he following contingency table shows inspection records for 720 units of a particular product. The records have been cross-classified by the inspector’s decision (Pass and Fail) and the inspector’s experience (Low, Medium, and High).
Decision Experience
Low Medium High
Pass 176 335 108
Fail 16 51 34
a-1. How many of the units passed inspection?
a-2. How many of the units failed inspection?
b. How many of the units were inspected by inspectors with high experience?
c. What proportion of the units were inspected by inspectors with low experience? (Round your answer to 2 decimal places.)
d. What proportion of the units were inspected by inspectors with medium experience and failed inspection? (Round your answer to 2 decimal places.)
a-1.
The number of units that passed inspection is 176 + 335 + 108 = 619.
a-2.
The number of units that failed inspection is 16 + 51 + 34 = 101.
b.
The number of units inspected by inspectors with high experience is 108 + 34 = 142.
c.
The proportion of units inspected by inspectors with low experience is (176 + 16) / (176 + 335 + 16 + 51 + 108 + 34) = 192 / 720 = 0.27 (rounded to 2 decimal places).
d.
The proportion of units inspected by inspectors with medium experience and failed inspection is 51 / (335 + 51) = 0.13 (rounded to 2 decimal places).
a-1. The number of units that passed inspection can be found by adding up the values in the "Pass" row:
176 + 335 + 108 = 619
Therefore, 619 units passed inspection.
a-2. The number of units that failed inspection can be found by adding up the values in the "Fail" row:
16 + 51 + 34 = 101
Therefore, 101 units failed inspection.
b. The number of units inspected by inspectors with high experience can be found by adding up the values in the "High" column:
108 + 34 = 142
Therefore, 142 units were inspected by inspectors with high experience.
c. To find the proportion of units inspected by inspectors with low experience, divide the total number of units inspected by inspectors with low experience by the total number of units:
(176 + 16) / 720 = 192 / 720 ≈ 0.267
Therefore, approximately 0.27 (or 26.7%) of the units were inspected by inspectors with low experience.
d. To find the proportion of units inspected by inspectors with medium experience and failed inspection, divide the number of units that failed inspection by the total number of units inspected by inspectors with medium experience:
51 / (335 + 51) ≈ 0.132
Therefore, approximately 0.13 (or 13.2%) of the units inspected by inspectors with medium experience failed inspection.