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 down the problem step by step:

We are given:
- Mean daily cost for food and lodging: $\mu = 247$ dollars/day
- Standard deviation of the daily cost: $\sigma = 60$ dollars/day
- Duration of the vacation: 10 days

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

The mean for a 10-day vacation can be computed by multiplying the per-day mean by the number of days:

\[
\text{Mean for 10 days} = \mu_{\text{10-day}} = \mu \times 10 = 247 \times 10 = 2470 \, \text{dollars}
\]

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

The standard deviation for a multi-day period is not simply the standard deviation multiplied by the number of days. Instead, it increases by a factor of the square root of the number of days:

\[
\text{Standard deviation for 10 days} = \sigma_{\text{10-day}} = \sigma \times \sqrt{10} = 60 \times \sqrt{10} \approx 60 \times 3.162 = 189.72 \, \text{dollars}
\]

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

Variance is the square of the standard deviation. Therefore, to find the variance for the 10-day vacation, we square the standard deviation calculated in part b:

\[
\text{Variance for 10 days} = (\sigma_{\text{10-day}})^2 = (189.72)^2 \approx 35991.6 \, \text{dollars}^2
\]

### Summary:
- a. The mean for food and lodging for a 10-day vacation is **2470 dollars**.
- b. The standard deviation of the distribution for the 10-day vacation is **189.72 dollars**.
- c. The variance of the distribution for the 10-day vacation is **35991.6 dollars²**.

Would you like further details or clarifications?

Here are 5 related questions:
1. How would these calculations change if the vacation lasted 14 days?
2. How do these results help in understanding the range of vacation costs?
3. What is the probability that the cost for 10 days exceeds a certain threshold, given the mean and standard deviation?
4. How would the variance change if the daily standard deviation were higher, say $75/day?
5. What assumptions underlie the use of the standard deviation in this scenario?

**Tip:** When dealing with multiple days or events, standard deviations grow with the square root of the number of events, while means grow 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. 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 of food and lodging costs for a 10-day family vacation in the U.S. Based on a daily average of $247 with a $60 standard deviation, this problem uses key statistical concepts and formulas. Suitable for Grades 9-11.

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