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The company ships the components in lots of 200. Lots containing more than 20 defective components may be returned. Find a 95% confidence interval for the proportion of lots that will be returned. Use the normal approximation to compute this proportion. Round the answers to four decimal places. The 95% confidence interval is

User Cnnr
by
7.4k points

1 Answer

4 votes

Answer:

[ 0.0584, 1.0416 ]

Explanation:

Data provided in the question:

sample size, n = 200

Number of defective components = 20

thus,

p =
(20)/(200)= 0.1

Now,

Confidence interval for p is given as:

⇒ p ±
z\sqrt(p(1-p))/(n)

here, z = 1.96 for 95% confidence interval

therefore,

⇒ 0.1 ±
1.96\sqrt(0.1(1-0.1))/(200)

or

⇒ 0.1 ± 0.0416

or

confidence interval = [ 0.1 - 0.0416, 0.1 + 0.0416 ]

or

confidence interval = [ 0.0584, 1.0416 ]

User Michaelliu
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