Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take

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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?

Mathematics

1 answer:

Svet_ta [12.7K] · Answer 1 · 2026-03-03T02:23:31+03:00

Response:

mean (μ) = 4.25

Detailed explanation:

Let p represent the likelihood of a defective component = $\frac{68}{400} = 0.17$ .

Let q represent the likelihood of a non-defective component = $\frac{332}{400} = 0.83$ .

The random sample size n = 25.

We will calculate the mean for the binomial distribution.

The mean for the binomial distribution is computed as np.

Here, 'n' stands for the sample size, and 'p' signals the proportion of defective components.

mean (μ) = 25 x 0.17 = 4.25

Conclusion:

mean (μ) = 4.25