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Expected value, often denoted as \(E(Y)\), in the context of probability distributions like the binomial distribution, represents the average or mean number of times an event occurs. It reflects what you would expect to happen on average, over a large number of trials.
In this specific exercise, we are interested in finding the expected number of American adults who would want their child to grow up to be president. We use the formula for expected value in a binomial distribution:

• \(E(Y) = n \times p\)
• where \(n\) is the number of trials (in our case, the number of American adults surveyed) and \(p\) is the probability of each trial resulting in a success (here, 32% or 0.32).

By plugging the given values into the formula, we calculate \(E(Y) = 120 \times 0.32 = 38.4\). This means that, on average, we would expect 38.4 Americans out of 120 to prefer that their child becomes president. It is crucial to understand that the expected value might not always be a whole number but represents a statistical measure of central tendency.

Standard deviation, denoted by \(\sigma\), is a measure that describes the amount of variation or dispersion in a set of values. In a binomial distribution, the standard deviation provides insight into how much the results of a survey or experiment are spread out from the expected value.
For this problem, the standard deviation helps us understand how the number of American adults who want their child to grow up to be president might vary from the expected 38.4. We use the following formula for standard deviation in binomial distributions:

• \(\sigma = \sqrt{n \times p \times (1-p)}\)
• where \(n\) is the sample size, \(p\) is the probability of success, and \(1-p\) is the probability of failure.

Substituting the given values, we find \(\sigma = \sqrt{120 \times 0.32 \times 0.68}\), which results in approximately 4.8. This indicates that, while the expected number is 38.4, it's common for the actual number to vary by about 4.8 either way. Knowing the standard deviation helps predict the likelihood of outcomes different from the expected value.

Probability deals with quantifying the likelihood of an event happening. In our exercise, we want to determine if more than 40 people preferring their child to be president is a reasonable expectation.
We use what’s called a Z-score in standard normal distribution to evaluate how likely it is to have more than 40 people express this preference. The Z-score measures how many standard deviations an observation is away from the mean. For our problem:

• First, compute the Z-score using the formula:
  \[Z = \frac{X - E(Y)}{\sigma}\]
• where \(X\) is 40, \(E(Y)\) is 38.4, and \(\sigma\) is 4.8.

Plugging in the numbers, we find \[Z = \frac{40 - 38.4}{4.8} \approx 0.33\]. A Z-score of 0.33 suggests that 40 is within a normal range of variation from the expected value. If a Z-score is close to or beyond the standard deviation (usually above 1.96 or below -1.96 represents a rare event), it indicates the event is not likely or unusual. Here, since our Z-score is relatively small, it means it's not unlikely for more than 40 people to prefer their child becoming president in the surveyed group of 120.