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When dealing with a binomial distribution, the expected value represents what you predict to happen on average over many trials. For example, in our problem, we're considering young adults consuming alcohol. With a total of 50 young people and a probability of 0.7 (or 70%) that any one of them drinks, we can calculate the expected number of drinkers. The expected value, denoted as \( E(x) \), is found using the formula \( n \times p \), where \( n \) is the number of trials, and \( p \) is the probability. In our example, this results in \( 50 \times 0.7 = 35 \). So, under these conditions, the typical expectation is that about 35 young adults in this sample consume alcohol.

The standard deviation in a binomial distribution helps us understand how much the number of successes (e.g., people drinking) might deviate from the expected value. A smaller standard deviation suggests the outcomes tend to be closer to the expected value, while a larger standard deviation indicates more spread or variability. For our sample of 50 people, with each having a 0.7 chance of drinking, the standard deviation \( \sigma \) is calculated using the formula \( \sqrt{n \times p \times (1-p)} \). Substituting in the values, we find \( \sqrt{50 \times 0.7 \times 0.3} = \sqrt{10.5} \approx 3.24 \). This means on average, the number of drinkers will deviate from the expected 35 by about 3.24 individuals.

Probability is the measure of how likely an event is to occur. In a binomial distribution, it allows us to calculate the likelihood of a certain number of outcomes happening. When trying to determine if 45 or more young people might drink, you would want to calculate the probability of this event. Generally, you would sum the probabilities of all scenarios involving 45 or more drinkers. However, with a large number like 50 trials, this is often done using approximations such as the normal distribution, which simplifies calculations. The specific probability of some exact figures can be very small when they're far from the expected value, indicating unlikely events.

A Z-score is a statistical measure that indicates how many standard deviations an element is from the mean in a distribution. For example, if we want to know whether having 45 drinkers is unusual, we calculate the Z-score: \( Z = \frac{X - \mu}{\sigma} \). Here, \( X \) is the observed outcome (45 drinkers), \( \mu \) is the mean (35), and \( \sigma \) is the standard deviation (3.24). Thus, \( Z = \frac{45 - 35}{3.24} \approx 3.09 \). A Z-score of 3.09 is high, as it suggests 45 is far from the mean. In the context of a normal distribution, this result would be uncommon, reinforcing our intuition that 45 out of 50 is indeed a surprising amount.

The normal approximation is a method used to simplify calculations in large sample binomial distributions using the normal distribution. When dealing with a high number of trials, such as 50, direct probability calculations can be cumbersome. By approximating the binomial distribution with a normal distribution that has the same mean and standard deviation, we can use Z-scores and normal distribution tables for calculations. For instance, when checking 45 or more drinkers, we convert 44.5 (due to continuity correction) to a Z-score and find the probability for greater values. With \( Z \approx 2.93 \), we determine this probability is small (below 0.2%). This approach simplifies complex calculations, providing a reasonable estimate in analyzing such scenarios.