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.

The mean food and lodging cost for a 10-day vacation can be easily computed by multiplying the daily average cost by 10. The standard deviation for the 10-day vacation can also be calculated, but we need to be careful in interpreting how it scales.

Here’s a step-by-step explanation:

1. **Mean calculation**: 
   The mean daily cost is $247. Over a 10-day vacation, the mean total cost is:

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

   So, the mean cost for a 10-day vacation is **$2,470**.

2. **Standard deviation calculation**:
   The standard deviation of the daily cost is given as $60. However, to find the standard deviation for the 10-day vacation, we do **not** simply multiply the standard deviation by 10. Instead, we use the fact that the standard deviation of the sum of independent variables scales with the square root of the number of days.

   The formula for the standard deviation for a 10-day period is:

   \[
   \sigma_{\text{10 days}} = 60 \times \sqrt{10} 
   \]

   Let’s calculate that:

   \[
   \sigma_{\text{10 days}} = 60 \times 3.162 = 189.72
   \]

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

### Summary:
- Mean cost for 10 days: **$2,470**.
- Standard deviation for 10 days: **$189.72**.

Would you like more details or clarifications?

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Here are some related questions for further exploration:

1. What is the probability that the vacation costs more than $2,500?
2. How would the standard deviation change for a 15-day vacation?
3. Can you compute the 95% confidence interval for the total 10-day cost?
4. How does the standard deviation affect budgeting for a vacation?
5. If the daily cost distribution is normal, what is the chance that a single day’s cost exceeds $300?

**Tip**: Standard deviation grows with the square root of time (or number of days), not linearly.

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.

Learn how to calculate the total mean cost and the standard deviation for a 10-day vacation based on a daily average of $247 and a standard deviation of $60. This example demonstrates key statistical concepts, including the scaling of the standard deviation with the square root of time. Suitable for Grades 11-12 or introductory college-level statistics.

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