Many women take oral contraceptives to prevent pregnancy. Under ideal conditions, 1% will become pregnant within one year. In typcal use, 5% become pregnant. Choose 20 random women taking the pill. How many become pregnant in the next year?

a. What is the probability that at least one woman becomes pregnant under ideal conditions?
b. What is the probability in typical use?

Part 2-Based on a random sample of 500 women taking the pill:
c. Based on the information about typical use in b, what is the probablity that at least 25 of these women become pregnant in the next year? Check that the Normal approximation is permissable and use it to find this probability.
d. We can't use the Normal approximation to the binomial distributions to find thisprobability under ideal conditions as described in Part I. Why not?

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To answer these questions, we need to understand the concept of probability and how it applies to this scenario.

For Part 1:
a. To find the probability that at least one woman becomes pregnant under ideal conditions, we need to subtract the probability that no woman becomes pregnant from 1. Under ideal conditions, the probability of a woman getting pregnant within a year is 1%. Therefore, the probability of no woman getting pregnant is 99% or 0.99. The probability that at least one woman becomes pregnant is then 1 - 0.99^20 = 0.181, or 18.1%.

b. In typical use, the probability of a woman getting pregnant within a year is 5%. The probability of no woman getting pregnant is 95% or 0.95. Using the same approach as above, the probability that at least one woman becomes pregnant is 1 - 0.95^20 = 0.641, or 64.1%.

For Part 2:
c. Based on the information about typical use in Part 1b, we can use the Normal approximation to estimate the probability that at least 25 out of 500 women become pregnant in the next year. The conditions for using the Normal approximation are that the sample size is large (n = 500) and the probability of success (getting pregnant) is not too close to 0 or 1 (5% or 0.05).

Using the Normal approximation, we can calculate the mean (μ) and standard deviation (σ) for the binomial distribution:

μ = n * p = 500 * 0.05 = 25
σ = sqrt(n * p * (1-p)) = sqrt(500 * 0.05 * 0.95) = 5

Now, we can use the Normal distribution to estimate the probability of at least 25 women becoming pregnant. We need to calculate the z-score for 25:

z = (x - μ) / σ = (25 - 25) / 5 = 0

Using a z-table or a calculator, we can find the probability associated with a z-score of 0, which is 0.500. However, since we are interested in at least 25 women becoming pregnant (not exactly 25), we need to consider the area to the right of 25. This is equal to 1 - 0.500 = 0.500.

Therefore, the probability that at least 25 out of 500 women become pregnant in the next year, based on the Normal approximation, is 0.500 or 50%.

d. We can't use the Normal approximation to the binomial distribution to find the probability under ideal conditions as described in Part 1a because the probability of success (getting pregnant) is very small (1% or 0.01), and we have a relatively small sample size (20). The conditions required for the Normal approximation are not met in this case, and therefore, the approximation may not be accurate.