Question

A recent survey by the American Automobile Association showed that a family of two adults and two children on vacation in the United States will pay an average of $247 per day for food and lodging with a standard deviation of $60 per day. Assuming the data are normally distributed, find, to the nearest hundredth, the z-scores for each of the following vacation expense amounts. $197 per day. $277 per day. $310 per day.

Answer 1

Answer: We are given:

$\mu=247,\sigma=60$

We need to find the z scores for the following vacation expense amounts:

$197, $277, $310

We know that z score formula is:

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

When $x = 197$, the z score is:

$z=\frac{197-247}{60}$

        $=\frac{-50}{60}$

        $=-0.83$

When $x = 277$, the z score is:

$z=\frac{277-247}{60}$

        $=\frac{30}{60}$

        $=0.5$

When $x = 310$, the z score is:

$z=\frac{310-247}{60}$

        $=\frac{63}{60}$

        $=1.05$

Therefore, the z scores for the vacation expense amounts $197 per day, $277 per day, and $310 per day are -0.83, 0.5 and 1.05 respectively

Answer 2

Answer:

We are given:

\mu=247,\sigma=60μ=247,σ=60

We need to find the z scores for the following vacation expense amounts:

197,197,277, $310

We know that z score formula is:

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

σ

x−μ

When x = 197x=197 , the z score is:

z=\frac{197-247}{60}z=

60

197−247

=\frac{-50}{60}=

60

−50

=-0.83=−0.83

When x = 277x=277 , the z score is:

z=\frac{277-247}{60}z=

60

277−247

=\frac{30}{60}=

60

30

=0.5=0.5

When x = 310x=310 , the z score is:

z=\frac{310-247}{60}z=

60

310−247

=\frac{63}{60}=

60

63

=1.05=1.05

Therefore, the z scores for the vacation expense amounts 197 per day,197perday,277 per day, and $310 per day are -0.83, 0.5 and 1.05 respective