Compute 95% confidence intervals for the proportion of defective items in a process when it is found that a sample of size 100 yields 8 defectives.

0.02683<x<0.133173

Step-by-step explanation:

The confidence interval formula is expressed as;

CI = p ± z√p(1-p)/n

p is the proportion

z is the confidence interval at 95% confidence

n is the sample size

Since it is found that a sample of size 100 yields 8 defectives, the sample proportion p = 8/100 = 0.08

Given

n = 100

z score = 1.96

p = 0.08

1-p = 1-0.08 = 0.92

Required

95% confidence interval

Substitute the given parameters into the formula as shown;

CI = p ± z√p(1-p)/n

CI = 0.08 ± 1.96√0.08(0.92)/100

CI = 0.08 ± 1.96√0.0736/100

CI = 0.08 ± 1.96(0.027129)

CI = 0.08 ± 0.053173

CI = (0.08-0.053173, 0.08+0.053173)

CI = (0.02683, 0.133173)

Hence the 95% confidence intervals for the proportion of defective items is 0.02683<x<0.133173

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so:

i pretty sure it is