Answer:
Option D is correct.
Out of every 1,000 cars assembled, the number of cars that should have one or more defects is about 330.
Explanation:
We can obtain the limits of the confidence interval for this problem and the mean must always be in that interval (the definition of confidence interval!)
Limits of the interval = (mean ± margin of error)
Mean = (sample size) × (proportion) = np
n = 1000
p = (40/100) = 0.4
Mean = 0.4 × 1000 = 400
Margin of error = (critical value) × (standard deviation of the sample mean)
Using a confidence interval of 99.999% to fully cover all grounds, critical value = z = 4.42
Standard deviation of the sample mean = √[np(1-p)] = √(1000×0.4×0.6) = 15.50
Margin of error = 4.42 × 15.50 = 68.51
Limits of the interval = (mean ± margin of error)
Limits of the interval = (400 ± 68.51)
Upper limit of the interval = 400 + 68.51 = 468.51
lower limit of the interval = 400 - 68.51 = 331.49
The confidence interval then is (331.49, 468.51).
From the options provided, about 330 is the closest to this confidence interval obtained.
Hope this Helps!!!