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: $7.78\%$

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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.

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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).

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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.

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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.