A salesperson has a 20% probability of making a sale to each customer who enters the store. Each sale is independent of all other sales.
a) Determine the mean number of sales on a day where 25 customers enter the store
b) Determine the standard deviation of the number of sales on a day where 25 customers enter the store
c) Determine the minimum number of customers who would have to enter the store to have an 88% chance or more of making at least one sale.

Given that each customer has a 20% probability of making a purchase, and each sale is independent of the others, we can model this scenario using a binomial distribution. Let's denote:

- \( n \) as the number of customers,
- \( p \) as the probability of making a sale to each customer, which is 0.20.

### Part (a): Mean Number of Sales

The mean \( \mu \) of a binomial distribution is given by:

\[ \mu = n \cdot p \]

For 25 customers:

\[ \mu = 25 \cdot 0.20 = 5 \]

Thus, the mean number of sales is 5.

### Part (b): Standard Deviation of the Number of Sales

The standard deviation \( \sigma \) of a binomial distribution is given by:

\[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \]

For 25 customers:

\[ \sigma = \sqrt{25 \cdot 0.20 \cdot 0.80} = \sqrt{25 \cdot 0.16} = \sqrt{4} = 2 \]

Thus, the standard deviation of the number of sales is 2.

### Part (c): Minimum Number of Customers for an 88% Chance of Making at Least One Sale

We need to find the smallest \( n \) such that the probability of making at least one sale is at least 88%. The probability of making at least one sale is:

\[ P(\text{at least one sale}) = 1 - P(\text{no sales}) \]

The probability of no sales when \( n \) customers enter is:

\[ P(\text{no sales}) = (1 - p)^n \]

We want:

\[ 1 - (1 - p)^n \geq 0.88 \]

This simplifies to:

\[ (1 - 0.20)^n \leq 0.12 \]

\[ 0.80^n \leq 0.12 \]

To solve for \( n \), take the natural logarithm of both sides:

\[ \ln(0.80^n) \leq \ln(0.12) \]

\[ n \ln(0.80) \leq \ln(0.12) \]

\[ n \geq \frac{\ln(0.12)}{\ln(0.80)} \]

Now, calculate the value:

\[ n \geq \frac{\ln(0.12)}{\ln(0.80)} \approx \frac{-2.1203}{-0.2231} \approx 9.50 \]

Since \( n \) must be an integer, we round up to the next whole number:

\[ n \geq 10 \]

Thus, a minimum of 10 customers must enter the store to have at least an 88% chance of making at least one sale.

Would you like any details or have any questions about this solution? 

Here are 5 more questions you can ask:
1. How would you solve a similar problem if the probability of making a sale changes?
2. Can you explain how the binomial distribution is derived?
3. How does the standard deviation change with different probabilities of making a sale?
4. What other distributions can be used for similar problems?
5. How can we use these calculations to optimize store performance?

**Tip:** Understanding the properties of different probability distributions can help in modeling real-world scenarios more accurately.

Understand the probability of sales in a store where each customer has a 20% chance of making a purchase. Calculate the mean number of sales, standard deviation, and minimum number of customers required for at least one sale with detailed step-by-step solutions.

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