Answer:
The percentage is 15%
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%