Question

A process engineer is implementing a quality assurance system on a breakfast cereal production line. A new sensor is installed on the line that tests the weight of the filled boxes and rejects any product outside the correct package weight. Historically, 90% of the product manufactured on this line is within the correct weight range. The sensor rejects boxes of incorrect weight 98% of the time. The sensor rejects boxes of correct weight 1% of the time. What is the probability that a correct weight box of cereal will get rejected by the sensor

Answer 1

Answer:

0.2%

Step-by-step explanation:

The given parameters are;

The percentage of product within the correct range = 95%

The percentage of the times the sensor reject boxes of incorrect weight = 98%

The percentage of the times the company reject boxes with correct weight = 1%

$\begin{array}{cccc}&P&N&T\\P&TP&0.9&90\\N&FP&9.8&10\end{array}$

We have, FN = 0.9, TN = 9.8, TP = 90 - 0.9 = 89.1, FP = 10 - 9.8 = 0.2

FNR = FN/(FN + TP) = 0.9/(0.9 + 89.1) = 0.01

FPR = 0.2/(0.2 + 9.8) = 0.02

TPR = 89.1/(89.1 + 0.9) = 0.99

TNR = 9.8/(9.8 + 0.2) = 0.98

P(C/A) = TPR × P(C)/((TPR × P(C) + FPR×(1 - P(C)))

Where;

P(C/A) = The probability that a correct weight package is accepted

∴ P(C/A) = 0.99*0.9/(0.99*0.9 + 0.02*0.1) ≈ 0.99776

The probability that a correctly weighed package is rejected, P(C/R), is given as follows;

P(C/R) = 1 - P(C/A)

∴ P(C/R) ≈ 1 - 0.99776 = 0.00224 ≈ 0.2%

The probability that a correctly weighed package is rejected, P(C/R) ≈ 0.2%