The average price of a personal computer is $949. If the computer distributed with a standard deviation of $100... the top 20% of personal computers cost more than what amount?
The average price of a personal computer is $949. If the computer distributed with a standard deviation of $100... the top 20% of personal computers cost more than what amount?
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 Personal Computer Pricing Distribution
**Problem Statement:**
3. The average price of a personal computer is $949. If the computer prices are approximately normally distributed with a standard deviation of $100:
a. The top 20% of personal computers cost more than what amount?
b. The least expensive 10% of personal computers cost less than what amount?
**Solution Approach:**
Given that the prices follow a normal distribution, we can use the properties of the normal distribution to answer these questions. Specifically, we will use the Z-score table to find the relevant percentiles:
**a. The Top 20% of Personal Computers:**
- To find the price where the top 20% begin, we need the z-score corresponding to the 80th percentile (100% - 20% = 80%).
- Using the standard normal distribution table, the z-score for the 80th percentile is approximately 0.84.
- Use the formula for converting a z-score to a value in the original distribution:
\[ X = \mu + (z \times \sigma) \]
Where:
- \( \mu \) (mean) = $949
- \( \sigma \) (standard deviation) = $100
- \( z \) (z-score) = 0.84
Therefore,
\[ X = 949 + (0.84 \times 100) \]
\[ X = 949 + 84 \]
\[ X = 1033 \]
So, the top 20% of personal computers cost more than approximately $1033.
**b. The Least Expensive 10% of Personal Computers:**
- To find the price for the least expensive 10%, we need the z-score corresponding to the 10th percentile.
- Using the standard normal distribution table, the z-score for the 10th percentile is approximately -1.28.
- Again, use the formula for converting a z-score to a value in the original distribution:
\[ X = \mu + (z \times \sigma) \]
Where:
- \( \mu \) (mean) = $949
- \( \sigma \) (standard deviation) = $100
- \( z \) (z-score) = -1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f623e5c-134e-4c64-8ce9-91adb026b8cd%2Ff308b89d-0cbe-4da5-abf4-95ef550631ef%2Fv6hj7d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Analysis of Personal Computer Pricing Distribution
**Problem Statement:**
3. The average price of a personal computer is $949. If the computer prices are approximately normally distributed with a standard deviation of $100:
a. The top 20% of personal computers cost more than what amount?
b. The least expensive 10% of personal computers cost less than what amount?
**Solution Approach:**
Given that the prices follow a normal distribution, we can use the properties of the normal distribution to answer these questions. Specifically, we will use the Z-score table to find the relevant percentiles:
**a. The Top 20% of Personal Computers:**
- To find the price where the top 20% begin, we need the z-score corresponding to the 80th percentile (100% - 20% = 80%).
- Using the standard normal distribution table, the z-score for the 80th percentile is approximately 0.84.
- Use the formula for converting a z-score to a value in the original distribution:
\[ X = \mu + (z \times \sigma) \]
Where:
- \( \mu \) (mean) = $949
- \( \sigma \) (standard deviation) = $100
- \( z \) (z-score) = 0.84
Therefore,
\[ X = 949 + (0.84 \times 100) \]
\[ X = 949 + 84 \]
\[ X = 1033 \]
So, the top 20% of personal computers cost more than approximately $1033.
**b. The Least Expensive 10% of Personal Computers:**
- To find the price for the least expensive 10%, we need the z-score corresponding to the 10th percentile.
- Using the standard normal distribution table, the z-score for the 10th percentile is approximately -1.28.
- Again, use the formula for converting a z-score to a value in the original distribution:
\[ X = \mu + (z \times \sigma) \]
Where:
- \( \mu \) (mean) = $949
- \( \sigma \) (standard deviation) = $100
- \( z \) (z-score) = -1
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