A premium candy manufacturer makes chocolate candies that, when finished, vary in color. The hue value of a randomly selected candy follows an approximately normal distribution with mean 30 and standard deviation 5. A quality inspector discards 12% of the candies due to unacceptable hues (which are equally likely to be too small or too large). What is the largest hue value that the inspector would find acceptable? Round your answer to two decimal places.
A premium candy manufacturer makes chocolate candies that, when finished, vary in color. The hue value of a randomly selected candy follows an approximately normal distribution with mean 30 and standard deviation 5. A quality inspector discards 12% of the candies due to unacceptable hues (which are equally likely to be too small or too large). What is the largest hue value that the inspector would find acceptable? Round your answer to two 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
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![### Quality Control in Chocolate Manufacturing: Managing Hue Variability
**Context:**
A premium candy manufacturer produces chocolate candies that exhibit a range of final hues. The hue value of a randomly chosen candy is modeled as an approximately normal distribution with a mean of 30 and a standard deviation of 5.
**Quality Control Challenge:**
A quality inspector eliminates 12% of the candies due to unacceptable hue variations, equally split between hues that are too low or too high. This implies that 6% of the candies are discarded for being too small (too low hue) and another 6% for being too large (too high hue).
**Problem Statement:**
Determine the largest hue value that the inspector would consider acceptable. Provide the answer rounded to two decimal places.
### Solution:
The problem requires identifying the hue value corresponding to the top 94% (since 6% is discarded for being too high) of the normal distribution with the given parameters.
**Steps to Solve:**
1. **Determine the Z-score:**
- For a standard normal distribution, the Z-score corresponding to the top 94% (or equivalently, the 94th percentile) needs to be found.
- Using standard normal distribution tables or statistical software, the Z-score for the 94th percentile is approximately 1.554.
2. **Calculate the hue value:**
Using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Here, \( \mu = 30 \) (mean) and \( \sigma = 5 \) (standard deviation). Solving for \( X \) (hue value):
\[
X = Z \sigma + \mu
\]
Substitute the values:
\[
X = 1.554 \times 5 + 30
\]
\[
X = 7.77 + 30
\]
\[
X = 37.77
\]
**Conclusion:**
The largest hue value that the inspector would find acceptable is **37.77**.
---
This educational scenario is aimed at helping students understand the application of normal distribution and Z-scores in quality control processes.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb989860-4d1c-4fb5-9e6b-42a4528dce9c%2F27dbbaa1-dae3-46aa-957e-1befbdb3151d%2Ftlv1x4h_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Quality Control in Chocolate Manufacturing: Managing Hue Variability
**Context:**
A premium candy manufacturer produces chocolate candies that exhibit a range of final hues. The hue value of a randomly chosen candy is modeled as an approximately normal distribution with a mean of 30 and a standard deviation of 5.
**Quality Control Challenge:**
A quality inspector eliminates 12% of the candies due to unacceptable hue variations, equally split between hues that are too low or too high. This implies that 6% of the candies are discarded for being too small (too low hue) and another 6% for being too large (too high hue).
**Problem Statement:**
Determine the largest hue value that the inspector would consider acceptable. Provide the answer rounded to two decimal places.
### Solution:
The problem requires identifying the hue value corresponding to the top 94% (since 6% is discarded for being too high) of the normal distribution with the given parameters.
**Steps to Solve:**
1. **Determine the Z-score:**
- For a standard normal distribution, the Z-score corresponding to the top 94% (or equivalently, the 94th percentile) needs to be found.
- Using standard normal distribution tables or statistical software, the Z-score for the 94th percentile is approximately 1.554.
2. **Calculate the hue value:**
Using the Z-score formula:
\[
Z = \frac{X - \mu}{\sigma}
\]
Here, \( \mu = 30 \) (mean) and \( \sigma = 5 \) (standard deviation). Solving for \( X \) (hue value):
\[
X = Z \sigma + \mu
\]
Substitute the values:
\[
X = 1.554 \times 5 + 30
\]
\[
X = 7.77 + 30
\]
\[
X = 37.77
\]
**Conclusion:**
The largest hue value that the inspector would find acceptable is **37.77**.
---
This educational scenario is aimed at helping students understand the application of normal distribution and Z-scores in quality control processes.
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