Answer :
Sure, let's go through each part of the question step-by-step:
### a. What is the distribution of \( X \)?
The amount of time that people spend at Grover Hot Springs is normally distributed with a mean (\(\mu\)) of 77 minutes and a standard deviation (\(\sigma\)) of 16 minutes. Thus, the distribution of \( X \) can be expressed as:
[tex]\[ X \sim \mathcal{N}(77, 16^2) \][/tex]
### b. Find the probability that a randomly selected person at the hot springs stays longer than 79 minutes.
We need to find \( P(X > 79) \).
To begin, we standardize the variable \( X \) using the Z-score formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
For \( X = 79 \):
[tex]\[ Z = \frac{79 - 77}{16} = \frac{2}{16} = 0.125 \][/tex]
Using Z-tables or a standard normal distribution calculator, we find the probability that \( Z \) is less than 0.125. This value is approximately 0.5498.
Thus, \( P(Z > 0.125) \) is:
[tex]\[ 1 - 0.5498 = 0.4502 \][/tex]
So the probability that a randomly selected person stays longer than 79 minutes is approximately:
[tex]\[ 0.4502 \][/tex]
### c. The park service is considering offering a discount for the 7% 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?
Let \( k \) be this longest amount of time. We want to find \( k \) such that:
[tex]\[ P(X < k) = 0.07 \][/tex]
We need to find the Z-score corresponding to the bottom 7% of the standard normal distribution. Using Z-tables or a standard normal distribution calculator, we get:
[tex]\[ Z_{\text{critical}} \approx -1.4758 \][/tex]
Now, we convert this Z-score back to the original scale:
[tex]\[ k = \mu + Z_{\text{critical}} \cdot \sigma \][/tex]
[tex]\[ k = 77 + (-1.4758) \cdot 16 \][/tex]
[tex]\[ k \approx 77 - 23.6128 \][/tex]
[tex]\[ k \approx 53.3872 \][/tex]
So, the longest amount of time a patron can spend at the hot springs and still receive the discount is approximately:
[tex]\[ 53.3872 \, \text{minutes} \][/tex]
### d. Find the Inter Quartile Range (IQR) for time spent at the hot springs.
To find the IQR, we need to determine the 1st quartile (Q1) and the 3rd quartile (Q3).
The 1st quartile corresponds to the 25th percentile:
[tex]\[ P(X \leq Q1) = 0.25 \][/tex]
The Z-score corresponding to the 25th percentile is approximately:
[tex]\[ Z_{0.25} \approx -0.6745 \][/tex]
Convert this Z-score back to the original scale:
[tex]\[ Q1 = \mu + Z_{0.25} \cdot \sigma \][/tex]
[tex]\[ Q1 = 77 + (-0.6745) \cdot 16 \][/tex]
[tex]\[ Q1 \approx 77 - 10.792 \][/tex]
[tex]\[ Q1 \approx 66.208 \][/tex]
The 3rd quartile corresponds to the 75th percentile:
[tex]\[ P(X \leq Q3) = 0.75 \][/tex]
The Z-score corresponding to the 75th percentile is approximately:
[tex]\[ Z_{0.75} \approx 0.6745 \][/tex]
Convert this Z-score back to the original scale:
[tex]\[ Q3 = \mu + Z_{0.75} \cdot \sigma \][/tex]
[tex]\[ Q3 = 77 + 0.6745 \cdot 16 \][/tex]
[tex]\[ Q3 \approx 77 + 10.792 \][/tex]
[tex]\[ Q3 \approx 87.792 \][/tex]
Inter Quartile Range (IQR) is:
[tex]\[ IQR = Q3 - Q1 \][/tex]
[tex]\[ IQR = 87.792 - 66.208 \][/tex]
[tex]\[ IQR \approx 21.584 \][/tex]
So, summarizing:
- Q1: \( 66.208 \) minutes
- Q3: \( 87.792 \) minutes
- IQR: [tex]\( 21.584 \)[/tex] minutes
### a. What is the distribution of \( X \)?
The amount of time that people spend at Grover Hot Springs is normally distributed with a mean (\(\mu\)) of 77 minutes and a standard deviation (\(\sigma\)) of 16 minutes. Thus, the distribution of \( X \) can be expressed as:
[tex]\[ X \sim \mathcal{N}(77, 16^2) \][/tex]
### b. Find the probability that a randomly selected person at the hot springs stays longer than 79 minutes.
We need to find \( P(X > 79) \).
