Question

QUESTION 7 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? OA. 0.2% OB. 0.9% OC. 1.8% OD. 90%

Answer 1

Solution-

Let the events are denoted as follows-

Correct weight box - C

Incorrect box weight - C'

Rejected by sensor - R

Now,

Since 90 % of the manufactured product on the line is within correct weight range.

So, probability of a cereal box of correct weight is

P(C) = 90% = 0.90 ....(1)

And probability of a cereal box of incorrect weight is

P(C') = 1- P(C) = 1- 0.90 = 0.10 ....(2)

Since sensor reject incorrect weight 98% of time.

So, Probability that sensor reject the incorrect weight (working properly) is

P(R/C') = 98% = 0.98 ...(3)

Since sensor reject correct weight 1% of time.

So, probability that sensor reject the correct weight (working not properly) is

P(R/C) = 1% = 0.01..(4)

Now, probability that a correct weight box will be rejected by the sensor is denoted by P(C/R) and is given by Bayes' Theoram as

$P(C/R)=\frac{P(C)*P(R/C)}{P(C)*P(R/C) + P(C')*P(R/C')}$

On putting the values from above equations we get

$P(C/R)=\frac{0. 90*0.01}{0.90*0.01 + 0.10*0.98}$

$P(C/R)=\frac{0.009}{0.009 + 0.098}$

$P(C/R)=\frac{0.009}{0.107}$

$P(C/R)=0.084$

Hence, probability that correct weight of box is rejected by the sensor is 0.084 or 8.40 % .

(no option is correct)