Final answer:
The question is about finding the approximate probability of more than 25 defective semiconductor chips in a lot of 1000, where each chip has a 2% chance of being defective. This can be solved using a normal approximation of the binomial distribution and then calculating the Z-score followed by referencing a standard normal distribution table.
Step-by-step explanation:
The student is asking about approximate probability in the context of quality control in semiconductor manufacturing. Specifically, we are to approximate the probability of having more than 25 defective chips in a lot of 1000 chips, given that 2% of chips are defective (which is essentially a binomial probability problem).
Firstly, we identify the parameters of the binomial distribution: n = 1000 (number of trials or chips) and p = 0.02 (probability of a defect in one chip). The mean (μ) of this distribution is np = 20, and the standard deviation (σ) is sqrt(np(1-p)).
Since the number of trials is large (n = 1000) and the probability of success is small (p = 0.02), we can approximate this binomial distribution with a normal distribution using the Central Limit Theorem. The Z-score for 25.5 (we use 25.5 because of continuity correction) will be:
Z = (25.5 - 20) / σ
Calculate this score, then use a standard normal distribution table or a calculator to find the probability of Z being greater than this calculated value. That will be the probability of more than 25 chips being defective. Based on the options provided, we are seeking the probability that corresponds closely to one of these options.
Without running the actual calculations, we cannot definitively choose between the options provided (a) 0.184, (b) 0.816, (c) 0.976, and (d) 0.024. However, we can state that the answer involves computing a Z-score and referring to a normal distribution table to find the required approximate probability.