Between 1974 and 2011, there were 460 NFL football games decided in overtime. Consider that to start overtime, there is a coin flip to determine possession. Assume that winning the overtime coin toss does not provide an advantage, and each team has a \( 50 \% \) chance of winning the game in overtime. a. Find the mean for the number of wins in the 460 games b. Find the variance c. Find the standard deviation

To find the mean, variance, and standard deviation for the number of wins in the 460 NFL overtime games where each team has a 50% chance of winning, we can use the properties of a binomial distribution. Let \( n = 460 \) (the total number of games) and \( p = 0.5 \) (the probability of winning for one team). a. The mean \( \mu \) for a binomial distribution is calculated as: \[ \mu = n \cdot p = 460 \cdot 0.5 = 230 \] So, the mean number of wins in the 460 games is **230**. b. The variance \( \sigma^2 \) for a binomial distribution is calculated as: \[ \sigma^2 = n \cdot p \cdot (1 - p) = 460 \cdot 0.5 \cdot 0.5 = 115 \] So, the variance of the number of wins is **115**. c. The standard deviation \( \sigma \) is the square root of the variance: \[ \sigma = \sqrt{\sigma^2} = \sqrt{115} \approx 10.72 \] Thus, the standard deviation of the number of wins in the overtime games is approximately **10.72**.