Question

the amount of time that people spend at grover hot springs is normally distributed with a mean of 73 minutes and a standard deviation of 18 minutes. suppose one person at the hot springs is randomly chosen. let x = the amount of time that person spent at grover hot springs. a. what is the distribution of x? x - n( ) b. find the probability that a randomly selected person at the hot springs stays longer than 97 minutes. (round answer to four decimal places) c. the park service is considering offering a discount for the 5% of their patrons who spend the least time at the hot springs. what is the longest amount of time a patron can spend at the hot springs and still receive the discount? minutes. (round answer to one decimal place) d. find the inter quartile range (iqr) for time spent at the hot springs. (round answers to one decimal place) q1: minutes q3: minutes iqr: minutes

Explanation:

Step1: Identify normal - distribution parameters

Given mean $\mu = 73$ and standard deviation $\sigma=18$, so $X\sim N(73,18)$.

Step2: Standardize for part b

We want $P(X > 97)$. First, calculate the z - score $z=\frac{x-\mu}{\sigma}=\frac{97 - 73}{18}=\frac{24}{18}\approx1.33$. Then $P(X > 97)=P(Z>1.33)=1 - P(Z\leq1.33)$. From the standard - normal table, $P(Z\leq1.33) = 0.9082$, so $P(X > 97)=1 - 0.9082 = 0.0918$.

Step3: Find the z - score for part c

We want to find $x$ such that $P(X\leq x)=0.05$. Looking up in the standard - normal table, the z - score corresponding to a left - tail probability of $0.05$ is approximately $z=-1.645$. Then, using the z - score formula $z=\frac{x-\mu}{\sigma}$, we have $-1.645=\frac{x - 73}{18}$. Solving for $x$ gives $x=73-1.645\times18=73 - 29.61\approx43.4$.

Step4: Find z - scores for quartiles in part d

For $Q1$, the left - tail probability is $0.25$. The z - score corresponding to $P(Z\leq z)=0.25$ is approximately $z=-0.674$. Using $z=\frac{x-\mu}{\sigma}$, we get $-0.674=\frac{x_{Q1}-73}{18}$, and $x_{Q1}=73-0.674\times18=73 - 12.132\approx60.9$.
For $Q3$, the left - tail probability is $0.75$. The z - score corresponding to $P(Z\leq z)=0.75$ is approximately $z = 0.674$. Using $z=\frac{x-\mu}{\sigma}$, we get $0.674=\frac{x_{Q3}-73}{18}$, and $x_{Q3}=73 + 0.674\times18=73+12.132\approx85.1$.
The inter - quartile range $IQR = Q3 - Q1=85.1 - 60.9 = 24.2$.

Answer:

a. $73,18$
b. $0.0918$
c. $43.4$
d. $Q1:60.9$
$Q3:85.1$
$IQR:24.2$