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

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

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

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 $\mu = 247$, $\sigma=60$.

Step2: Calculate z - score for part a

For $x = 150$, $z=\frac{150 - 247}{60}=\frac{- 97}{60}\approx - 1.62$

Step3: Rearrange z - score formula for part b

From $z=\frac{x-\mu}{\sigma}$, we can solve for $x$: $x=\mu+z\sigma$. Substituting $\mu = 247$, $z = 2.1$, and $\sigma = 60$, we get $x=247+2.1\times60=247 + 126=373$

Step4: Calculate z - score for part c

For $x = 307$, $z=\frac{307 - 247}{60}=\frac{60}{60}=1$. Looking up the value in the standard - normal distribution table, the area to the left of $z = 1$ is approximately $0.8413$, so the percentage is $84.13%$

Step5: Calculate variance for part d

The variance $\sigma^{2}$ is the square of the standard deviation. Since $\sigma = 60$, $\sigma^{2}=60^{2}=3600$

Step6: Calculate 7 - day mean for part e

The mean per day 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