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Data Interview Question

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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)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

      where (nk)\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 (μ\mu) = np = 1000 * 0.5 = 500
      • Standard deviation (σ\sigma) = np(1p)=10000.50.5=15.81\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)P(X > 550).

    • Convert this to a Z-score using:

      Z=XμσZ = \frac{X - \mu}{\sigma}

      For X = 550:

      Z=55050015.81=3.16Z = \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)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)0.0008P(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.