Homework Engr 102: Individual Assignment Data And Statistics

Homework Engr 102bindividual Assignment Data And Statisticsdue On Th

1. (30pts) An ENGR 102B class has forty students enrolled in it. The exam scores for six students within this class are 86, 97, 76, 83, 92, and 55. For this data sample, calculate the following (and show all of your work):

• The mean
• The median
• The range
• The unbiased sample variance
• The standard deviation

2. (10pts) Your boss puts you in charge of collecting data that shows the costs of the five major components and subassemblies that go into manufacturing a piece of equipment. What type of data display chart would you select to most effectively convey this information to your boss, and why would that type of chart be a good choice?

3. (25 pts total) A semiconductor manufacturing company is experiencing difficulties with devices that are rejected because they fail to work properly after being manufactured. Data are collected, and it is found that the errors causing the defective semiconductors can be categorized into seven different classes:

Cause of Failure	Number of Occurrences
Contamination	134
Corrosion	207
Doping	33
Oxide defect	92
Silicon defect	122
Metallization	145
Miscellaneous	pts

First, construct a Pareto Chart to show failure types in descending order by their frequency. You can use another application such as Excel to construct your chart or draw it by hand. Then, identify the categories that make up 80% of the cases of failure to advise the manufacturing company where to concentrate their efforts to improve—that is, which specific categories account for 80% of the failures.

4. (10 pts) Fill in the blanks: The number of hours it takes for a certain metal part to wear out is assumed to be from a population with a normal probability distribution with a mean of 1,200 hours and a standard deviation of 130 hours. If this is true, then approximately two-thirds, or 68 percent, of the parts are predicted to wear out between _______ and _______ hours of operation.

5. (25pts) Suppose that the scores on a mathematics aptitude test are normally distributed. If the test results have a mean score of 80 points and a standard deviation of 10 points, what is the probability that a student from this population scored 90 points or lower on this particular test? (Hint: How many standard deviations above the mean is the score for 90? If you know this, you can find the area covered by the normal distribution.)

Paper For Above instruction

This assignment encompasses foundational data analysis and statistical concepts crucial for engineering applications, including descriptive statistics, data visualization, Pareto analysis, and probability calculations with normal distributions. It aims to bolster students' ability to interpret real-world engineering data, select appropriate visualization tools, and apply statistical reasoning to practical problems.

1. Descriptive Statistics for Exam Scores

The dataset comprises exam scores from six students in an ENGR 102B class: 86, 97, 76, 83, 92, and 55. Calculating the mean involves summing these scores and dividing by the number of observations:

Mean = (86 + 97 + 76 + 83 + 92 + 55) / 6 = 489 / 6 ≈ 81.5

The median is the middle value when scores are ordered from lowest to highest: 55, 76, 83, 86, 92, 97. As there is an even number of observations, median = (83 + 86) / 2 = 84.5.

The range is the difference between the maximum and minimum scores: 97 - 55 = 42.

To compute the unbiased sample variance, first find the deviations from the mean, square each deviation, sum them, and divide by n - 1:

Deviations:

(86 - 81.5) = 4.5, (97 - 81.5) = 15.5, (76 - 81.5) = -5.5,

(83 - 81.5) = 1.5, (92 - 81.5) = 10.5, (55 - 81.5) = -26.5

Squared deviations:

20.25, 240.25, 30.25, 2.25, 110.25, 702.25

Sum of squared deviations = 20.25 + 240.25 + 30.25 + 2.25 + 110.25 + 702.25 = 1105.5

Sample variance = 1105.5 / (6 - 1) = 1105.5 / 5 ≈ 221.1

Standard deviation = √221.1 ≈ 14.87

2. Choice of Data Display Chart for Component Costs

A bar chart would be the most effective data display for illustrating the costs of different components and subassemblies. Bar charts provide an intuitive visual comparison of categorical data, enabling the boss to quickly identify which components are the most expensive. Because costs are categorical and discrete, a bar chart effectively highlights differences in magnitude and allows easy comparison across the five components. Additionally, bar charts handle data with varying magnitudes efficiently, making them suitable for presenting manufacturing cost data to managerial stakeholders.

3. Pareto Analysis of Semiconductor Failures

Constructing the Pareto chart requires listing the causes in descending order:

• Corrosion: 207
• Metallization: 145
• Contamination: 134
• Silicon defect: 122
• Oxide defect: 92
• Doping: 33
• Miscellaneous: pts (assuming a placeholder for a numeric value; if actual data is missing, it should be corrected accordingly)

Calculating the total number of failures:

Total = 207 + 145 + 134 + 122 + 92 + 33 + pts. Assuming "pts" represents an actual value, for example, 50, total would be 783. The categories contributing the most are corrosion, metallization, contamination, silicon defect, and oxide defect.

Identifying the top categories that comprise roughly 80% involves calculating the cumulative contributions:

• Corrosion: 207
• Metallization: 145 (cumulative: 352)
• Contamination: 134 (cumulative: 486)
• Silicon defect: 122 (cumulative: 608)

If the total was, for example, 783, then the cumulative sum of top categories is 608, which is approximately 77.7% of total failures, just under 80%. Including the oxide defect (92) increases total to 700, bringing the cumulative to 700/783 ≈ 89.4%, which exceeds 80%. Therefore, focusing on the top five categories including oxide defect covers over 80%. The firm should prioritize addressing corrosion, metallization, contamination, silicon defects, and oxide defects.

4. Normal Distribution of Wear-Out Hours

Given a mean (μ) = 1200 hours and standard deviation (σ) = 130 hours, about 68% of the parts are expected to wear out within one standard deviation of the mean:

Lower bound: μ - σ = 1200 - 130 = 1070 hours

Upper bound: μ + σ = 1200 + 130 = 1330 hours

Thus, approximately 68% of the parts are predicted to fail between 1070 and 1330 hours.

5. Probability of Scoring 90 or Lower on a Normal Test

The test scores follow a normal distribution with mean μ = 80 and standard deviation σ = 10. To find the probability that a randomly selected student scores 90 or less, first calculate the z-score:

z = (X - μ) / σ = (90 - 80) / 10 = 1.0

Using standard normal distribution tables or software, the probability of z ≤ 1.0 is approximately 0.8413, or 84.13%. This indicates that roughly 84.13% of students score 90 or below.

In conclusion, these calculations demonstrate essential statistical tools that aid in evaluating data trends, making informed decisions, and understanding variability within engineering contexts.

References

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• Ott, R. L., & Longnecker, M. (2010). An Introduction to Statistical Methods and Data Analysis (6th ed.). Cengage Learning.
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