A factory bought a new machine. It is not configured well yet, so the probability of defects is 0.05. Estimate the probability that there are at most 2 spoiled details in the random sample of 20.

Pr(at most 2 are defective) = 0.38

Step-by-step explanation:

probability of defective = 0.05

probability of not  defective =1- 0.05 = 0.95

Pr(at most 2 are defective) = pr (0 defective) +  pr (1 defective) + pr (2defective)

pr (0 defective) = (0.05^0) * (0.95^20) = 0.36

pr (1 defective) =  (0.05^1) * (0.95^19) = 0.019

pr (2 defective) =  (0.05^2) * (0.95^18) = 0.0025 * 0.397 = 0.00099

Pr(at most 2 are defective) =  0.36 +  0.019 +  0.00099

Pr(at most 2 are defective) = 0.37999 = 0.38

1) c 2143.6 cm

2) c 113.0 m

3) a 33.5 ft

4)   d 2 cm

5) c 20 mm

6) 180 in

7) d 120 ft