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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?

Respuesta :

Answer:

mean (μ) = 4.25

Step-by-step explanation:

Let p = probability of a defective computer components = [tex]\frac{68}{400} = 0.17[/tex]

let q = probability of a non-defective computer components = [tex]\frac{332}{400} = 0.83[/tex]

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