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.