A travel agency did a survey and found that the average local family spends $1900 on a summer vacation. The distribution is normally distributed with standard deviation $390. A. What percent of the families took vacations that cost under $1500. Round to the nearest percent.

The percentage is 15%

Step-by-step explanation:

* Lets revise how to find the z-score

- The rule the z-score is z = (x - μ) / σ, where

# x is the score

# μ is the mean

# σ is the standard deviation

* Lets solve the problem

- The average local family spends $1900 on a summer vacation

- The distribution is normally distributed with standard deviation $390

- We need to find what percent of the families took vacations that

cost under $1500

∵ The mean is $ 1900

∴ μ = 1900

∵ The standard deviation is $390

∴ σ = 390

∵ The vacation cost is under $1500

∴ x = 1500

∵ z-score = (x - μ) / σ

- Substitute the values above in the rule

∴ z-score = (1500 - 1900) / 390 = - 1.0256

- Lets use the normal distribution table of z

∵ P (z < 1500) = 0.1515

∵ P (x < 1500) = P (z < 1500)

∴ P (x < 1500) = 0.1515

∴ The percentage of the families took vacations that cost under

$1500 is ⇒ 0.15 * 100% = 15%

∴ The percentage is 15%