K ** de 065 galone per month Consider a random sample of people
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![### Analysis of Bottled Water Consumption
#### Problem Statement
According to a particular marketing consultant, the per capita consumption of bottled water is 3.3 gallons per month. Assume the standard deviation for this population is 0.4 gallons per month. Consider a random sample of 81 people.
1. **What is the probability that the sample mean will be less than 3.4 gallons per month?**
2. **What is the probability that the sample mean will be more than 3.2 gallons per month?**
3. **Identify the symmetrical interval that includes 78% of the sample means if the population mean is 3.3 gallons per month.**
#### Solution
1. **The probability that the sample mean will be less than 3.4 gallons per month is:**
\[ P(\bar{x} < 3.4) = 0.8944 \]
*Type your answer as a decimal rounded to four decimal places as needed.*
2. **The probability that the sample mean will be more than 3.2 gallons per month is:**
\[ P(\bar{x} > 3.2) = 0.8944 \]
*Type an integer or decimal rounded to four decimal places as needed.*
#### Explanation
For questions 1 and 2, the solutions involve the use of the Z-score formula which is:
\[ Z = \frac{(\bar{x} - \mu)}{\left(\frac{\sigma}{\sqrt{n}}\right)} \]
Where:
- \( \bar{x} \) is the sample mean,
- \( \mu \) is the population mean,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
Using the Z-score tables or a statistical software, the probabilities are determined based on the computed Z-scores.
#### Graphical Representation (if applicable)
There is no graph or image in the document. However, if a graph were to be included, it might represent the normal distribution of the population mean, with highlighted areas under the curve corresponding to the probabilities calculated in questions 1 and 2. The symmetrical interval corresponding to 78% of the sample means around the population mean of 3.3 gallons would also be illustrated.
*For detailed solutions or additional questions related to this problem, please refer to the statistical methods for probability distributions, particularly the Normal](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7c47eb58-11c8-4f3d-a446-35857f47e26f%2Fb9b4f121-9807-419c-b39e-9d4dcd9baea9%2F7dde2vk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Analysis of Bottled Water Consumption
#### Problem Statement
According to a particular marketing consultant, the per capita consumption of bottled water is 3.3 gallons per month. Assume the standard deviation for this population is 0.4 gallons per month. Consider a random sample of 81 people.
1. **What is the probability that the sample mean will be less than 3.4 gallons per month?**
2. **What is the probability that the sample mean will be more than 3.2 gallons per month?**
3. **Identify the symmetrical interval that includes 78% of the sample means if the population mean is 3.3 gallons per month.**
#### Solution
1. **The probability that the sample mean will be less than 3.4 gallons per month is:**
\[ P(\bar{x} < 3.4) = 0.8944 \]
*Type your answer as a decimal rounded to four decimal places as needed.*
2. **The probability that the sample mean will be more than 3.2 gallons per month is:**
\[ P(\bar{x} > 3.2) = 0.8944 \]
*Type an integer or decimal rounded to four decimal places as needed.*
#### Explanation
For questions 1 and 2, the solutions involve the use of the Z-score formula which is:
\[ Z = \frac{(\bar{x} - \mu)}{\left(\frac{\sigma}{\sqrt{n}}\right)} \]
Where:
- \( \bar{x} \) is the sample mean,
- \( \mu \) is the population mean,
- \( \sigma \) is the population standard deviation,
- \( n \) is the sample size.
Using the Z-score tables or a statistical software, the probabilities are determined based on the computed Z-scores.
#### Graphical Representation (if applicable)
There is no graph or image in the document. However, if a graph were to be included, it might represent the normal distribution of the population mean, with highlighted areas under the curve corresponding to the probabilities calculated in questions 1 and 2. The symmetrical interval corresponding to 78% of the sample means around the population mean of 3.3 gallons would also be illustrated.
*For detailed solutions or additional questions related to this problem, please refer to the statistical methods for probability distributions, particularly the Normal
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