To begin, we standardize the variable \( X \) using the Z-score formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
For \( X = 79 \):
[tex]\[ Z = \frac{79 - 77}{16} = \frac{2}{16} = 0.125 \][/tex]
Using Z-tables or a standard normal distribution calculator, we find the probability that \( Z \) is less than 0.125. This value is approximately 0.5498.
Thus, \( P(Z > 0.125) \) is:
[tex]\[ 1 - 0.5498 = 0.4502 \][/tex]
So the probability that a randomly selected person stays longer than 79 minutes is approximately:
[tex]\[ 0.4502 \][/tex]
### c. The park service is considering offering a discount for the 7% 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?
Let \( k \) be this longest amount of time. We want to find \( k \) such that:
[tex]\[ P(X < k) = 0.07 \][/tex]
We need to find the Z-score corresponding to the bottom 7% of the standard normal distribution. Using Z-tables or a standard normal distribution calculator, we get:
[tex]\[ Z_{\text{critical}} \approx -1.4758 \][/tex]
Now, we convert this Z-score back to the original scale:
[tex]\[ k = \mu + Z_{\text{critical}} \cdot \sigma \][/tex]
[tex]\[ k = 77 + (-1.4758) \cdot 16 \][/tex]
[tex]\[ k \approx 77 - 23.6128 \][/tex]
[tex]\[ k \approx 53.3872 \][/tex]
So, the longest amount of time a patron can spend at the hot springs and still receive the discount is approximately:
[tex]\[ 53.3872 \, \text{minutes} \][/tex]
### d. Find the Inter Quartile Range (IQR) for time spent at the hot springs.
To find the IQR, we need to determine the 1st quartile (Q1) and the 3rd quartile (Q3).
The 1st quartile corresponds to the 25th percentile:
[tex]\[ P(X \leq Q1) = 0.25 \][/tex]
The Z-score corresponding to the 25th percentile is approximately:
[tex]\[ Z_{0.25} \approx -0.6745 \][/tex]
Convert this Z-score back to the original scale:
[tex]\[ Q1 = \mu + Z_{0.25} \cdot \sigma \][/tex]
[tex]\[ Q1 = 77 + (-0.6745) \cdot 16 \][/tex]
[tex]\[ Q1 \approx 77 - 10.792 \][/tex]
[tex]\[ Q1 \approx 66.208 \][/tex]
The 3rd quartile corresponds to the 75th percentile:
[tex]\[ P(X \leq Q3) = 0.75 \][/tex]
The Z-score corresponding to the 75th percentile is approximately:
[tex]\[ Z_{0.75} \approx 0.6745 \][/tex]
Convert this Z-score back to the original scale:
[tex]\[ Q3 = \mu + Z_{0.75} \cdot \sigma \][/tex]
[tex]\[ Q3 = 77 + 0.6745 \cdot 16 \][/tex]
[tex]\[ Q3 \approx 77 + 10.792 \][/tex]
[tex]\[ Q3 \approx 87.792 \][/tex]
Inter Quartile Range (IQR) is:
[tex]\[ IQR = Q3 - Q1 \][/tex]
[tex]\[ IQR = 87.792 - 66.208 \][/tex]
[tex]\[ IQR \approx 21.584 \][/tex]
So, summarizing:
- Q1: \( 66.208 \) minutes
- Q3: \( 87.792 \) minutes
- IQR: [tex]\( 21.584 \)[/tex] minutes
