International travel is usually more expensive than domestic travel. A recent survey found that the average per-person cost of a 12-day international vacation is $1,755. This includes transportation, food, lodging, and entertainment.
a. If the data are normally distributed with standard deviation $390, find the percentage of vacationers who spent less than $1,200 per 12-day vacation. Round to the nearest hundredth of a percent. 

b. Find the per-day expense for one of these travelers who had a z-score of –2.1. 

c. If the data are normally distributed with standard deviation $490 instead of $390, would the answer to the above question become lower or higher? 


Let's tackle each part of the problem one by one.

### Part (a) - Finding the Percentage of Vacationers Who Spent Less Than $1,200

Given dataLet's tackle each part of the problem one by one.

### Part (a) - Finding the Percentage of Vacationers Who Spent Less Than $1,200

Given 
- Mean $\mu = 1755$ dollars
- Standard deviation $\sigma = 390$ dollars
- Cost of vacation $X = 1200$ dollars

We want to find the percentage of vacationers who spent less than $1,200.

First, we calculate the z-score using the formula:
\[
z = \frac{X - \mu}{\sigma}
\]
Substitute the values:
\[
z = \frac{1200 - 1755}{390} = \frac{-555}{390} \approx -1.4231
\]

Now, we use the z-score to find the corresponding percentage from the standard normal distribution table (or using a calculator):

For $z \approx -1.42$, the cumulative probability (area to the left of z) is approximately 0.0778.

So, the percentage of vacationers who spent less than $1,200 is:
\[
\text{Percentage} = 0.0778 \times 100 \approx 7.78\%
\]
Rounded to the nearest hundredth of a percent, the answer is **7.78%**.

### Part (b) - Finding the Per-Day Expense for a Traveler with a Z-Score of -2.1

Given:
- Mean $\mu = 1755$ dollars
- Standard deviation $\sigma = 390$ dollars
- Z-score $z = -2.1$

We can use the z-score formula to find the corresponding cost $X$:
\[
X = \mu + z \times \sigma
\]
Substitute the values:
\[
X = 1755 + (-2.1) \times 390 = 1755 - 819 = 936 \text{ dollars}
\]
This is the total cost for 12 days. To find the per-day expense, divide by 12:
\[
\text{Per-day expense} = \frac{936}{12} = 78 \text{ dollars}
\]

### Part (c) - Effect of Increasing the Standard Deviation on the Per-Day Expense Calculation

In part (b), we used a standard deviation of $390. If the standard deviation increases to $490, we use the same z-score formula:

\[
X = 1755 + (-2.1) \times 490 = 1755 - 1029 = 726 \text{ dollars}
\]

Then, the per-day expense would be:
\[
\text{Per-day expense} = \frac{726}{12} = 60.5 \text{ dollars}
\]

Since the standard deviation increased, the resulting total cost and therefore the per-day expense **decreases**.

### Summary:
a. **7.78%** of vacationers spent less than $1,200.
b. The per-day expense for a traveler with a z-score of -2.1 is **$78**.
c. If the standard deviation increased to $490, the per-day expense would be **lower**.

Would you like further details on any part? Here are some related questions:

1. What would be the per-day expense if the z-score was -1.5 instead?
2. How does a higher standard deviation affect the distribution curve?
3. What is the percentage of vacationers who spent more than $2,000?
4. How do you calculate the z-score if only the percentage is given?
5. What if the survey included different vacation lengths?

**Tip:** Understanding how z-scores relate to normal distributions can help you quickly assess probabilities and percentiles in many statistical scenarios.

International travel is usually more expensive than domestic travel. A recent survey found that the average per-person cost of a 12-day international vacation is $1,755. This includes transportation, food, lodging, and entertainment.

a. If the data are normally distributed with a standard deviation of $390, find the percentage of vacationers who spent less than $1,200 per 12-day vacation. Round to the nearest hundredth of a percent.

b. Find the per-day expense for one of these travelers who had a z-score of –2.1.

c. If the data are normally distributed with a standard deviation of $490 instead of $390, would the answer to the above question become lower or higher?

Explore how to calculate the percentage of vacationers who spent less than $1,200 on a 12-day international vacation and find the per-day expense for travelers with specific z-scores. This problem involves key statistical concepts such as normal distribution and z-score calculations, suitable for students in Grades 10-12.

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