     A recent survey by the American Automobile Association showed that a family of two adults and two children on vacation in the United States will pay an average of $247 per day for food and lodging with a standard deviation of $60 per day. You can compute the mean food and lodging cost of a 10-day vacation by multiplying the per-day mean by 10. The standard deviation for the 10-day vacation can be found by multiplying by 10 also.
a. What is the mean for food and lodging for a 10-day vacation?                                                                                                                                                                
                   b. What is the standard deviation of the distribution for the 10-day vacation? 
 
c. What is the variance of the distribution for the 10-day vacation?

Let's break this down step by step.

### Given:
- **Mean daily cost** for food and lodging = $247
- **Standard deviation daily cost** = $60
- **Duration** = 10 days

---

### Part (a): Mean for food and lodging for a 10-day vacation

To calculate the mean for a 10-day vacation, we multiply the per-day mean by 10:

\[
\text{Mean for 10 days} = 247 \times 10 = 2470
\]

So, the mean food and lodging cost for the 10-day vacation is **$2470**.

---

### Part (b): Standard deviation of the distribution for the 10-day vacation

The standard deviation of the total cost for 10 days is **not** found by multiplying the daily standard deviation by 10. Instead, the standard deviation for the sum of 10 days is calculated by multiplying the daily standard deviation by the square root of 10:

\[
\text{Standard deviation for 10 days} = 60 \times \sqrt{10} \approx 60 \times 3.162 = 189.72
\]

So, the standard deviation for the 10-day vacation is approximately **$189.72**.

---

### Part (c): Variance of the distribution for the 10-day vacation

The variance is the square of the standard deviation. So, using the standard deviation from part (b):

\[
\text{Variance} = (189.72)^2 \approx 35992.74
\]

So, the variance for the 10-day vacation is approximately **$35,992.74**.

---

### Summary:
- (a) Mean for 10 days: **$2470**
- (b) Standard deviation for 10 days: **$189.72**
- (c) Variance for 10 days: **$35,992.74**

Would you like more details or explanations on any part of this? Here are some related questions for further practice:

1. What would be the mean for a 7-day vacation under the same conditions?
2. How would the variance change if the trip was 14 days instead of 10?
3. If the standard deviation was $50 per day, what would the new 10-day standard deviation be?
4. What would the variance be for a 7-day vacation?
5. How does increasing the number of days impact the mean and standard deviation?

**Tip**: Variance increases linearly with the number of days, but standard deviation grows by the square root of the number of days.

A recent survey by the American Automobile Association showed that a family of two adults and two children on vacation in the United States will pay an average of $247 per day for food and lodging with a standard deviation of $60 per day. You can compute the mean food and lodging cost of a 10-day vacation by multiplying the per-day mean by 10. The standard deviation for the 10-day vacation can be found by multiplying by 10 also.
a. What is the mean for food and lodging for a 10-day vacation?
b. What is the standard deviation of the distribution for the 10-day vacation?
c. What is the variance of the distribution for the 10-day vacation?

Learn how to calculate the mean, standard deviation, and variance for a 10-day vacation with an average daily cost of $247 and a standard deviation of $60. This example illustrates key statistical concepts and is suitable for students in Grades 9-12.

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