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

[ 0.0584, 1.0416 ]

Step-by-step explanation:

Data provided in the question:

sample size, n = 200

Number of defective components = 20

thus,

p = = 0.1

Now,

Confidence interval for p is given as:

⇒ p ±

here, z = 1.96 for 95% confidence interval

therefore,

⇒ 0.1 ±

or

⇒ 0.1 ± 0.0416

or

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

or

confidence interval = [ 0.0584, 1.0416 ]

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation: