Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P61, the 61-percentile. This is the temperature reading separating the bottom 61% from the top 39%. P61 = C Submit Question
Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P61, the 61-percentile. This is the temperature reading separating the bottom 61% from the top 39%. P61 = C Submit Question
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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![### Educational Module: Understanding Percentiles in Normal Distribution
**Problem Statement:**
Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find \( P_{61} \), the 61-percentile. This is the temperature reading separating the bottom 61% from the top 39%.
\[ P_{61} = \_\_\_\_\_\_^\circ C \]
**Instructions:**
1. To find \( P_{61} \), you need to determine the z-score that corresponds to the 61st percentile in the standard normal distribution.
2. Utilize z-tables or standard normal distribution calculators to find the z-score.
3. Convert the z-score to the actual temperature using the mean and standard deviation.
**Submit Answer:**
There is a submission box followed by the unit (°C) to input the calculated percentile value and a "Submit Question" button to record your answer.
**Explanation of Graphs and Diagrams:**
- No specific graphs or diagrams are present in this problem statement. In a typical normal distribution graph, the area to the left of the 61-percentile would be shaded to represent the 61% of the data.
The known parameters:
- Mean (μ) = 0°C
- Standard Deviation (σ) = 1.00°C
By solving, you can better understand the application and concept of percentiles in the normal distribution which is essential in statistical analysis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77379e33-cd78-4bf1-a88e-5489bb4fa9c3%2F2757a158-80be-4d08-adb4-a224a84df1ab%2F204eh9f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Educational Module: Understanding Percentiles in Normal Distribution
**Problem Statement:**
Assume that the readings at freezing on a bundle of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find \( P_{61} \), the 61-percentile. This is the temperature reading separating the bottom 61% from the top 39%.
\[ P_{61} = \_\_\_\_\_\_^\circ C \]
**Instructions:**
1. To find \( P_{61} \), you need to determine the z-score that corresponds to the 61st percentile in the standard normal distribution.
2. Utilize z-tables or standard normal distribution calculators to find the z-score.
3. Convert the z-score to the actual temperature using the mean and standard deviation.
**Submit Answer:**
There is a submission box followed by the unit (°C) to input the calculated percentile value and a "Submit Question" button to record your answer.
**Explanation of Graphs and Diagrams:**
- No specific graphs or diagrams are present in this problem statement. In a typical normal distribution graph, the area to the left of the 61-percentile would be shaded to represent the 61% of the data.
The known parameters:
- Mean (μ) = 0°C
- Standard Deviation (σ) = 1.00°C
By solving, you can better understand the application and concept of percentiles in the normal distribution which is essential in statistical analysis.
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