Answer :
Alright, let's work through filling in the missing frequencies and then interpreting the table appropriately with a step-by-step solution.
Let's consider the given problem: We have data for the amount of gasoline in gallons and their corresponding frequencies. This data is divided into several intervals, and we need to understand key metrics such as probability within specified bounds as well as computing z-scores and understanding their implications.
We are given:
- A sample size of 85.
- A population mean (µ) of 22 gallons.
- A population standard deviation (σ) of 13 gallons.
- We want to find the probability that the amount of gasoline falls between 19 to 23 gallons.
First, let's define the intervals and fill in the missing frequencies. However, we don't have the specific frequencies here, so we'll leave them generic and focus on the main question at hand.
### Step-by-Step Solution:
1. Define the Problem:
We need to find the probability that the amount of gasoline falls between 19 and 23 gallons.
2. Convert to Z-scores:
The z-score for a value [tex]\(X\)[/tex] in a normal distribution is given by:
[tex]\[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} \][/tex]
where [tex]\(\mu\)[/tex] is the mean, [tex]\(\sigma\)[/tex] is the standard deviation, and [tex]\(n\)[/tex] is the sample size.
3. Calculate the z-scores for 19 and 23 gallons:
For the lower bound (19 gallons):
[tex]\[ Z_{lower} = \frac{19 - 22}{13 / \sqrt{85}} \approx -2.1276 \][/tex]
For the upper bound (23 gallons):
[tex]\[ Z_{upper} = \frac{23 - 22}{13 / \sqrt{85}} \approx 0.7092 \][/tex]
4. Find the Probability using the Cumulative Distribution Function (CDF):
We find the cumulative probability up to each z-score using the CDF of the standard normal distribution:
- The CDF value at [tex]\(Z_{lower} \approx -2.1276\)[/tex] provides the probability that [tex]\(X\)[/tex] is less than 19 gallons.
- The CDF value at [tex]\(Z_{upper} \approx 0.7092\)[/tex] provides the probability that [tex]\(X\)[/tex] is less than 23 gallons.
5. Compute the Probability:
The probability that the amount of gasoline falls between 19 and 23 gallons is the difference between these cumulative probabilities:
[tex]\[ P(19 < X < 23) = \text{CDF}(Z_{upper}) - \text{CDF}(Z_{lower}) \approx 0.7442 \][/tex]
### Summary:
- Z-scores: [tex]\(-2.1276\)[/tex] for 19 gallons, [tex]\(0.7092\)[/tex] for 23 gallons.
- Probability: Approximately [tex]\(0.7442\)[/tex] or 74.42%.
In essence, there's a 74.42% chance that the amount of gasoline falls between 19 and 23 gallons given the mean and standard deviation provided.
Let's consider the given problem: We have data for the amount of gasoline in gallons and their corresponding frequencies. This data is divided into several intervals, and we need to understand key metrics such as probability within specified bounds as well as computing z-scores and understanding their implications.
We are given:
- A sample size of 85.
- A population mean (µ) of 22 gallons.
- A population standard deviation (σ) of 13 gallons.
- We want to find the probability that the amount of gasoline falls between 19 to 23 gallons.
First, let's define the intervals and fill in the missing frequencies. However, we don't have the specific frequencies here, so we'll leave them generic and focus on the main question at hand.
### Step-by-Step Solution:
1. Define the Problem:
We need to find the probability that the amount of gasoline falls between 19 and 23 gallons.
2. Convert to Z-scores:
The z-score for a value [tex]\(X\)[/tex] in a normal distribution is given by:
[tex]\[ Z = \frac{X - \mu}{\sigma / \sqrt{n}} \][/tex]
where [tex]\(\mu\)[/tex] is the mean, [tex]\(\sigma\)[/tex] is the standard deviation, and [tex]\(n\)[/tex] is the sample size.
3. Calculate the z-scores for 19 and 23 gallons:
For the lower bound (19 gallons):
[tex]\[ Z_{lower} = \frac{19 - 22}{13 / \sqrt{85}} \approx -2.1276 \][/tex]
For the upper bound (23 gallons):
[tex]\[ Z_{upper} = \frac{23 - 22}{13 / \sqrt{85}} \approx 0.7092 \][/tex]
4. Find the Probability using the Cumulative Distribution Function (CDF):
We find the cumulative probability up to each z-score using the CDF of the standard normal distribution:
- The CDF value at [tex]\(Z_{lower} \approx -2.1276\)[/tex] provides the probability that [tex]\(X\)[/tex] is less than 19 gallons.
- The CDF value at [tex]\(Z_{upper} \approx 0.7092\)[/tex] provides the probability that [tex]\(X\)[/tex] is less than 23 gallons.
5. Compute the Probability:
The probability that the amount of gasoline falls between 19 and 23 gallons is the difference between these cumulative probabilities:
[tex]\[ P(19 < X < 23) = \text{CDF}(Z_{upper}) - \text{CDF}(Z_{lower}) \approx 0.7442 \][/tex]
### Summary:
- Z-scores: [tex]\(-2.1276\)[/tex] for 19 gallons, [tex]\(0.7092\)[/tex] for 23 gallons.
- Probability: Approximately [tex]\(0.7442\)[/tex] or 74.42%.
In essence, there's a 74.42% chance that the amount of gasoline falls between 19 and 23 gallons given the mean and standard deviation provided.
