Question

constructing a two-way table using driving data
distribution of mileage efficiencies across different weight classes
based on the table, what is the missing value for medium - weight cars in the high - mileage category?

weight	high mileage	medium mileage	low mileage	total
medium	?	2	1	3
heavy	0	?	2	2
total	2	3	3	8

options: 3, 1, 2, 0

Explanation:

Step1: Recall two - way table total rule

In a two - way table, the sum of the values in a row should equal the total for that row. For the "Medium" weight class row, the total is given as 3. The values in the "Medium" weight class row for "High Mileage", "Medium Mileage", and "Low Mileage" are \(?\), 2, and 1 respectively. So we can use the formula: \( \text{High Mileage (Medium)}+\text{Medium Mileage (Medium)}+\text{Low Mileage (Medium)}=\text{Total (Medium)} \)

Step2: Solve for the missing value

Let the missing value (high - mileage for medium - weight cars) be \( x \). Then \( x + 2+1=3 \). Simplifying the left - hand side gives \( x + 3 = 3 \). Subtracting 3 from both sides, we get \( x=3 - 3=0 \)? Wait, no, wait the total for medium weight is 3? Wait, no, looking at the table again, the "Total" column for "Medium" weight is 3? Wait, no, the "Total" row for "Medium Mileage" is 3, and the "Total" column for "Medium" weight: Wait, the "Light" weight total is 3, "Medium" weight total: let's recalculate. Wait, "Light" row: 2 (high)+1 (medium)+0 (low) = 3 (total), correct. "Medium" row: \( x \) (high)+2 (medium)+1 (low) = total. The "Total" column: "Light" total 3, "Medium" total, "Heavy" total 2, and overall total 8. So 3 + medium total+2 = 8, so medium total is 3. So for "Medium" row: \( x+2 + 1=3 \), so \( x=0 \)? But that contradicts? Wait, no, maybe I misread the table. Wait, the "High Mileage" total is 2 (light:2, medium: \( x \), heavy:0). So 2 + \( x \)+0 = 2, so \( x = 0 \)? Wait, the "High Mileage" total is 2 (from the "Total" row, "High Mileage" column is 2). So light (2) + medium (\( x \))+heavy (0) = 2. So 2 + \( x+0 = 2 \), so \( x = 0 \)? But the options have 0 as an option. Wait, the options are 3,1,2,0. So the missing value for medium - weight cars in high - mileage category is 0? Wait, no, let's check again.

Wait, the "High Mileage" column total is 2 (light:2, medium: \( x \), heavy:0). So 2 + \( x+0 = 2 \), so \( x = 0 \). Also, the "Medium" weight row: \( x \) (high)+2 (medium)+1 (low) = total. The total for "Medium" weight row: 3 (since 3 (light) + 3 (medium)+2 (heavy) = 8). So \( x+2 + 1=3 \), so \( x = 0 \). So the missing value is 0.

Answer:

0