To solve this problem, we need to understand the standard deviation and its relation to the expected value (mean) of a probability distribution.
1. Expected Value ([tex]$\mu$[/tex]):
- The expected value, also known as the mean, of a random variable [tex]$X$[/tex] represents the long-run average number of tickets purchased in each transaction.
2. Variance ([tex]$\sigma^2$[/tex]):
- The variance measures the spread of the number of tickets purchased around the mean. It is calculated as the average of the squared differences from the expected value.
3. Standard Deviation ([tex]$\sigma$[/tex]):
- The standard deviation is the square root of the variance. It provides a measure of the average distance of each ticket count from the expected value.
Given that the expected value of the number of tickets purchased is 2.08, the variance is 0.8936, and the standard deviation is 0.9453041838477179, we interpret these results as follows:
- The expected value of 2.08 indicates that, on average, customers purchase about 2.08 tickets per transaction.
- The variance of 0.8936 suggests the spread or variability in the number of tickets purchased.
- The standard deviation of approximately 0.95 tickets indicates that the number of tickets purchased typically deviates from the mean by around 0.95 tickets. In other words, most ticket purchases vary from the mean (2.08 tickets) by about 0.95 tickets.
Thus, the correct interpretation of the standard deviation among the provided options is:
The number of tickets purchased typically varies from the expected value by 0.95 tickets.
