Question

cises1. A travel agency did a survey and found that the average local family spends $1,900 on a summervacation. The distribution is normally distributed with standard deviation $390.a. What percent of the families took vacations that cost under $1,500? Round to the nearestpercent.b. What percent of the families took vacations that cost over $2.800? Round to the nearestpercent.c. Find the amount a family would have spent to be the 60th percentile. Round to the nearestdollar.2. A distribution is normal with mean 6

Answer 1

To solve this problem we first need to calculate the z-score of the sample:

$z\text{ = }\frac{x-\mu}{\sigma}$

Where x is the value we want to calculate the percent, the greek leter mu is the mean and the letter greek alpha is the standard deviation. For the first problem we have:

a.

$\begin{gathered} z\text{ = }\frac{1500-1900}{390} \\ z=\frac{-400}{390} \\ z\text{ = 1.}02 \end{gathered}$

We need to now check the z-table to find the percentage. In this case the percentage is:

$\alpha\text{ = 0.1539}$

Which is the same as 15.39%.

b.

$\begin{gathered} z\text{ = }\frac{2800-1500}{390} \\ z\text{ = }\frac{1300}{390} \\ z\text{ = }3.3334 \end{gathered}$

We need to check the z-table to find the percentage. In this case the percentage is:

$\alpha\text{ = 0.0004}$

Which is the same as 0.04%

c.

Now the problem gaves us the percentile and we need to use:

$\alpha\text{ = 0.6}$

Checking at the z-table we have a z equal to z=0.255. We can now find the amount they would need to spend.

$\begin{gathered} 0.255\text{ = }\frac{X\text{ - 1900}}{390} \\ 99.45\text{ = }X\text{ - 1900} \\ X\text{ = 99.45+1900} \\ X\text{ = }1999.45 \end{gathered}$

They would need to spend 1999.45