3. a family of two adults and two children on vacation in the united states will pay an average of $247 per…

3. 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.\n a. find, to the nearest hundredth, the z - score for $150 for vacation food and lodging expenses.\n b. if a vacationer had a z - score of 2.1, what were their daily expenses for food and lodging?\n c. if the data is normally distributed, find the percent of these vacationers who spent less than $307 per day.\n d. what is the variance?\n e. what is the mean expense for food and lodging for a 7 - day vacation?\n4. 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.\n 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.\n b. find the z - score for a rating of 5. round to the nearest hundredth.\n c. find the percentile for a rating of 7.5. round to the nearest percent.\n d. a campsite had a z - score of 2. what was its rating?\n5. 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.\n6. what z - score on the normal curve table has an area of 0.8849 to its left?\n7. what z - score on the normal curve table has an area of 0.6808 to its right?

3. 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.\n a. find, to the nearest hundredth, the z - score for $150 for vacation food and lodging expenses.\n b. if a vacationer had a z - score of 2.1, what were their daily expenses for food and lodging?\n c. if the data is normally distributed, find the percent of these vacationers who spent less than $307 per day.\n d. what is the variance?\n e. what is the mean expense for food and lodging for a 7 - day vacation?\n4. 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.\n 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.\n b. find the z - score for a rating of 5. round to the nearest hundredth.\n c. find the percentile for a rating of 7.5. round to the nearest percent.\n d. a campsite had a z - score of 2. what was its rating?\n5. 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.\n6. what z - score on the normal curve table has an area of 0.8849 to its left?\n7. what z - score on the normal curve table has an area of 0.6808 to its right?

Answer

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. $1729$