Question

50 points

A finance magazine did a survey and found that the average American family spends $1,600 on a summer vacation. If the distribution is normal with standard deviation $411, find the amount a family would have spent to be the 80th percentile.

Answer 1

$1,945.03

Explanation:

To find the amount a family would have spent to be at the 80th percentile, we need to find the z-score that corresponds to the 80th percentile and then use that z-score to calculate the corresponding amount.

First, we find the z-score using the standard normal distribution table or a calculator:

z = InvNorm(0.8) ≈ 0.8416

Next, we use the formula for transforming a z-score to a raw score:

z = (x - μ) / σ

where x is the raw score, μ is the mean, and σ is the standard deviation.

Substituting the given values and solving for x, we get:

0.8416 = (x - 1600) / 411

x - 1600 = 0.8416 * 411

x - 1600 = 345.0256

x = 1945.0256

Therefore, a family would have spent about $1,945.03 to be at the 80th percentile.

Answer 2

$1,945.91

Explanation:

To find the amount a family would have spent to be at the 80th percentile, we can use the normal distribution.

If a continuous random variable X is normally distributed with mean μ and variance σ², it is written as:

`\boxed{X \sim\text{N}(\mu,\sigma^2)}`

Given that the average spending on a summer vacation is $1,600 and the standard deviation is $411:

• Mean μ = $1,600
• Standard deviation σ = $411

We need to convert the problem to the standard normal distribution by calculating the z-score.

The z-score represents the number of standard deviations a data point is away from the mean. To find the z-score corresponding to the 80th percentile, we need to find the critical value or z-score that corresponds to an area of 0.80 under the standard normal curve.

Using a standard normal distribution table or a calculator, the z-score corresponding to an area of 0.80 is 0.84162084...

Next, we use the formula for the z-score:

`\boxed{z = (x - \mu)/(\sigma)}`

where z is the z-score, x is the data point, μ is the mean, and σ is the standard deviation.

Substitute z = 0.84162084..., μ = 1600, σ = 411 into the z-score formula:

`0.84162084... = (x - 1600)/(411)`

Solving for x:

`x=1945.90616...`

Therefore, a family would have spent approximately $1,945.91 to be at the 80th percentile of spending on a summer vacation.