A distribution of values is normal with a mean of 187.9 and a standard deviation of 71.2. Find P80, which is the score separating the bottom 80% from the top 20%. P80 = Enter your answer as a number accurate to 4 decimal places.
A distribution of values is normal with a mean of 187.9 and a standard deviation of 71.2. Find P80, which is the score separating the bottom 80% from the top 20%. P80 = Enter your answer as a number accurate to 4 decimal places.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![### Understanding Percentiles in a Normal Distribution
In this exercise, we consider a normally distributed set of values characterized by a mean (average) of 187.9 and a standard deviation of 71.2.
To solve the question, we are asked to identify **P₈₀**, which is the value below which 80% of the observations fall (also known as the 80th percentile). This score will effectively separate the bottom 80% of the data from the top 20%.
Given:
- Mean (μ) = 187.9
- Standard Deviation (σ) = 71.2
To find **P₈₀**, follow these steps:
1. **Determine the Z-score for the 80th percentile**: The Z-score indicates how many standard deviations a value is from the mean. For the 80th percentile (P₈₀), the Z-score can be found in a Z-table or using statistical software or a calculator. The Z-score for the 80th percentile is approximately **0.8416**.
2. **Convert the Z-score to the actual value**: Use the formula for converting a Z-score to an X value:
\[
X = μ + (Z * σ)
\]
where:
- \( μ \) is the mean
- \( Z \) is the Z-score
- \( σ \) is the standard deviation
Plugging in the values:
\[
P₈₀ = 187.9 + (0.8416 * 71.2)
\]
3. **Calculate**:
\[
P₈₀ = 187.9 + 59.87 ≈ 247.7700
\]
### Answer
Hence, **P₈₀** ≈ 247.7700
Enter this value accurate to 4 decimal places.
---
This example illustrates the process of finding a specific percentile in a normally distributed dataset. It's a key concept in the study of statistics, particularly useful for understanding data distribution and probability.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e97f72a-1c8d-4466-a361-8c71e85dddcc%2F74834141-5795-46fe-8a27-57d9ee1972b6%2Fa7gixca_processed.png&w=3840&q=75)
Transcribed Image Text:### Understanding Percentiles in a Normal Distribution
In this exercise, we consider a normally distributed set of values characterized by a mean (average) of 187.9 and a standard deviation of 71.2.
To solve the question, we are asked to identify **P₈₀**, which is the value below which 80% of the observations fall (also known as the 80th percentile). This score will effectively separate the bottom 80% of the data from the top 20%.
Given:
- Mean (μ) = 187.9
- Standard Deviation (σ) = 71.2
To find **P₈₀**, follow these steps:
1. **Determine the Z-score for the 80th percentile**: The Z-score indicates how many standard deviations a value is from the mean. For the 80th percentile (P₈₀), the Z-score can be found in a Z-table or using statistical software or a calculator. The Z-score for the 80th percentile is approximately **0.8416**.
2. **Convert the Z-score to the actual value**: Use the formula for converting a Z-score to an X value:
\[
X = μ + (Z * σ)
\]
where:
- \( μ \) is the mean
- \( Z \) is the Z-score
- \( σ \) is the standard deviation
Plugging in the values:
\[
P₈₀ = 187.9 + (0.8416 * 71.2)
\]
3. **Calculate**:
\[
P₈₀ = 187.9 + 59.87 ≈ 247.7700
\]
### Answer
Hence, **P₈₀** ≈ 247.7700
Enter this value accurate to 4 decimal places.
---
This example illustrates the process of finding a specific percentile in a normally distributed dataset. It's a key concept in the study of statistics, particularly useful for understanding data distribution and probability.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps with 3 images
Recommended textbooks for you
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
Elementary Statistics: Picturing the World (7th E…
Statistics
ISBN:
9780134683416
Author:
Ron Larson, Betsy Farber
Publisher:
PEARSON
The Basic Practice of Statistics
Statistics
ISBN:
9781319042578
Author:
David S. Moore, William I. Notz, Michael A. Fligner
Publisher:
W. H. Freeman
Introduction to the Practice of Statistics
Statistics
ISBN:
9781319013387
Author:
David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:
W. H. Freeman