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 break down the problem step-by-step to answer each part clearly.

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

The average cost (mean) for the 12-day vacation is given as:
\[
\mu = 1755
\]
The standard deviation is:
\[
\sigma = 390
\]
We need to find the percentage of vacationers who spent **less than $1,200**. This is equivalent to finding the **cumulative probability** for $1,200$, i.e., $ P(X < 1200) $.

1. **First, calculate the z-score** for $1,200$:
\[
z = \frac{X - \mu}{\sigma} = \frac{1200 - 1755}{390} = \frac{-555}{390} \approx -1.42
\]

2. **Now, look up the cumulative probability for a z-score of -1.42**. Using the standard normal distribution table (or a calculator), we find:
\[
P(Z < -1.42) \approx 0.0778
\]
So, about 7.78% of vacationers spent less than $1,200. Rounded to the nearest hundredth:
\[
7.78\%
\]

---

### Part (b): Per-Day Expense for a Traveler with a Z-score of -2.1

The z-score formula is:
\[
z = \frac{X - \mu}{\sigma}
\]
We are given that the z-score is $ z = -2.1 $, and we want to find the corresponding total expense $ X $ for the 12-day vacation.

1. **Rearrange the formula** to solve for $ X $:
\[
X = z \cdot \sigma + \mu = (-2.1) \cdot 390 + 1755 = -819 + 1755 = 936
\]
So, the traveler spent $936 in total.

2. **Now, calculate the per-day expense** for this traveler:
\[
\text{Per-day expense} = \frac{936}{12} = 78
\]
Thus, the per-day expense is **$78**.

---

### Part (c): Effect of a Higher Standard Deviation

If the standard deviation changes from $390 to $490, we want to determine how this affects the answer to Part (b) (where the traveler had a z-score of -2.1).

1. Using the new standard deviation $ \sigma = 490 $, the total expense $ X $ can be recalculated using the formula:
\[
X = z \cdot \sigma + \mu = (-2.1) \cdot 490 + 1755 = -1029 + 1755 = 726
\]
So, the new total expense is $726.

2. **Calculate the per-day expense**:
\[
\text{Per-day expense} = \frac{726}{12} = 60.5
\]
With a higher standard deviation, the per-day expense drops to **$60.50**, which is **lower** than the original value calculated in Part (b).

---

### Summary of Answers:
- **(a)** 7.78% of vacationers spent less than $1,200.
- **(b)** The per-day expense for the traveler with a z-score of -2.1 was $78.
- **(c)** With a higher standard deviation, the per-day expense becomes **lower**, dropping to $60.50.

---

Would you like more details or have any questions?

Here are five related questions to expand your understanding:
1. How would the calculation in part (a) change if the vacation cost $2,000 instead of $1,200?
2. What is the z-score for a traveler who spent $2,100 on their vacation?
3. How do changes in the mean affect the z-score and overall distribution of costs?
4. How would you interpret a z-score of 0 in the context of this vacation expense?
5. What percentage of vacationers spent more than $2,100 if the standard deviation is $390?

**Tip:** Z-scores help compare individual values to the mean of a distribution by measuring how many standard deviations away the value is from the mean.

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?

Explore how to solve a statistics problem involving vacation costs using normal distribution and z-scores. Learn how to find the percentage of vacationers spending less than $1,200 and calculate per-day expenses for a traveler with a z-score of -2.1. Ideal for college-level statistics learners.

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