To solve the problem of finding the probability of obtaining at most 55 heads in 100 tosses of a fair coin, we can model this situation using the binomial distribution. The binomial distribution describes the number of successes (in this case, heads) in a fixed number of independent trials (the 100 coin tosses), where each trial has two possible outcomes (heads or tails) with a constant probability of success.

Parameters:

• The probability of getting a head (success) in one toss is $p = 0.5$ (since the coin is fair).
• The number of trials (tosses) is $n = 100$.
• The number of successes (heads) is denoted by $X$.

We want to find the probability of obtaining at most 55 heads, which is expressed as:

$P(X \leq 55)$

where $X$ is binomially distributed as $X \sim \text{Binomial}(n = 100, p = 0.5)$.

Approximation using the Normal Distribution:

For large $n$, we can approximate the binomial distribution using the normal distribution by applying the Central Limit Theorem. The binomial distribution $X \sim \text{Binomial}(100, 0.5)$ has:

• Mean: $\mu = np = 100 \times 0.5 = 50$
• Standard deviation: $\sigma = \sqrt{np(1-p)} = \sqrt{100 \times 0.5 \times 0.5} = \sqrt{25} = 5$

So, $X$ can be approximated by a normal distribution $X \sim N(50, 5^2)$.

To improve the accuracy of the normal approximation, we apply the continuity correction by adjusting for the fact that the binomial distribution is discrete while the normal distribution is continuous. Therefore, we need to calculate:

$P(X \leq 55) \approx P\left( X \leq 55.5 \right) \text{ for the normal distribution.}$

Standardization:

We convert $X = 55.5$ to the corresponding z-score:

$z = \frac{X - \mu}{\sigma} = \frac{55.5 - 50}{5} = \frac{5.5}{5} = 1.1$

Now, we look up the cumulative probability for $z = 1.1$ in the standard normal distribution table, or use a calculator:

$P(Z \leq 1.1) \approx 0.8643$

Thus, the probability of obtaining at most 55 heads in 100 tosses is approximately 0.8643 or 86.43%.

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Would you like further clarification or more details on any part of this solution? Here are five related questions to explore further:

1. How do you derive the normal distribution approximation from the binomial distribution?
2. What is the Central Limit Theorem and why is it useful for approximating binomial probabilities?
3. How does continuity correction improve normal approximation for discrete distributions?
4. What are other methods to calculate exact binomial probabilities without normal approximation?
5. How would the probability change if the coin were biased instead of fair?

Tip: When solving problems involving large numbers of trials in a binomial setting, the normal approximation can simplify calculations significantly, but it's important to always use continuity correction!