Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take a random sample of 25. When sampling with replacement (so that the p = probability of success does not change), note that a success in this case is selecting a defective part. The mean of this situation is?

mean (μ) = 4.25

Step-by-step explanation:

Let p = probability of a defective computer components =

let q = probability of a non-defective computer components =

Given random sample n = 25

we will find mean value in binomial distribution

The mean of binomial distribution = np

here 'n' is sample size and 'p' is defective components

mean (μ) = 25 X 0.17 = 4.25

Conclusion:-

mean (μ) =  4.25

unless if u could do my questions, if not i wouldn't want to

b. 18

step-by-step explanation:

so the equation for this is:

ab^2=ac⋅ad

ab being the top length of 6 and ac being the length of 2 and ad being the length of x. let's plug into the equation:

6^2 = 2x

now lets just solve for x:

36 = 2x

18 = x

i hope that !