(a) Write the exact binomial expression for the probability of 5000 heads in \(10^{4}\) tosses of a true coin. (b) Use the normal approximation and tables or calculator to evaluate (a). (c) Use the normal approximation and tables to find the probability of between 4900 and 5075 heads.

The normal approximation is a useful method when dealing with large sample sizes in binomial distribution problems. Instead of working with the binomial formula, which can be complex for large values, we approximate the binomial distribution by a normal distribution. For this exercise, we approximated the binomial distribution of 10,000 coin tosses with a normal distribution that has the same mean and variance. The normal distribution is given by: \(N(\text{mean}, \text{variance})\).In this problem, the mean is the expected value of heads, which is \(np\) where \(n = 10,000\) and \(p = 0.5\). So, the mean becomes: \( \text{mean} = 10,000 \times 0.5 = 5000\).The variance is \(np(1-p)\). Here, it turns out to be: \( \text{variance} = 10,000 \times 0.5 \times 0.5 = 2500\). Hence, the standard deviation is: \( \text{stdev} = \textrm{sqrt}(2500) = 50\).\

The Z-score helps in finding the probability corresponding to a value in a normal distribution. It represents how many standard deviations an element is from the mean. The formula to calculate the Z-score is: \( Z = \frac{X - \text{mean}}{\text{stdev}}\).For instance, if we want to find the probability of exactly 5000 heads, we calculate the Z-score: \(Z = \frac{5000 - 5000}{50} = 0\).Using the Z-score tables, we find that the probability for Z = 0 is 0.5. This implies that the probability of getting exactly 5000 heads (without continuity correction) is 0.5.\

Continuity correction is applied when a discrete distribution (like binomial) is approximated using a continuous distribution (like normal). Since the normal distribution is continuous, we adjust for the discrete nature by adding or subtracting 0.5. For example, to find the probability of exactly 5000 heads using a continuity correction, we actually find: \(P(4999.5 < X < 5000.5)\).For this, we compute Z-scores for 4999.5 and 5000.5: \(Z = \frac{4999.5-5000}{50} = -0.01\), \(Z = \frac{5000.5-5000}{50} = 0.01\).From Z-tables, we find probabilities for Z = -0.01 and Z = 0.01. This small range around Z-score 0 accounts for the exact position in the normal curve and the probabilities close to 0.5.

The binomial coefficient, represented as \(\binom{n}{k}\), is a crucial element in calculating the exact binomial probability. It gives us the number of ways to choose \(k\) successes (heads) out of \(n\) trials (tosses). The formula is: \( \binom{n}{k} = \frac{n!}{k!(n-k)!}\).In this problem, \(n = 10,000\) and \(k = 5000\). While this results in a massive number, it is typically computed using software or a scientific calculator. The exact probability of obtaining 5000 heads out of 10,000 tosses without approximation would be: \(P(X = 5000) = \binom{10000}{5000}(0.5)^{5000}(0.5)^{5000} = \binom{10000}{5000}(0.5)^{10000}\).\

With the normal approximation, finding probabilities like those between 4900 and 5075 heads can be simpler. First, we use continuity correction: \(P(4900 < X < 5075)\) becomes \(P(4899.5 < X < 5075.5)\).We calculate Z-scores: \(Z = \frac{4899.5-5000}{50} = -2.01\) and \(Z = \frac{5075.5-5000}{50} = 1.51\).Using Z-tables, find the cumulative probabilities: for Z = -2.01, it is approximately 0.0222, and for Z = 1.51, it is approximately 0.9345. Subtracting these gives: \(P(4900 < X < 5075) \textrm{approximately} = 0.9345 - 0.0222 = 0.9123\). This means there's about a 91.23% probability of getting between 4900 and 5075 heads in 10,000 tosses.