Problem 15 - Calculus for Biology and Medicine

Question

Toss a fair coin 400 times. Use the central limit theorem and the histogram correction to find an approximation for the probability of getting at most 190 heads.

Step-by-Step Solution

Answer

The probability of getting at most 190 heads is approximately 0.1711.

Step 1: Define the Random Variable

Let the random variable \( X \) represent the number of heads in 400 coin tosses. Since the coin is fair, \( X \) follows a binomial distribution with parameters \( n = 400 \) and \( p = 0.5 \). Thus, \( X \sim \text{Binomial}(400, 0.5) \).

Step 2: Calculate Mean and Standard Deviation

For a binomial distribution, the mean \( \mu \) is given by \( np \) and the standard deviation \( \sigma \) is \( \sqrt{np(1-p)} \). Calculating these values:\[ \mu = 400 \times 0.5 = 200 \]\[ \sigma = \sqrt{400 \times 0.5 \times 0.5} = 10 \]

Step 3: Apply Central Limit Theorem (CLT)

The CLT states that for large \( n \), the binomial distribution can be approximated by a normal distribution with the same mean and standard deviation. Therefore, \( X \) can be approximated by \( N(200, 10^2) \).

Step 4: Convert to Standard Normal Variable

To find the probability of getting at most 190 heads, we convert \( X = 190 \) to a standard normal variable \( Z \) using z-score:\[ Z = \frac{X - \mu}{\sigma} = \frac{190 - 200}{10} = -1 \]

Step 5: Use Histogram Correction (Continuity Correction)

Because we are approximating a discrete distribution with a continuous one, use continuity correction. So, we adjust 190 to 190.5. Calculate the new z-score:\[ Z = \frac{190.5 - 200}{10} = -0.95 \]

Step 6: Find Probability Using Standard Normal Distribution

Consult the standard normal distribution table (or use a calculator) to find the probability that \( Z \leq -0.95 \). The table or calculator will show \( P(Z \leq -0.95) \approx 0.1711 \).

Key Concepts

Binomial Distribution

When addressing problems involving probability, the binomial distribution is a fundamental concept. It describes the number of successes in a fixed number of trials, with each trial having two possible outcomes: success or failure. In this instance, tossing a fair coin 400 times and counting the number of heads is a perfect example of a binomial scenario.

• The trials are fixed at 400.
• Each trial is independent, meaning the result of one toss doesn't affect the other.
• The probability of getting a head in one toss is constant at 0.5.

Here, the random variable, represented as \( X \), follows a binomial distribution. It is denoted as \( X \sim \text{Binomial}(400, 0.5) \). The parameters \( n = 400 \) and \( p = 0.5 \) indicate there are 400 trials and each trial has a 50% probability of resulting in heads. Using these parameters, we calculate two critical pieces of statistical information: the mean \( \mu \) and the standard deviation \( \sigma \).

• The mean \( \mu = np = 400 \times 0.5 = 200 \).
• The standard deviation \( \sigma = \sqrt{np(1-p)} = \sqrt{400 \times 0.5 \times 0.5} = 10 \).

Normal Distribution

The normal distribution is crucial for understanding how data behaves around a central value, often shaped like a symmetrical bell curve. In the context of the binomial distribution, when the number of trials \( n \) is large, it can be approximated by a normal distribution. This is thanks to the central limit theorem (CLT).
The CLT allows us to use the normal distribution as an approximation for the binomial distribution when \( n \) is large, which simplifies calculations of probabilities for complex scenarios. For the coin-tossing exercise, since \( n = 400 \) is extensive, we can reasonably approximate the distribution of the number of heads using a normal distribution.

• Under CLT, \( X \sim \text{Binomial}(400, 0.5) \) can be approximated by \( N(200, 10^2) \).
• This means the distribution of \( X \) is centered around 200 with a standard deviation of 10, forming a normal curve.

To find probabilities using this normal distribution, we can convert specific outcomes into z-scores, which standardize results, making it easier to find probabilities using standard normal distribution tables. For instance, the z-score for 190 heads is \( Z = \frac{190 - 200}{10} = -1 \).

Continuity Correction

When moving from a discrete to a continuous model, the continuity correction becomes essential. This correction addresses the difference between a discrete outcome (like the number of heads, which is a count and can only be an integer) and the continuous nature of the normal curve.
When calculating probabilities for a discrete distribution using a continuous normal approximation, we adjust the discrete value by 0.5 to cater to its continuous counterpart; this small adjustment is known as continuity correction.

• For example, estimating the probability of at most 190 heads involves using 190.5 in our calculations instead of 190.
• This adjustment changes the z-score calculation slightly, leading to new probabilities when consulting the standard normal distribution.

In the exercise, using this corrected value, the new z-score is \( Z = \frac{190.5 - 200}{10} = -0.95 \). This slight shift provides a more accurate estimation of the probability, enhancing the model's precision, showing the probability in practice as \( P(Z \leq -0.95) \approx 0.1711 \).