I need help with parts B & C.

Suppose that about 75% of 18-20 year olds consumed alcoholic beverages in any given year. We now consider a random sample of sixty 18–20 year olds.

(a) How many people would you expect to have consumed alcoholic beverages?
Answer: 45 (Correct)

What is the standard deviation? (Round your answer to two decimal places.)
Answer: 3.35 (Correct)

(b) Would you be surprised if there were 55 or more people who have consumed alcoholic beverages? (Round your answer to two decimal places.)
Since the z-score is 51.7 (Incorrect), which is greater than 2 standard deviations away from the mean, it would be surprising for there to be 55 or more people who have consumed alcoholic beverages. (Correct)

(c) What is the probability that 55 or more people in this sample have consumed alcoholic beverages? (Round your answer to four decimal places.)
Answer: 41.25 (Incorrect)

How does this probability relate to your answer to part (b)?
In part (b) we determined it would be unusual to observe 55 or more people who have consumed an alcoholic beverage based on the z-score. In part (c), we found the probability of observing this event to be quite small, implying the event would be unusual. Thus, the results are consistent. (Correct)

It would be surprising if 55 or more people in a random sample of 60 18-20 year olds have consumed alcoholic beverages, as the probability of this event occurring is only 0.0015.

(a) To calculate the number of people you would expect to have consumed alcoholic beverages, you need to multiply the sample size (60) by the percentage of 18-20 year olds who consumed alcoholic beverages (75%).

Expected number of people = Sample size * Percentage

Expected number of people = 60 * 0.75

Expected number of people = 45

So, you would expect 45 people to have consumed alcoholic beverages.

(b) To determine if it would be surprising to have 55 or more people who have consumed alcoholic beverages, we need to consider the standard deviation. The standard deviation measures the spread of the data around the mean.

Given that the standard deviation is 3.35, we can calculate the z-score for 55 people.

z = (x - mean) / standard deviation

z = (55 - 45) / 3.35

z = 10 / 3.35

z ≈ 2.99

The z-score of 2.99 indicates that 55 people is about 2 standard deviations away from the mean. Generally, events that are more than 2 standard deviations away from the mean are considered unusual or surprising. Therefore, it would be surprising to have 55 or more people who have consumed alcoholic beverages.

(c) To find the probability that 55 or more people in this sample have consumed alcoholic beverages, we can use the standard normal distribution table or a calculator. However, the provided probability of 41.25% is incorrect. The correct probability cannot be determined without knowing the specific values in the standard normal distribution table.

In summary, based on the given information, we would expect 45 people to have consumed alcoholic beverages out of a random sample of 60 18-20 year olds. It would be surprising to have 55 or more people who have consumed alcoholic beverages, as it is more than 2 standard deviations away from the mean. However, the exact probability cannot be determined without further information.

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