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
