Question

1. 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, according to a recent survey by a national travel association.

a. find, to the nearest hundredth, the z - score for $150 for vacation food and lodging expenses.
b. if a vacationer had a z - score of 2.1, what were their daily expenses for food and lodging?
c. if the data is normally distributed, find the percent of these vacationers who spent less than $307 per day.
d. what is the variance?
e. what is the mean expense for food and lodging for a 7 - day vacation?

1. the vacation times website rates recreational vehicle campgrounds using integers from 0 to 15. last year they rated over 1,000 campsites. the ratings were normally distributed with mean 7.6 and standard deviation 1.7.

a. how high would a campsites rating have to be for it to be considered in the top 10% of rated campsites? round to the nearest hundredth.
b. find the z - score for a rating of 5. round to the nearest hundredth.
c. find the percentile for a rating of 7.5. round to the nearest percent.
d. a campsite had a z - score of 2. what was its rating?

1. a certain amusement park ride requires riders to be at least 48 inches tall. if the heights of children in a summer camp are normally distributed with mean 52 and standard deviation 2.5, how many of the 140 campers will be allowed on the ride? round to the nearest integer.
2. what z - score on the normal curve table has an area of 0.8849 to its left?
3. what z - score on the normal curve table has an area of 0.6808 to its right?

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 247$ and $\sigma=60$.
For part a, to find the z - score when $x = 150$:
$z=\frac{150 - 247}{60}=\frac{- 97}{60}\approx - 1.62$

Step2: Rearrange z - score formula for part b

The z - score formula $z=\frac{x-\mu}{\sigma}$ can be rearranged to $x=\mu+z\sigma$. Given $z = 2.1$, $\mu = 247$, and $\sigma = 60$.
$x=247+2.1\times60=247 + 126=373$

Step3: Find z - score and use z - table for part c

First, find the z - score when $x = 307$. $z=\frac{307 - 247}{60}=\frac{60}{60}=1$.
Looking up the value of $z = 1$ in the standard normal distribution table, the area to the left of $z = 1$ is $0.8413$. So the percent of vacationers who spent less than $307$ per day is $84.13\%$

Step4: Recall variance formula for part d

The variance $\sigma^{2}$ is related to the standard deviation $\sigma$ by the formula $\sigma^{2}=\sigma\times\sigma$. Given $\sigma = 60$, then $\sigma^{2}=60^{2}=3600$

Step5: Calculate 7 - day mean for part e

The mean daily expense is $\mu = 247$. For a 7 - day vacation, the mean expense is $7\times\mu=7\times247 = 1729$

Answer:

a. $-1.62$
b. $373$
c. $84.13\%$
d. $3600$
e. $1729$