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Let's analyze and solve each part of this problem.

Given dataLet's analyze and solve each part of this problem.

Given 
- An average of \(4\) out of every \(10\) customers ask for water, so the probability \( p = 0.4 \).
- A sample of \(100\) customers is selected, so \( n = 100 \).

This situation follows a **binomial distribution** since we have a fixed number of trials (100 customers), each trial has two outcomes (ask for water or not), and the probability of success (asking for water) is the same for each trial.

### Part (i)
We need to find the probability that **more than 35 customers ask for water**. 

1. **Defining the Binomial Random Variable**: 
   - Let \( X \) be the number of customers who ask for water.
   - \( X \sim \text{Binomial}(n=100, p=0.4) \).

2. **Using a Normal Approximation**:
   Since \( n \) is large, we can use the **Normal approximation** to the binomial distribution:
   - The mean \( \mu = np = 100 \times 0.4 = 40 \).
   - The variance \( \sigma^2 = np(1-p) = 100 \times 0.4 \times 0.6 = 24 \).
   - The standard deviation \( \sigma = \sqrt{24} \approx 4.9 \).

3. **Setting Up the Probability**:
   We want \( P(X > 35) \). Using the continuity correction, this is approximately:
   \[
   P(X > 35) \approx P\left(Z > \frac{35.5 - 40}{4.9}\right)
   \]
   where \( Z \) is a standard normal variable.

4. **Calculating**:
   - Compute the Z-score:
     \[
     Z = \frac{35.5 - 40}{4.9} \approx -0.92
     \]
   - Using the standard normal table, \( P(Z > -0.92) \approx 0.8212 \).

   **Answer for (i):** The probability that more than 35 customers ask for water is approximately \( 0.8212 \).

---

### Part (ii)
We need to find the smallest value of \( n \) such that the probability of **less than \( n \)** customers asking for water is **larger than 0.8**.

1. **Setting Up the Inequality**:
   - Let \( P(X < n) > 0.8 \).
   - Using the normal approximation, we want to find \( n \) such that:
     \[
     P\left(Z < \frac{n + 0.5 - 40}{4.9}\right) > 0.8
     \]
   - From the standard normal table, we know that \( P(Z < 0.84) \approx 0.8 \).

2. **Solving for \( n \)**:
   - Set up the equation for the Z-score:
     \[
     \frac{n + 0.5 - 40}{4.9} = 0.84
     \]
   - Solving for \( n \):
     \[
     n + 0.5 = 40 + 0.84 \times 4.9
     \]
     \[
     n + 0.5 \approx 44.116
     \]
     \[
     n \approx 43.6
     \]

   Since \( n \) must be an integer, we round up to \( n = 44 \).

   **Answer for (ii):** The smallest value of \( n \) such that the probability of less than \( n \) customers asking for water is greater than 0.8 is \( n = 44 \).

---

Would you like a deeper explanation of any part? 

### Further Related Questions:
1. How would the probability change if the probability of asking for water increased to 0.5?
2. Why is the continuity correction necessary when approximating a binomial distribution with a normal distribution?
3. What is the probability that exactly 40 customers ask for water?
4. How would we solve this without using the normal approximation?
5. What would happen to the probability if the sample size changed to 200 customers?

**Tip:** When dealing with binomial distributions, normal approximations are useful when \( n \) is large and \( p \) is not too close to 0 or 1, which helps simplify calculations.

In a large restaurant, an average of 4 out of every 10 customers ask for water with their meal. A random sample of 100 customers is selected.

(i) Find the probability that more than 35 customers ask for water. State your assumptions and/or approximations.

(ii) Find the smallest value of n such that the probability of less than n customers ask for water is larger than 0.8.

This problem explores the probability that more than 35 out of 100 customers in a restaurant ask for water, given that 4 in 10 usually do. It involves binomial distribution and normal approximation, applying continuity correction to find probabilities and determine sample sizes for specific probability thresholds.

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