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According to a poll of adults, about 47% work during their summer vacation. Assume that the true proportion of all adults that work during summer vacation is p = 0.47. Now consider a random sample of 400 adults. Complete parts a and b below. a. What is the probability that between 43% and 51% of the sampled adults work during summer vacation? The probability is (Round to three decimal places as needed.) b. What is the probability that over 65% of the sampled adults work during summer vacation? The probability is (Round to three decimal places as needed.)

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Final answer:

To find the probability that between 43% and 51% of the sampled adults work during summer vacation, calculate the z-scores for both proportions and find the corresponding probabilities using the z-score table. The probability of over 65% of the sampled adults working during summer vacation is 1.

Step-by-step explanation:

To determine the probability that between 43% and 51% of the sampled adults work during summer vacation, we need to calculate the z-scores for both proportions and use the z-score table to find the corresponding probabilities.

First, we calculate the z-score for 43%:

z = (43 - 47) / sqrt(0.47 * (1 - 0.47) / 400) = -2

Next, we calculate the z-score for 51%:

z = (51 - 47) / sqrt(0.47 * (1 - 0.47) / 400) = 2

Using the z-score table, we find the corresponding probabilities for z = -2 and z = 2:

Probability for z = -2: 0.0228

Probability for z = 2: 0.9772

To find the probability between 43% and 51%, we subtract the probability for z = 2 from the probability for z = -2:

Probability = 0.9772 - 0.0228 = 0.9544

Therefore, the probability that between 43% and 51% of the sampled adults work during summer vacation is 0.9544 (rounded to three decimal places).

To determine the probability that over 65% of the sampled adults work during summer vacation, we need to calculate the z-score for 65%:

z = (65 - 47) / sqrt(0.47 * (1 - 0.47) / 400) = 4.977

Using the z-score table, we find the corresponding probability for z = 4.977:

Probability = 1 (since it is beyond the limit of the z-score table)

Therefore, the probability that over 65% of the sampled adults work during summer vacation is 1 (rounded to three decimal places).

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