A company purchases shipments of machine components and uses this acceptance sampling plan: Randomly select and test 20 components and accept the whole batch if there are fewer than 3 defectives. If a particular shipment of thousands of components actually has a 8% rate of defects, what is the probability that this whole shipment will be accepted? Round to three decimal places.

Answer: Probability that this whole shipment will be accepted is 0.629

Step-by-step explanation:

The company randomly select and test 20 components and accept the whole batch if there are fewer than 3 defectives.

Applying normal distribution,

z = (x-u) / s

Where

n = number of components that were tested.

s = standard deviation

u = mean = np

p = probability that the component has defect

q = probability that the component does not have defect

x = r = the number of components

From the information given,

p = 8/100 = 0.08

q = 1-q = 1-0.08 = 0.92

n = 20

u = 20*0.08 = 1.6

s = √20 * 0.08 * 0.92 = √1.472

s = 1.21

Probability that this whole shipment will be accepted will be probability that fewer than 3 components have defects. This means that it must be lesser than or equal to 2 components.

P (x lesser than 3) = P (x lesser than or equal to 2). Therefore,

x = 2

z = (2-1.6) / 1.21 = 0.33

Looking at the normal distribution table, the corresponding z score is 0.629