Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go"), so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that 52% of its customers order their food to go. If this proportion is correct, what is the probability that, in a

random sample of 5 customers at Anita's, exactly 3 order their food to go?

Round your response to at least three decimal places.

To solve this problem, we need to calculate the probability that exactly 3 out of 5 customers order their food to go. This scenario can be modeled using the binomial distribution.

### Binomial Distribution
The binomial distribution describes the number of successes in a fixed number of independent trials, where each trial has two possible outcomes (success or failure) and the probability of success is constant. The probability mass function (PMF) for the binomial distribution is given by:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}
\]

Where:
- \( n \) is the number of trials (in this case, 5 customers).
- \( k \) is the number of successes (in this case, 3 customers ordering their food to go).
- \( p \) is the probability of success on a single trial (in this case, 0.52).
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \).

### Step-by-Step Solution
1. **Identify the parameters:**
   \[
   n = 5, \quad k = 3, \quad p = 0.52
   \]

2. **Calculate the binomial coefficient:**
   \[
   \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{20}{2} = 10
   \]

3. **Calculate the probability:**
   \[
   P(X = 3) = 10 \cdot (0.52)^3 \cdot (1 - 0.52)^{5-3}
   \]
   \[
   P(X = 3) = 10 \cdot (0.52)^3 \cdot (0.48)^2
   \]

4. **Compute the values:**
   \[
   (0.52)^3 = 0.140608, \quad (0.48)^2 = 0.2304
   \]
   \[
   P(X = 3) = 10 \cdot 0.140608 \cdot 0.2304 \approx 10 \cdot 0.032396
   \]
   \[
   P(X = 3) \approx 0.32396
   \]

5. **Round the result:**
   \[
   P(X = 3) \approx 0.324
   \]

Thus, the probability that exactly 3 out of 5 customers order their food to go is approximately **0.324**.

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

### Related Questions:
1. What is the probability that all 5 customers order their food to go?
2. How does the probability change if the sample size increases to 10 customers?
3. What is the probability that at least 3 out of 5 customers order their food to go?
4. How would the probability change if the proportion of customers ordering to go was 60%?
5. Can the same approach be used if we were dealing with a continuous distribution?
6. What is the expected number of customers who order their food to go out of 5?
7. How does the binomial distribution differ from the normal distribution in this context?
8. If the sample size were very large, how would you approximate this probability?

### Tip:
When working with binomial probabilities, it can be helpful to use statistical software or a calculator with binomial functions to quickly compute these probabilities, especially for larger sample sizes.

Learn how to calculate the probability that exactly 3 out of 5 customers order their food to go at Anita's using the binomial distribution. Detailed step-by-step solution with calculations. Suitable for advanced high school students and those studying probability.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation