During the 2013 Major League Baseball season, the St. Louis Cardinals averaged 41,602 fans per home game. Suppose attendance during the season follows the normal probability distribution with a standard deviation of 9,440 per game. What is the probability that a randomly selected game during the 2013 season had an attendance between 25,000 and 35,000 people?

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To find the probability that a randomly selected game had an attendance between 25,000 and 35,000 people, we use the Z-score formula: $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value for which we're finding the Z-score, $\mu$ is the mean, and $\sigma$ is the standard deviation. For $X = 25,000$, $Z = \frac{25,000 - 41,602}{9,440} = -1.76$. For $X = 35,000$, $Z = \frac{35,000 - 41,602}{9,440} = -0.70$. Looking up these Z-scores on the standard normal distribution table gives the probabilities of 0.0392 and 0.2420, respectively. The probability of attendance being between 25,000 and 35,000 is the difference, which is $0.2420 - 0.0392 = 0.2028$.

During the 2013 Major League Baseball Season, the St