = 10\) and \(\sigma^2 = 25\), so \[ P(|X - 50| \geq 10) \leq \frac{25}{10^2} = 0.25 \].

Conclusion of the probability bound

Using Chebyshev's Inequality, the probability that the number of heads differs from the mean by 10 or more is at most 0.25.

Key concepts

Mean and Variance Calculation

When dealing with a binomial distribution, the mean and variance give us vital information about the expected outcomes and the spread of the data around the mean.
The mean, often represented by \( \mu \), tells us the expected value if we repeat an experiment many times. In the context of flipping a fair coin 100 times, the mean indicates the expected number of heads.
For a binomial distribution where \( n \) is the number of trials (in this case, 100 coin flips) and \( p \) is the probability of success on each trial (0.5 for heads), the mean is calculated as follows:

• \( \mu = np \)

Given that \( n = 100 \) and \( p = 0.5 \), the calculation is straightforward:

• \( \mu = 100 \times 0.5 = 50 \)

This means that we expect 50 heads on average.
The variance provides a measure of how much the outcomes will differ from the mean. For a binomial distribution, the variance is calculated using:

• \( \sigma^2 = np(1-p) \)

Substituting our values, we have:

• \( \sigma^2 = 100 \times 0.5 \times 0.5 = 25 \)

So, the variance is 25, indicating the degree of variation around the mean, helping us understand the data's dispersion.

Random Variables

The concept of random variables is fundamental to understanding probability and statistics. A random variable is essentially a rule or function that assigns numerical values to the outcomes of a random phenomenon.
In our example, each flip of the coin is a trial, and the outcome (heads or tails) contributes to determining the random variable \( X \), which counts the total number of heads across 100 flips.
This transformation from a set of outcomes to numbers allows for meaningful statistical analysis. Random variables can be:

• Discrete: Countable values (like the number of heads).
• Continuous: Any value within a range (like temperature).

Since the coin flips result in a countable number of heads, \( X \) is a discrete random variable following a binomial distribution.
This type of modeling is powerful as it helps simplify complex real-world problems into a format that’s easy to handle mathematically, enabling us to compute things like mean, variance, and probability distributions effectively.

Chebyshev's Inequality

Chebyshev's Inequality provides a conservative estimate of the probability that a random variable deviates from its mean. It is particularly useful when dealing with distributions that are not well understood or when we lack complete information about the distribution's shape.
The inequality states that, for any random variable with a known mean \( \mu \) and variance \( \sigma^2 \), the probability of the variable differing from \( \mu \) by more than a certain value \( k \) is at most \( \frac{\sigma^2}{k^2} \).
Applying this to our coin flip problem:

• We want to know the probability that the number of heads, \( X \), is at least 10 heads away from the mean of 50.
• Using Chebyshev's formula, \( P(|X - 50| \geq 10) \leq \frac{25}{10^2} \).
• Hence, the bound is given as \( 0.25 \).

This means that the probability of \( X \) deviating from its mean by 10 or more is no greater than 0.25.
While it may not give the exact likelihood, the inequality provides a bound that helps understand and manage the uncertainty associated with the distribution of outcomes.