Over 550 Heads

Solution & Explanation

To solve the problem of finding the probability that a fair coin lands heads-up more than 550 times out of 1000 flips, we can use the properties of the binomial distribution and apply the normal approximation due to the large number of trials (n = 1000).

Step-by-Step Explanation:

1. Understanding the Binomial Distribution:

   • The number of heads in 1000 flips of a fair coin follows a binomial distribution with parameters:

     • n = 1000 (number of trials)
     • p = 0.5 (probability of getting heads in a single flip)

   • The probability mass function (PMF) of a binomial distribution is given by:

     $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

     where $\binom{n}{k}$ is the binomial coefficient.

2. Normal Approximation to the Binomial Distribution:

   • For large n, the binomial distribution can be approximated by a normal distribution with:
     • Mean ($\mu$) = np = 1000 * 0.5 = 500
     • Standard deviation ($\sigma$) = $\sqrt{np(1-p)} = \sqrt{1000 * 0.5 * 0.5} = 15.81$
   • This is because the binomial distribution becomes symmetric and bell-shaped for large n.

3. Standardizing the Variable:

   • We need to find $P(X > 550)$.

   • Convert this to a Z-score using:

     $Z = \frac{X - \mu}{\sigma}$

     For X = 550:

     $Z = \frac{550 - 500}{15.81} = 3.16$

4. Using the Standard Normal Distribution:

   • The Z-score of 3.16 corresponds to a probability value in the standard normal distribution table.
   • $P(Z > 3.16)$ is the area to the right of Z = 3.16.
   • From standard normal distribution tables or using statistical software, $P(Z > 3.16) \approx 0.0008$.

5. Conclusion:

   • Therefore, the probability that the coin lands heads-up more than 550 times in 1000 flips is approximately 0.0008, or 0.08%.

Additional Tools:

• Using Python's SciPy Library:

  • You can calculate this probability using Python's scipy.stats module:

  from scipy.stats import norm
  
  # Parameters
  n = 1000
  p = 0.5
  
  # Mean and standard deviation
  mu = n * p
  sigma = (n * p * (1 - p))**0.5
  
  # Calculate Z
  Z = (550 - mu) / sigma
  
  # Probability
  probability = 1 - norm.cdf(Z)
  print(probability)  # Output: 0.0008
  

This method provides a quick and efficient way to calculate probabilities for large binomial distributions using normal approximation.