Suppose Martin is a very talented used-car salesman. Whenever Martin talks to a new customer, there is a \( 70 \% \) chance thi he convinces the customer to purchase one of his used cars. Brian, Martin's boss, is envious that Martin sells many more cars than he does. Because of his jealousy, Brian institutes a new rule that Martin is only allowed to talk to 45 customers per day. Thus, Martin continues to work each day until he speaks to 45 customers, at which point Brian sends him home. Let \( X \) represent the number of used cars that Martin sells on a given day. What are the mean, \( \mu \), and variance, \( \sigma^{2} \), of \( X \) ? Please round your answers to the nearest two decimal places.

To find the mean (\(\mu\)) and variance (\(\sigma^{2}\)) of the number of used cars that Martin sells (\(X\)), we can model this scenario using the binomial distribution. Here, Martin has a fixed number of trials (customers spoken to), a constant probability of success (convincing a customer), and each customer interaction is independent. 1. **Mean (\(\mu\))**: The mean of a binomial distribution is calculated using the formula: \[ \mu = n \cdot p \] where \(n\) is the number of trials (customers) and \(p\) is the probability of success (selling a car). In this case, \(n = 45\) and \(p = 0.7\). \[ \mu = 45 \cdot 0.7 = 31.5 \] 2. **Variance (\(\sigma^{2}\))**: The variance of a binomial distribution is calculated using the formula: \[ \sigma^{2} = n \cdot p \cdot (1 - p) \] Again, substituting the values: \[ \sigma^{2} = 45 \cdot 0.7 \cdot (1 - 0.7) = 45 \cdot 0.7 \cdot 0.3 = 9.45 \] Thus, rounding the results to the nearest two decimal places: - Mean (\(\mu\)) = **31.50** - Variance (\(\sigma^{2}\)) = **9.45**