The table below presents quarterly operational metrics for 10 semiconductor fabrication plants: | Plant ID | Region | Wafer Size | Defect Density (defects/cm²) | Operational Yield (%) | |---|---|---|---|---| | Fab 101 | Asia-Pacific | 300 mm | 1.8 | 94.2 | | Fab 102 | Asia-Pacific | 300 mm | 2.4 | 9

The table below presents quarterly operational metrics for 10 semiconductor fabrication pl… — GMAT

The table below presents quarterly operational metrics for 10 semiconductor fabrication plants:

Plant ID	Region	Wafer Size	Defect Density (defects/cm²)	Operational Yield (%)
Fab 101	Asia-Pacific	300 mm	1.8	94.2
Fab 102	Asia-Pacific	300 mm	2.4	92.6
Fab 103	Asia-Pacific	300 mm	3.1	89.5
Fab 104	Asia-Pacific	300 mm	3.7	88.0
Fab 105	Asia-Pacific	300 mm	4.3	86.4
Fab 106	Asia-Pacific	300 mm	5.1	83.1
Fab 107	North America	300 mm	2.2	93.0
Fab 108	North America	200 mm	4.8	85.0
Fab 109	Europe	300 mm	3.5	87.5
Fab 110	Europe	200 mm	5.5	81.0

Statement: For semiconductor fabrication plants with a $300\text{ mm}$ wafer size located in the Asia-Pacific region, the median defect density is greater than $3.3\text{ defects/cm}^2$ .

Cevap: Cevap

Cevap

True

The correct response is True because the filtered subset of 6 plants in the Asia-Pacific region with 300 mm wafer sizes has defect densities of 1.8, 2.4, 3.1, 3.7, 4.3, and 5.1 defects/cm². Averaging the two middle values (3.1 and 3.7) yields a median of 3.4 defects/cm², which exceeds 3.3 defects/cm².

Adım Adım Çözüm

Filter the dataset according to specified criteria

Identified 6 relevant plants: Fab 101, Fab 102, Fab 103, Fab 104, Fab 105, and Fab 106 (Region = Asia-Pacific AND Wafer Size = 300 mm).

The question requires analyzing descriptive statistics specifically for 300 mm wafer plants in the Asia-Pacific region.

Extract and sort the defect density data for the filtered subset

Sorted list of defect densities (

N = 6

1.8, 2.4, 3.1, 3.7, 4.3, 5.1

Calculating a median requires arranging the data points in ascending order.

Calculate the median of the even-count dataset

The 3rd element is

3.1

and the 4th element is

3.7

. The median is

\frac{3.1 + 3.7}{2} = 3.4\text{ defects/cm}^2

When

N

is even, the median is defined as the mean of the two middle numbers at positions

\frac{N}{2}

and

\frac{N}{2} + 1

Evaluate the statement

The calculated median of

3.4\text{ defects/cm}^2

is strictly greater than

3.3\text{ defects/cm}^2

. Therefore, the statement is True.

Comparing the calculated value (

3.4

) directly against the threshold in the stem (

3.3

) confirms the statement.

Anahtar Kavram

Descriptive Statistics Interpretation (Median of Even-Count Subsets)

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

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